Technical Aptitude Questions Ebook-172 Pages
Technical Aptitude Questions Ebook-172 Pages
Technical Aptitude Questions Ebook-172 Pages
Table of Contents
Data Structures
Aptitude
Data
Structures Aptitude
1. What is data structure?
A data structure is a way of organizing data that considers not only the items
stored, but also their relationship to each other. Advance knowledge about the
relationship between data items allows designing of efficient algorithms for the
manipulation of data.
2.
List out the areas in which data structures are applied extensively?
Compiler Design,
Operating System,
Database Management System,
Statistical analysis package,
Numerical Analysis,
Graphics,
Artificial Intelligence,
Simulation
3. What are the major data structures used in the following areas : RDBMS, Network
data model & Hierarchical data model.
RDBMS
Array (i.e. Array of structures)
Network data model
Graph
Hierarchical data model Trees
4. If you are using C language to implement the heterogeneous linked list, what pointer
type will you use?
The heterogeneous linked list contains different data types in its nodes and we
need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So
we go for void pointer. Void pointer is capable of storing pointer to any type as it is a
generic pointer type.
5. Minimum number of queues needed to implement the priority queue?
Two. One queue is used for actual storing of data and another for storing
priorities.
6. What is the data structures used to perform recursion?
Stack. Because of its LIFO (Last In First Out) property it remembers its caller so
knows whom to return when the function has to return. Recursion makes use of system
stack for storing the return addresses of the function calls.
Every recursive function has its equivalent iterative (non-recursive) function.
Even when such equivalent iterative procedures are written, explicit stack is to be used.
3
7. What are the notations used in Evaluation of Arithmetic Expressions using prefix and
postfix forms?
Polish and Reverse Polish notations.
8. Convert the expression ((A + B) * C (D E) ^ (F + G)) to equivalent Prefix and
Postfix notations.
Prefix Notation:
^ - * +ABC - DE + FG
Postfix Notation:
AB + C * DE - - FG + ^
9. Sorting is not possible by using which of the following methods?
(a) Insertion
(b) Selection
(c) Exchange
(d) Deletion
(d) Deletion.
Using insertion we can perform insertion sort, using selection we can perform
selection sort, using exchange we can perform the bubble sort (and other similar sorting
methods). But no sorting method can be done just using deletion.
10. A binary tree with 20 nodes has
null branches?
21
Let us take a tree with 5 nodes (n=5)
Null Branches
ii
iii
iv
In general:
If there are n nodes, there exist 2n-n different trees.
13. List out few of the Application of tree data-structure?
The manipulation of Arithmetic expression,
Symbol Table construction,
Syntax analysis.
14. List out few of the applications that make use of Multilinked Structures?
Sparse matrix,
Index generation.
15. In tree construction which is the suitable efficient data structure?
(a) Array
(b) Linked list
(c) Stack
(d) Queue (e) none
(b) Linked list
16. What is the type of the algorithm used in solving the 8 Queens problem?
Backtracking
17. In an AVL tree, at what condition the balancing is to be done?
If the pivotal value (or the Height factor) is greater than 1 or less than 1.
18. What is the bucket size, when the overlapping and collision occur at same time?
One. If there is only one entry possible in the bucket, when the collision occurs,
there is no way to accommodate the colliding value. This results in the overlapping of
values.
19. Traverse the given tree using Inorder, Preorder and Postorder traversals.
Given tree:
A
B
D
5 Inorder :
D H B EHA F C I G J
Preorder: A B D H E C F G I J
Postorder: H D E B F I J G C A
20. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could
have formed a full binary tree?
15.
In general:
There are 2n-1 nodes in a full binary tree.
By the method of elimination:
Full binary trees contain odd number of nodes. So there cannot be full
binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete
binary tree but not a full binary tree. So the correct answer is 15.
Note:
Full and Complete binary trees are different. All full binary trees are complete
binary trees but not vice versa.
21. In the given binary tree, using array you can store the node 4 at which location?
1
4
5
At location 6
1
Root
LC1
RC1
LC2
RC2
LC3
RC3
LC4
RC4
where LCn means Left Child of node n and RCn means Right Child of node n
22. Sort the given values using Quick Sort?
65
70
75
80
85
60
55
50
45
Sorting takes place from the pivot value, which is the first value of the given
elements, this is marked bold. The values at the left pointer and right pointer are indicated
using L and R respectively.
65
70L
75
80
85
60
55
50
45R
Since
pivot is not yet changed the same process is continued after interchanging the
6
values at L and R positions
65
45
75 L
80
85
60
55
50 R
70
65
45
50
80 L
85
60
55 R
75
70
65
45
50
55
85 L
60 R
80
75
70
65
45
50
55
60 R
85 L
80
75
70
When the L and R pointers cross each other the pivot value is interchanged with the value
at right pointer. If the pivot is changed it means that the pivot has occupied its original
position in the sorted order (shown in bold italics) and hence two different arrays are
formed, one from start of the original array to the pivot position-1 and the other from
pivot position+1 to end.
60 L
45
50
55 R
65
85 L
80
75
70 R
55 L
45
50 R
60
65
70 R
80 L
75
85
50 L
45 R
55
60
65
70
80 L
75 R
85
70
75
80
85
50
55
60
65
23. For the given graph, draw the DFS and BFS?
H
E
G
BFS:
AXGHPEMYJ
DFS:
AXHPEYMJG
24. Classify the Hashing Functions based on the various methods by which the key value
is found.
Direct method,
Subtraction method,
7 Modulo-Division method,
Digit-Extraction method,
Mid-Square method,
Folding method,
Pseudo-random method.
25. What are the types of Collision Resolution Techniques and the methods used in each
of the type?
Open addressing (closed hashing),
The methods used include:
Overflow block,
Closed addressing (open hashing)
The methods used include:
Linked list,
Binary tree
26. In RDBMS, what is the efficient data structure used in the internal storage
representation?
B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes
searching easier. This corresponds to the records that shall be stored in leaf nodes.
27. Draw the B-tree of order 3 created by inserting the following data arriving in
sequence 92 24 6 7 11 8 22 4 5 16 19 20 78
11
19
16
24
20
22
78
92
28.Of the following tree structure, which is, efficient considering space and
time complexities?
(a) Incomplete Binary Tree
(b) Complete Binary Tree
(c) Full Binary Tree
(b) Complete Binary Tree.
By the method of elimination:
Full binary tree loses its nature when operations of insertions and deletions
are done. For incomplete binary trees, extra storage is required and overhead of NULL
node checking takes place. So complete binary tree is the better one since the property of
complete binary tree is maintained even after operations like additions and deletions are
done on it.
29. What is a spanning Tree?
8
A spanning tree is a tree associated with a network. All the nodes of the graph
appear on the tree once. A minimum spanning tree is a spanning tree organized so that the
total edge weight between nodes is minimized.
30. Does the minimum spanning tree of a graph give the shortest distance between any 2
specified nodes?
No.
Minimal spanning tree assures that the total weight of the tree is kept at its
minimum. But it doesnt mean that the distance between any two nodes involved in the
minimum-spanning tree is minimum.
31. Convert the given graph with weighted edges to minimal spanning tree.
1
410
600
612
310
2985
200
5
400
1421
3
310
612
410
200
35. For the following COBOL code, draw the Binary tree?
01 STUDENT_REC.
02 NAME.
03 FIRST_NAME PIC X(10).
03 LAST_NAME PIC X(10).
02 YEAR_OF_STUDY.
03 FIRST_SEM PIC XX.
03 SECOND_SEM PIC XX.
01
STUDENT_REC
02
02
NAME
03
FIRST_NAME
10
YEAR_OF_STUDY
03
LAST_NAME
03
FIRST_SEM
03
SECOND_SEM
C Aptitude
C Aptitude
Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of
the "constant integer".
2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally array name is the base address for that array. Here s is the base address. i is the
index number/displacement from the base address. So, indirecting it with * is same as
s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
3. main()
11
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot
be predicted exactly. Depending on the number of bytes, the precession with of the value
represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9
with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers
with relational operators (== , >, <, <=, >=,!= ) .
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
54321
Explanation:
When static storage class is given, it is initialized once. The change in the
value of a static variable is retained even between the function calls. Main is also treated
like any other ordinary function, which can be called recursively.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2222223465
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only
q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is
incremented. So the values 2 3 4 6 5 will be printed.
12
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and
that address will be given to the current program at the time of linking. But linker finds
that no other variable of name i is available in any other program with memory space
allocated for it. Hence a linker error has occurred .
7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
00131
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical
AND (&&) operator has higher priority over the logical OR (||) operator. So the
expression i++ && j++ && k++ is executed first. The result of this expression is 0
(-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for 0 || 0 combination- for which it gives 0). So the value
of m is 1. The values of other variables are also incremented by 1.
8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer:
12
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a
character pointer, which needs one byte for storing its value (a character). Hence
sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character
pointer sizeof(p) gives 2.
9. main()
{
int i=3;
13
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed
only when all other cases doesn't match.
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least
significant 4 bits are filled with 0's.The %x format specifier specifies that the integer
value be printed as a hexadecimal value.
11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler
doesn't know anything about the function display. It assumes the arguments and return
types to be integers, (which is the default type). When it sees the actual function display,
the arguments and type contradicts with what it has assumed previously. Hence a compile
time error occurs.
12. main()
{
int c=- -2;
printf("c=%d",c);
}
14
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules
applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be
applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14. main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than >
symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false
(zero).
15. #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing
to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then
incremented
to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is
15
incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of
yy are to be accessed through the instance of structure xx, which needs an instance of yy
to be known. If the instance is created after defining the structure the compiler will not
know about the instance relative to xx. Hence for nested structure yy you have to declare
member.
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to
right. The evaluation is by popping out from the stack. and the evaluation is from right to
left, hence the result.
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes
i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4
i.e. 16*4 = 64
22. main()
{
17
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is h, which is changed to i by
executing ++*p and pointer moves to point, a which is similarly changed to b and so
on. Similarly blank space is converted to !. Thus, we obtain value in p becomes ibj!
gsjfoet and since p reaches \0 and p1 points to p thus p1doesnot print anything.
23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So
the most recently assigned value will be taken.
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the
compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler
looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any
problem
18
25. main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal numbers.
27)
main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In
the second clrscr(); is a function declaration (because it is not inside any
function).
28)
29)
void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
30)
19
main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any
number of printf's may be given. All of them take only the first two values.
If more number of assignments given in the program,then printf will take
garbage values.
31)
main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be
applied any number of times provided it is meaningful. Here p points to
the first character in the string "Hello". *p dereferences it and so its value
is H. Again & references it to an address and * dereferences it to the value
H.
32)
main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is
limited to functions . The label 'here' is available in function fun() Hence it
is not visible in function main.
33)
main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
20
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
34)
void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
35)
void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal
combination of operators.
36)
#include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that
we cannot use variable names directly so an error).
21
Note:
Enumerated types can be used in case statements.
37)
main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is
given as input which should have been scanned successfully. So number of
items read is 1.
38)
39)
main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated".
Here it evaluates to 0 (false) and comes out of the loop, and i is
incremented (note the semicolon after the for loop).
40)
22
#include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
#include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure
declaration
42)
#include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
43)
main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
23
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else.
The compiler passes the external variable to be resolved by the linker. So
compiler doesn't find an error. During linking the linker searches for the
definition of i. Since it is not found the linker flags an error.
44)
main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration.
Even though a is a global variable, it is not available for main. Hence an
error.
45)
main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
46)
main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about
it. So the default return type (ie, int) is assumed. But when compiler sees
the actual definition of show mismatch occurs since it is declared as void.
Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
24
47)
main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(%u %u %u %d \n,a,*a,**a,***a);
printf(%u %u %u %d \n,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4
7 8 3
4 2
2 2
3
3
4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element .
since the indirection ***a gives the value. Hence, the first line of the
output.
for the second printf a+1 increases in the third dimension thus points to
value at 114, *a+1 increments in second dimension thus points to 104, **a
+1 increments the first dimension thus points to 102 and ***a+1 first gets
the value at first location and then increments it by 1. Hence, the output.
48)
main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(%d ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(%d ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and
may be of any of scalar type for the any operator, array name only when
subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
49)
25
main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(\n %d %d %d, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(\n %d %d %d, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(\n %d %d %d, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(\n %d %d %d, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0
1
2
3
4
100
102
104
106
108
p
100
102
104
106
108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if
scaling factor for integer is 2 bytes. Now ptr p is value in ptr starting
location of array p, (1002 1000) / (scaling factor) = 1, *ptr a = value at
address pointed by ptr starting value of array a, 1002 has a value 102 so
the value is (102 100)/(scaling factor) = 1, **ptr is the value stored in
the location pointed by the pointer of ptr = value pointed by value pointed
by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is
1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the second printf are ptr
p = 2, *ptr a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the third printf are ptr
p = 3, *ptr a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed
by the value is incremented by the scaling factor. So the value in array p at
location 1006 changes from 106 10 108,. Hence, the outputs for the fourth
printf are ptr p = 1006 1000 = 3, *ptr a = 108 100 = 4, **ptr = 4.
50)
26
main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(%s ,(q+j));
for (j=0; j<3; j++) printf(%c ,*(q+j));
main( )
{
void *vp;
char ch = g, *cp = goofy;
int j = 20;
vp = &ch;
printf(%c, *(char *)vp);
vp = &j;
printf(%d,*(int *)vp);
vp = cp;
printf(%s,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer.
vp = &ch stores address of char ch and the next statement prints the value
stored in vp after type casting it to the proper data type pointer. the output
is g. Similarly the output from second printf is 20. The third printf
statement type casts it to print the string from the 4 th value hence the
output is fy.
52)
main ( )
{
static char *s[ ] = {black, white, yellow, violet};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(%s,*--*++p + 3);
}
Answer:
27
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4
strings. Then we have ptr which is a pointer to a pointer of type char and a
variable p which is a pointer to a pointer to a pointer of type char. p hold
the initial value of ptr, i.e. p = s+3. The next statement increment value in
p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is
executed and we get s+1 1 = s . the indirection operator now gets the
value from the array of s and adds 3 to the starting address. The string is
printed starting from this position. Thus, the output is ck.
53)
54)
28
main()
{
int i, n;
char *x = girl;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf(%s\n,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value girl. The
strlen function returns the length of the string, thus n has a value 4. The
next statement assigns value at the nth location (\0) to the first location.
Now the string becomes \0irl . Now the printf statement prints the string
after each iteration it increments it starting position. Loop starts from 0 to
4. The first time x[0] = \0 hence it prints nothing and pointer value is
incremented. The second time it prints from x[1] i.e irl and the third
time it prints rl and the last time it prints l and the loop terminates.
int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain conditions are
satisfied. If assertion fails, the program will terminate reporting the same.
After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.
55)
main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can
just ignore it just because it has no effect in the expressions (hence the
name dummy operator).
56)
What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches
with a quotation mark and then it reads all character upto another
quotation mark.
59)
29
60)
main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called
its return address is stored in the call stack. Since there is no condition to
terminate the function call, the call stack overflows at runtime. So it
terminates the program and results in an error.
61)
main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is
an empty type. In the second line you are creating variable vptr of type
void * and v of type void hence an error.
62)
main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
255
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the
pointer variable. In second sizeof the name str2 indicates the name of the
array whose size is 5 (including the '\0' termination character). The third
sizeof is similar to the second one.
63)
main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
30
#define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes.
The check by if condition is boolean value false so it goes to else. In
second if -1 is boolean value true hence "TRUE" is printed.
65)
main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they
are concatenated (this is called as "stringization" operation). So the string
is as if it is given as "%d==1 is %s". The conditional operator( ?: )
evaluates to "TRUE".
66)
main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
31
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
67)
#define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to
declare the variable name of the type arr2. But it is not the case of arr1.
Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for
declaring new types.
68)
int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is
declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In
the next block, i has value 20 and so printf prints 20. In the outermost
block, i is declared as extern, so no storage space is allocated for it. After
32
main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that
block only. But the lifetime of i is lifetime of the function so it lives upto
the exit of main function. Since the i is still allocated space, *j prints the
value stored in i since j points i.
70)
main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first
you just print the value of i. After that the value of the expression -i = -(-1)
is printed.
71)
#include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
72)
#include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
33
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to
access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. now q is
pointing to starting address of a.if you print *q meAnswer:it will print first
element of 3D array.
73)
#include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it
will take integer value. i value may be stored either in register or in
memory.
74)
main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
76)
struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
34
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
77)
struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
78)
main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be
returned.
79)
main()
{
char *p;
35
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to
their corresponding data-types.
80)
main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
81)
82)
# include <stdio.h>
int one_d[]={1,2,3};
main()
36
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
83)
# include<stdio.h>
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
85)
#include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against
NULL.
37
86)
main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth
value of the boolean expression. So the statement following the if
statement is not executed. The values of i and j remain unchanged and get
printed.
87)
main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information
from the accumulator. Here _AH is the pseudo global variable denoting
the accumulator. Hence, the value of the accumulator is set 1000 so the
function returns value 1000.
88)
int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t
i
x
4
0
-4
3
1
-2
2
2
0
38
89)
main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the
rightmost value is returned and the other values are evaluated and ignored.
Thus the value of last variable y is returned to check in if. Since it is a non
zero value if becomes true so, "hello" will be printed.
90)
main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition becomes false and control
comes out of the loop.
91)
In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
92)
39
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
93)
main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again
decremented by 2, it points to '%d\n' and 300 is printed.
94)
main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b'
then after incrementing to 'c' so bc will be printed.
95)
func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
40
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2
and 3, integers. When this function is invoked from main, the following
substitutions for formal parameters take place: func for pf, 3 for val1 and 6
for val2. This function returns the result of the operation performed by the
function 'func'. The function func has two integer parameters. The formal
parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6,
a==b returns 0. therefore the function returns 0 which in turn is returned
by the function 'process'.
96)
void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0000
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated
for only once, as it encounters the statement. The function main() will be called
recursively unless I becomes equal to 0, and since main() is recursively called, so
the value of static I ie., 0 will be printed every time the control is returned.
97)
void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be
the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is
passed, after the first expression the value in ret will be 6, as ret is integer hence
the value stored in ret will have implicit type conversion from float to int. The ret
is returned in main() it is printed after and preincrement.
98)
41
void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be
counted from 0 till the null character. Hence the 'I' will hold the value equal to 5,
after the pre-increment in the printf statement, the 6 will be printed.
99)
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100)
void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing
a null character returns 1 which makes the if statement true, thus
"Ok here" is printed.
101)
void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be
done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
42
3. Used when the exact pointer type will be known at a later point of
time.
102)
void main()
{
int i=i++,j=j++,k=k++;
printf(%d%d%d,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its
declaration.
So expressions such as i = i++ are valid statements. The i, j and k are
automatic variables and so they contain some garbage value. Garbage in
is garbage out (GIGO).
103)
void main()
{
static int i=i++, j=j++, k=k++;
printf(i = %d j = %d k = %d, i, j, k);
}
Answer:
i=1j=1k=1
Explanation:
Since static variables are initialized to zero by default.
104)
void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf
returns no of characters printed and this value also cannot be predicted.
Still the outer printf prints something and so returns a non-zero value. So
it encounters the break statement and comes out of the while statement.
104)
main()
{
43
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534..
Explanation:
Since i is an unsigned integer it can never become negative. So the
expression i-- >=0 will always be true, leading to an infinite loop.
105)
#include<conio.h>
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces
to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
106)
main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
107)
108)
main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
44
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes
0 it comes out of while loop. Due to post-increment on i the value of i
while printing is 1.
109)
main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the
expression and now the while loop is,
while(i--!=0) which is false
and so breaks out of while loop. The value 1 is printed due to the postdecrement operator.
113)
main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
110)
main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
45
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
113) main()
{
47
The name error is used in the two meanings. One means that it is a
enumerator constant with value 1. The another use is that it is a type name
(due to typedef) for enum errorType. Given a situation the compiler cannot
distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not
issue error (in pure technical terms, names can only be overloaded in
different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for
programmers convenience.
117)
Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at
any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable
name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword
as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is
perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real
programming dont use such overloading of names. It reduces the readability of
the code. Possible doesnt mean that we should use it!
48
118)
#ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The
name something is not already known to the compiler making the
declaration
int some = 0;
effectively removed from the source code.
119)
#if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
00
Explanation
This code is to show that preprocessor expressions are not the
same as the ordinary expressions. If a name is not known the
preprocessor treats it to be equal to zero.
arr2D[1]
arr2D[2]
arr2D[3]
The name arr2D refers to the beginning of all the 3 arrays. *arr2D
refers to the start of the first 1D array (of 3 integers) that is the
same address as arr2D. So the expression (arr2D == *arr2D) is true
(1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesnt change the value/meaning. Again arr2D[0] is the another
way of telling *(arr2D + 0). So the expression (*(arr2D + 0) ==
arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is
true(1) and the same is printed.
121) void main()
{
if(~0 == (unsigned int)-1)
printf(You can answer this if you know how values are represented in
memory);
}
Answer
You can answer this if you know how values are represented in
memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to
produce all ones to fill the space for an integer. 1 is represented in
unsigned value as all 1s and so both are equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
123)
50
main()
{
char *p = ayqm;
printf(%c,++*(p++));
}
Answer:
b
124)
main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is
required.
125)
main()
{
char *p = ayqm;
char c;
c = ++*p++;
printf(%c,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and +
+*p++. Parenthesis just works as a visual clue for the reader to see
which expression is first evaluated.
126)
int aaa() {printf(Hi);}
int bbb(){printf(hello);}
iny ccc(){printf(bye);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
51
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes
no arguments and returns the type int. By the assignment ptr[0] = aaa; it
means that the first function pointer in the array is initialized with the
address of the function aaa. Similarly, the other two array elements also
get initialized with the addresses of the functions bbb and ccc. Since ptr[2]
contains the address of the function ccc, the call to the function ptr[2]() is
same as calling ccc(). So it results in printing "bye".
127)
main()
{
int i=5;
printf(%d,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher
precedence than = operator. In the inner expression, ++i is equal to 6
yielding true(1). Hence the result.
128)
main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = c the string becomes, %c\n. Since this
string becomes the format string for printf and ASCII value of 65 is A,
the same gets printed.
129)
130)
52
131)
main()
{
while (strcmp(some,some\0))
printf(Strings are not equal\n);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So
some and some\0 are equivalent. So, strcmp returns 0 (false) hence
breaking out of the while loop.
main()
{
char str1[] = {s,o,m,e};
char str2[] = {s,o,m,e,\0};
while (strcmp(str1,str2))
printf(Strings are not equal\n);
}
Answer:
Strings are not equal
Strings are not equal
.
Explanation:
If a string constant is initialized explicitly with characters, \0 is not
appended automatically to the string. Since str1 doesnt have null
termination, it treats whatever the values that are in the following positions
as part of the string until it randomly reaches a \0. So str1 and str2 are
not the same, hence the result.
132)
main()
{
int i = 3;
for (;i++=0;) printf(%d,i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it
cannot appear on the left hand side of an assignment operation.
133)
void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(%d,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(%d,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc
returns the allocated memory space initialized to zeros.
134)
void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(%d, i);
}
Answer:
53
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional
operator evaluates to false, executing i--. This continues till the integer
value rotates to positive value (32767). The while condition becomes false
and hence, comes out of the while loop, printing the i value.
135)
main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the
question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136)
137)
54
main()
{
int i=5,j=10;
i=i&=j&&10;
: legal
: illegal
printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the
result.
138)
main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
41
Explanation:
The boolean expression needs to be evaluated only till the truth value of
the expression is not known. j is not equal to zero itself means that the
expressions truth value is 1. Because it is followed by || and true ||
(anything) => true where (anything) will not be evaluated. So the
remaining expression is not evaluated and so the value of i remains the
same.
Similarly when && operator is involved in an expression, when any of the
operands become false, the whole expressions truth value becomes false
and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
139)
main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.
140)
55
main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.
141)
main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no
use in resolving it.
142)
main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1
and f2 ultimately affects only the value of a.
143)
main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p),
sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
56
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array
and it is not the same as the sizeof the pointer variable. Here the sizeof(a)
where a is the character array and the size of the array is 5 because the
space necessary for the terminating NULL character should also be taken
into account.
144)
145)
146)
main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
57
}
Answer:
1
2
3
4
5
6
7
8
9
4
7
2
5
8
3
6
9
2
3
4
5
6
7
8
9
4
7
2
5
8
3
6
9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].
147)
main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary
variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take
any number of arguments (not no arguments) and returns nothing. So this
doesnt issue a compiler error by the call swap(&x,&y); that has two
arguments.
This convention is historically due to pre-ANSI style (referred to as
Kernighan and Ritchie style) style of function declaration. In that style, the
swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap
will look like, void swap() which means the swap can take any number of
arguments.
148)
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
58
}
Answer:
11
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so
the individual bytes are taken by casting it to char * and get printed.
149)
main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
21
Explanation:
The integer value 257 can be represented in binary as, 00000001
00000001. Remember that the INTEL machines are small-endian
machines. Small-endian means that the lower order bytes are stored in the
higher memory addresses and the higher order bytes are stored in lower
addresses. The integer value 258 is stored in memory as: 00000001
00000010.
150)
main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is
stored in memory (small-endian) as: 00101100 00000001. Result of the
expression *++ptr = 2 makes the memory representation as: 00101100
00000010. So the integer corresponding to it is 00000010 00101100 =>
556.
151)
59
#include <stdio.h>
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr<least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After ptr reaches the end of the string the value pointed by str is \0.
So the value of str is less than that of least. So the value of least
finally is 0.
152)
153)
main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno,
&student.dob.month,
&student.dob.year);
&student.dob.day,
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of student the member of type struct date is
given. The compiler doesnt have the definition of date structure (forward
reference is not allowed in C in this case) so it issues an error.
154)
main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month,
&student.dob.year);
60
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition
of student but to have a variable of type struct date the definition of the
structure is required.
155)
There were 10 records stored in somefile.dat but the following program printed
11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(somefile.dat,r);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will
return EOF only when fread tries to read another record and fails
reading EOF (and returning EOF). So it prints the last record
again. After this only the condition feof(fp) becomes false, hence
comes out of the while loop.
156)
157)
158)
61
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The
check s != NULL is for error/exception handling and for that assert
shouldnt be used. A plain if and the corresponding remedy statement has
to be given.
159)
160)
void main()
{
int *i = 0x400; // i points to the address 400
*i = 0;
// set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to
some location whose value may not be available for modification. This
type of pointer in which the non-availability of the implementation of the
referenced location is known as 'incomplete type'.
161)
62
void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert
macro. Hence nothing is printed.
163)
164)
63
165)
166)
167)
void main()
{
printf(sizeof (void *) = %d \n, sizeof( void *));
printf(sizeof (int *) = %d \n, sizeof(int *));
printf(sizeof (double *) = %d \n, sizeof(double *));
printf(sizeof(struct unknown *) = %d \n, sizeof(struct unknown *));
}
Answer
:
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *) = 2
sizeof(struct unknown *) = 2
Explanation:
The pointer to any type is of same size.
168)
64
170)
void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(%d,k);
}
Answer:
Compiler Error: Unexpected end of file in comment started in line 5.
Explanation:
The programmer intended to divide two integers, but by the
maximum munch rule, the compiler treats the operator
sequence / and * as /* which happens to be the starting of
comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171)
void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(%c %d \n, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed then
ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is
always smaller than 127.
172)
65
Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location 0x400.
173)
main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to get
stored.
174)
main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string
HELL and doesnt have enough space to store the terminating null
character. So it prints the HELL correctly and continues to print garbage
values till it accidentally comes across a NULL character.
175)
main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u %u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the type
is not known and as an intermediate address storage type. No pointer
arithmetic can be done on it and you cannot apply indirection operator (*)
on void pointers.
176)
main()
{
66
{
{
extern int i;
int i=20;
char *someFun1()
{
char temp[ ] = string";
return temp;
}
char *someFun2()
{
char temp[ ] = {s, t,r,i,n,g};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1()
temp is a character array and so the space for it is allocated in heap and is initialized with
character string string. This is created dynamically as the function is called, so is also
deleted dynamically on exiting the function so the string data is not available in the
calling
function main() leading to print some garbage values. The function someFun2()
67
also suffers from the same problem but the problem can be easily identified in this case.
C++ Aptitude
and OOPS
C++ Aptitude and OOPS
Note : All the programs are tested under Turbo C++ 3.0, 4.5 and Microsoft VC++ 6.0
compilers.
It is assumed that,
Programs run under Windows environment,
The underlying machine is an x86 based system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
1) class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is " << *ptr;
}
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in someFunc
Null pointer assignment(Run-time error)
Explanation:
68
someFunc calls it with an array of bae objects, so it works correctly and prints the bval of
all the objects. When Somefunc is called the second time the argument passed is the
pointeer to an array of derived class objects and not the array of base class objects. But
that is what the function expects to be sent. So the derived class pointer is promoted to
base class pointer and the address is sent to the function. SomeFunc() knows nothing
about this and just treats the pointer as an array of base class objects. So when arr++ is
met, the size of base class object is taken into consideration and is incremented by
sizeof(int) bytes for bval (the deri class objects have bval and dval as members and so is
of size >= sizeof(int)+sizeof(int) ).
4) class base
{
public:
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from base
Explanation:
As we have seen in the previous case, SomeFunc expects a pointer to a base class.
Since a pointer to a derived class object is passed, it treats the argument only as a base
class pointer and the corresponding base function is called.
5) class base
{
public:
virtual void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void baseFun(){ cout<< "from derived"<<endl;}
};
void
SomeFunc(base *baseObj)
70
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from derived
Explanation:
Remember that baseFunc is a virtual function. That means that it supports runtime polymorphism. So the function corresponding to the derived class object is called.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
/*
Answer :
Compiler Error: 'ra',reference must be initialized
Explanation :
Pointers are different from references. One of the main
differences is that the pointers can be both initialized and assigned,
whereas references can only be initialized. So this code issues an error.
*/
const int size = 5;
void print(int *ptr)
{
cout<<ptr[0];
}
void print(int ptr[size])
{
cout<<ptr[0];
}
void main()
{
int a[size] = {1,2,3,4,5};
int *b = new int(size);
print(a);
print(b);
71
}
/*
Answer:
Compiler Error : function 'void print(int *)' already has a body
Explanation:
Arrays cannot be passed to functions, only pointers (for arrays, base addresses)
can be passed. So the arguments int *ptr and int prt[size] have no difference
as function arguments. In other words, both the functoins have the same signature and
so cannot be overloaded.
*/
class some{
public:
~some()
{
cout<<"some's destructor"<<endl;
}
};
void main()
{
some s;
s.~some();
}
/*
Answer:
some's destructor
some's destructor
Explanation:
Destructors can be called explicitly. Here 's.~some()' explicitly calls the
destructor of 's'. When main() returns, destructor of s is called again,
hence the result.
*/
#include <iostream.h>
class fig2d
{
int dim1;
int dim2;
public:
fig2d() { dim1=5; dim2=6;}
virtual void operator<<(ostream & rhs);
};
void fig2d::operator<<(ostream &rhs)
{
rhs <<this->dim1<<" "<<this->dim2<<" ";
72
}
//
}
/*
Answer :
56
Explanation:
In this program, the << operator is overloaded with ostream as argument.
This enables the 'cout' to be present at the right-hand-side. Normally, 'cout'
is implemented as global function, but it doesn't mean that 'cout' is not possible
to be overloaded as member function.
Overloading << as virtual member function becomes handy when the class in which
it is overloaded is inherited, and this becomes available to be overrided. This is as
opposed
to global friend functions, where friend's are not inherited.
*/
class opOverload{
public:
bool operator==(opOverload temp);
};
bool opOverload::operator==(opOverload temp){
if(*this == temp ){
cout<<"The both are same objects\n";
return true;
}
else{
cout<<"The both are different\n";
return false;
73
}
}
void main(){
opOverload a1, a2;
a1= =a2;
}
Answer :
Runtime Error: Stack Overflow
Explanation :
Just like normal functions, operator functions can be called recursively. This
program just illustrates that point, by calling the operator == function recursively, leading
to an infinite loop.
class complex{
double re;
double im;
public:
complex() : re(1),im(0.5) {}
bool operator==(complex &rhs);
operator int(){}
};
bool complex::operator == (complex &rhs){
if((this->re == rhs.re) && (this->im == rhs.im))
return true;
else
return false;
}
int main(){
complex c1;
cout<< c1;
}
Answer : Garbage value
Explanation:
The programmer wishes to print the complex object using output
re-direction operator,which he has not defined for his lass.But the compiler instead of
giving an error sees the conversion function
and converts the user defined object to standard object and prints
some garbage value.
class complex{
double re;
double im;
public:
74
complex() : re(0),im(0) {}
complex(double n) { re=n,im=n;};
complex(int m,int n) { re=m,im=n;}
void print() { cout<<re; cout<<im;}
};
void main(){
complex c3;
double i=5;
c3 = i;
c3.print();
}
Answer:
5,5
Explanation:
Though no operator= function taking complex, double is defined, the double on
the rhs is converted into a temporary object using the single argument constructor taking
double and assigned to the lvalue.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
Answer :
Compiler Error: 'ra',reference must be initialized
Explanation :
Pointers are different from references. One of the main
differences is that the pointers can be both initialized and assigned,
whereas references can only be initialized. So this code issues an error.
Try it Yourself
1) Determine the output of the 'C++' Codelet.
class base
{
public :
out()
{
cout<<"base ";
}
};
class deri{
public : out()
{
cout<<"deri ";
75
}
};
void main()
{
deri dp[3];
base *bp = (base*)dp;
for (int i=0; i<3;i++)
(bp++)->out();
}
2) Justify the use of virtual constructors and destructors in C++.
3) Each C++ object possesses the 4 member fns,(which can be declared by the
programmer explicitly or by the implementation if they are not available). What are
those 4 functions?
4) What is wrong with this class declaration?
class something
{
char *str;
public:
something(){
st = new char[10]; }
~something()
{
delete str;
}
};
5) Inheritance is also known as -------- relationship. Containership as
relationship.
________
76
1. What is a modifier?
Answer:
A modifier, also called a modifying function is a member function that changes the
value of at least one data member. In other words, an operation that modifies the state of
an object. Modifiers are also known as mutators.
2. What is an accessor?
Answer:
An accessor is a class operation that does not modify the state of an object. The
accessor functions need to be declared as const operations
3. Differentiate between a template class and class template.
Answer:
Template class:
A class template specifies how individual classes can be constructed much like the
way a class specifies how individual objects can be constructed. Its jargon for plain
classes.
4. When does a name clash occur?
Answer:
A name clash occurs when a name is defined in more than one place. For
example., two different class libraries could give two different classes the same name. If
you try to use many class libraries at the same time, there is a fair chance that you will be
unable to compile or link the program because of name clashes.
5. Define namespace.
Answer:
It is a feature in c++ to minimize name collisions in the global name space. This
namespace keyword assigns a distinct name to a library that allows other libraries to use
the same identifier names without creating any name collisions. Furthermore, the
compiler uses the namespace signature for differentiating the definitions.
6. What is the use of using declaration.
Answer:
A using declaration makes it possible to use a name from a namespace without the
scope operator.
7. What is an Iterator class?
Answer:
A class that is used to traverse through the objects maintained by a container
class. There are five categories of iterators:
input iterators,
output iterators,
forward iterators,
bidirectional iterators,
random access.
An iterator is an entity that gives access to the contents of a container object
77
without violating encapsulation constraints. Access to the contents is granted on a one-at-
a-time basis in order. The order can be storage order (as in lists and queues) or some
arbitrary order (as in array indices) or according to some ordering relation (as in an
ordered binary tree). The iterator is a construct, which provides an interface that, when
called, yields either the next element in the container, or some value denoting the fact that
there are no more elements to examine. Iterators hide the details of access to and update
of the elements of a container class.
The simplest and safest iterators are those that permit read-only access to the
contents of a container class. The following code fragment shows how an iterator might
appear in code:
cont_iter:=new cont_iterator();
x:=cont_iter.next();
while x/=none do
...
s(x);
...
x:=cont_iter.next();
end;
In this example, cont_iter is the name of the iterator. It is created on the first line by
instantiation of cont_iterator class, an iterator class defined to iterate over some container
class, cont. Succesive elements from the container are carried to x. The loop terminates
when x is bound to some empty value. (Here, none)In the middle of the loop, there is s(x)
an operation on x, the current element from the container. The next element of the
container is obtained at the bottom of the loop.
9. List out some of the OODBMS available.
Answer:
ONTOS of Ontos.
Object Oriented Analysis and Design (OOA/D) (Coad and Yourdon 1991).
*i=0;
//set the value of memory location pointed by i.
Incomplete types are otherwise called uninitialized pointers.
12. What is a dangling pointer?
Answer:
A dangling pointer arises when you use the address of an object after its lifetime
is over.
This may occur in situations like returning addresses of the automatic variables from a
function or using the address of the memory block after it is freed.
13. Differentiate between the message and method.
Answer:
Message
Method
Objects communicate by sending messages Provides response to a message.
to each other.
A message is sent to invoke a method.
It is an implementation of an operation.
14. What is an adaptor class or Wrapper class?
Answer:
A class that has no functionality of its own. Its member functions hide the use of a
third party software component or an object with the non-compatible interface or a nonobject- oriented implementation.
15. What is a Null object?
Answer:
It is an object of some class whose purpose is to indicate that a real object of that
class does not exist. One common use for a null object is a return value from a member
function that is supposed to return an object with some specified properties but cannot
find such an object.
16. What is class invariant?
Answer:
A class invariant is a condition that defines all valid states for an object. It is a
logical condition to ensure the correct working of a class. Class invariants must hold
when an object is created, and they must be preserved under all operations of the class. In
particular all class invariants are both preconditions and post-conditions for all operations
or member functions of the class.
17. What do you mean by Stack unwinding?
Answer:
It is a process during exception handling when the destructor is called for all local
objects between the place where the exception was thrown and where it is caught.
18. Define precondition and post-condition to a member function.
Answer:
Precondition:
A precondition is a condition that must be true on entry to a member function. A
class is used correctly if preconditions are never false. An operation is not responsible for
doing anything sensible if its precondition fails to hold.
79
For example, the interface invariants of stack class say nothing about pushing yet
another element on a stack that is already full. We say that isful() is a precondition of the
push operation.
Post-condition:
A post-condition is a condition that must be true on exit from a member function
if the precondition was valid on entry to that function. A class is implemented correctly if
post-conditions are never false.
For example, after pushing an element on the stack, we know that isempty() must
necessarily hold. This is a post-condition of the push operation.
19. What are the conditions that have to be met for a condition to be an invariant of the
class?
Answer:
The condition should hold at the end of every constructor.
The condition should hold at the end of every mutator(non-const) operation.
20. What are proxy objects?
Answer:
Objects that stand for other objects are called proxy objects or surrogates.
Example:
template<class T>
class Array2D
{
public:
class Array1D
{
public:
T& operator[] (int index);
const T& operator[] (int index) const;
...
};
Array1D operator[] (int index);
const Array1D operator[] (int index) const;
...
};
The following then becomes legal:
Array2D<float>data(10,20);
........
cout<<data[3][6]; // fine
Here data[3] yields an Array1D object and the operator [] invocation on that
object yields the float in position(3,6) of the original two dimensional array. Clients of the
Array2D class need not be aware of the presence of the Array1D class. Objects of this
latter class stand for one-dimensional array objects that, conceptually, do not exist for
clients of Array2D. Such clients program as if they were using real, live, two-dimensional
arrays. Each Array1D object stands for a one-dimensional array that is absent from a
conceptual model used by the clients of Array2D. In the above example, Array1D is a
proxy
class. Its instances stand for one-dimensional arrays that, conceptually, do not
80
exist.
}
Given this, we can write code say a member that can store actions for later
execution without using pointers to functions, without knowing anything about the
objects involved, and without even knowing the name of the operation it invokes. For
example:
class write_file : public Action
{
File& f;
public:
int do_it(int)
{
return fwrite( ).suceed( );
}
};
class error_message: public Action
{
response_box db(message.cstr( ),"Continue","Cancel","Retry");
switch (db.getresponse( ))
{
case 0: return 0;
case 1: abort();
case 2: current_operation.redo( );return 1;
}
};
A user of the Action class will be completely isolated from any knowledge of
derived classes such as write_file and error_message.
31. When can you tell that a memory leak will occur?
Answer:
A memory leak occurs when a program loses the ability to free a block of
dynamically allocated memory.
32.What is a parameterized type?
Answer:
A template is a parameterized construct or type containing generic code that can
use or manipulate any type. It is called parameterized because an actual type is a
parameter of the code body. Polymorphism may be achieved through parameterized
types. This type of polymorphism is called parameteric polymorphism. Parameteric
polymorphism is the mechanism by which the same code is used on different types
passed as parameters.
33. Differentiate between a deep copy and a shallow copy?
Answer:
Deep copy involves using the contents of one object to create another instance of
the same class. In a deep copy, the two objects may contain ht same information but the
target object will have its own buffers and resources. the destruction of either object will
not affect the remaining object. The overloaded assignment operator would create a deep
copy of objects.
Shallow copy involves copying the contents of one object into another instance of
83
the same class thus creating a mirror image. Owing to straight copying of references and
pointers, the two objects will share the same externally contained contents of the other
object to be unpredictable.
Explanation:
Using a copy constructor we simply copy the data values member by member.
This method of copying is called shallow copy. If the object is a simple class, comprised
of built in types and no pointers this would be acceptable. This function would use the
values and the objects and its behavior would not be altered with a shallow copy, only the
addresses of pointers that are members are copied and not the value the address is
pointing to. The data values of the object would then be inadvertently altered by the
function. When the function goes out of scope, the copy of the object with all its data is
popped off the stack.
If the object has any pointers a deep copy needs to be executed. With the deep
copy of an object, memory is allocated for the object in free store and the elements
pointed to are copied. A deep copy is used for objects that are returned from a function.
34. What is an opaque pointer?
Answer:
A pointer is said to be opaque if the definition of the type to which it points to is
not included in the current translation unit. A translation unit is the result of merging an
implementation file with all its headers and header files.
35. What is a smart pointer?
Answer:
A smart pointer is an object that acts, looks and feels like a normal pointer but
offers more functionality. In C++, smart pointers are implemented as template classes
that encapsulate a pointer and override standard pointer operators. They have a number of
advantages over regular pointers. They are guaranteed to be initialized as either null
pointers or pointers to a heap object. Indirection through a null pointer is checked. No
delete is ever necessary. Objects are automatically freed when the last pointer to them has
gone away. One significant problem with these smart pointers is that unlike regular
pointers, they don't respect inheritance. Smart pointers are unattractive for polymorphic
code. Given below is an example for the implementation of smart pointers.
Example:
X& operator *( );
const X& operator*( ) const;
X* operator->() const;
smart_pointer(const smart_pointer <X> &);
const smart_pointer <X> & operator =(const smart_pointer<X>&);
~smart_pointer();
private:
//...
};
84
This class implement a smart pointer to an object of type X. The object itself is
located on the heap. Here is how to use it:
smart_pointer <employee> p= employee("Harris",1333);
Like other overloaded operators, p will behave like a regular pointer,
cout<<*p;
p->raise_salary(0.5);
36. What is reflexive association?
Answer:
The 'is-a' is called a reflexive association because the reflexive association permits
classes to bear the is-a association not only with their super-classes but also with
themselves. It differs from a 'specializes-from' as 'specializes-from' is usually used to
describe the association between a super-class and a sub-class. For example:
Printer is-a printer.
37. What is slicing?
Answer:
Slicing means that the data added by a subclass are discarded when an object of
the subclass is passed or returned by value or from a function expecting a base class
object.
Explanation:
Consider the following class declaration:
class base
{
...
base& operator =(const base&);
base (const base&);
}
void fun( )
{
base e=m;
e=m;
}
As base copy functions don't know anything about the derived only the base part
of the derived is copied. This is commonly referred to as slicing. One reason to pass
objects of classes in a hierarchy is to avoid slicing. Other reasons are to preserve
polymorphic behavior and to gain efficiency.
38. What is name mangling?
Answer:
Name mangling is the process through which your c++ compilers give each
function in your program a unique name. In C++, all programs have at-least a few
functions with the same name. Name mangling is a concession to the fact that linker
always insists on all function names being unique.
Example:
In general, member names are made unique by concatenating the name of the
member with that of the class e.g. given the declaration:
class Bar
{
public:
85
int ival;
...
};
ival becomes something like:
// a possible member name mangling
ival__3Bar
Consider this derivation:
class Foo : public Bar
{
public:
int ival;
...
}
The internal representation of a Foo object is the concatenation of its base and
derived class members.
// Pseudo C++ code
// Internal representation of Foo
class Foo
{
public:
int ival__3Bar;
int ival__3Foo;
...
};
Unambiguous access of either ival members is achieved through name mangling.
Member functions, because they can be overloaded, require an extensive mangling to
provide each with a unique name. Here the compiler generates the same name for the two
overloaded instances(Their argument lists make their instances unique).
39. What are proxy objects?
Answer:
Objects that points to other objects are called proxy objects or surrogates. Its an
object that provides the same interface as its server object but does not have any
functionality. During a method invocation, it routes data to the true server object and
sends back the return value to the object.
40. Differentiate between declaration and definition in C++.
Answer:
A declaration introduces a name into the program; a definition provides a unique
description of an entity (e.g. type, instance, and function). Declarations can be repeated in
a given scope, it introduces a name in a given scope. There must be exactly one definition
of every object, function or class used in a C++ program.
A declaration is a definition unless:
it declares a function without specifying its body,
it contains an extern specifier and no initializer or function body,
it is the declaration of a static class data member without a class definition,
it is a class name definition,
it is a typedef declaration.
A definition is a declaration unless:
it defines a static class data member,
it defines a non-inline member function.
86
}
};
This function returns a pointer to a Widget object that's constructed within the
buffer passed to the function. Such a function might be useful for applications using
shared memory or memory-mapped I/O, because objects in such applications must be
placed at specific addresses or in memory allocated by special routines.
OOAD
1. What do you mean by analysis and design?
Analysis:
Basically, it is the process of determining what needs to be done before
how it should be done. In order to accomplish this, the developer refers the existing
systems and documents. So, simply it is an art of discovery.
Design:
It is the process of adopting/choosing the one among the many, which best
accomplishes the users needs. So, simply, it is compromising mechanism.
2. What are the steps involved in designing?
Before getting into the design the designer should go through the SRS prepared
by the System Analyst.
The main tasks of design are Architectural Design and Detailed Design.
In Architectural Design we find what are the main modules in the problem
domain.
In Detailed Design we find what should be done within each module.
3. What are the main underlying concepts of object orientation?
Objects, messages, class, inheritance and polymorphism are the main concepts of
object orientation.
4. What do u meant by "SBI" of an object?
SBI stands for State, Behavior and Identity. Since every object has the above
three.
State:
It is just a value to the attribute of an object at a particular time.
Behaviour:
It describes the actions and their reactions of that object.
Identity:
An object has an identity that characterizes its own existence. The identity
makes it possible to distinguish any object in an unambiguous way, and independently
from its state.
5. Differentiate persistent & non-persistent objects?
Persistent refers to an object's ability to transcend time or space. A persistent
object stores/saves its state in a permanent storage system with out losing the information
represented by the object.
A non-persistent object is said to be transient or ephemeral. By default objects are
considered as non-persistent.
88
Active objects are one which instigate an interaction which owns a thread and
they are responsible for handling control to other objects. In simple words it can be
referred as client.
Passive objects are one, which passively waits for the message to be processed. It
waits for another object that requires its services. In simple words it can be referred as
server.
Diagram:
client server
(Active) (Passive)
7. What is meant by software development method?
Software development method describes how to model and build software
systems in a reliable and reproducible way. To put it simple, methods that are used to
represent ones' thinking using graphical notations.
8. What are models and meta models?
Model:
It is a complete description of something (i.e. system).
Meta model:
It describes the model elements, syntax and semantics of the notation that
allows their manipulation.
9. What do you meant by static and dynamic modeling?
Static modeling is used to specify structure of the objects that exist in the problem
domain. These are expressed using class, object and USECASE diagrams.
But Dynamic modeling refers representing the object interactions during runtime.
It is represented by sequence, activity, collaboration and statechart diagrams.
10. How to represent the interaction between the modeling elements?
Model element is just a notation to represent (Graphically) the entities that exist
in the problem domain. e.g. for modeling element is class notation, object notation etc.
Relationships are used to represent the interaction between the modeling
elements.
The following are the Relationships.
Association: Its' just a semantic connection two classes.
e.g.:
uses
class A
class B
Aggregation: Its' the relationship between two classes which are related in the fashion
that master and slave. The master takes full rights than the slave. Since the slave
works under the master. It is represented as line with diamond in the master area.
ex:
car contains wheels, etc.
car
car
wheels
89 Containment: This relationship is applied when the part contained with in the whole
part, dies when the whole part dies.
class B
class B
class C
Booch: In this method classes are represented as "Clouds" which are not very easy
to draw as for as the developer's view is concern.
Diagram:
91
:obj1
:obj2
In the above representation I, obj1 sends message to obj2. But in the case of II the
data is transferred from obj1 to obj2.
22. USECASE is an implementation independent notation. How will the designer give the
implementation details of a particular USECASE to the programmer?
This can be accomplished by specifying the relationship called "refinement
which talks about the two different abstraction of the same thing.
Or example,
calculate pay
calculate
class1 class2 class3
92
23. Suppose a class acts an Actor in the problem domain, how to represent it in the static
model?
In this scenario you can use stereotype. Since stereotype is just a string that
gives extra semantic to the particular entity/model element. It is given with in the << >>.
class A
<< Actor>>
attributes
methods.
24. Why does the function arguments are called as "signatures"?
The arguments distinguish functions with the same name (functional
polymorphism). The name alone does not necessarily identify a unique function.
However, the name and its arguments (signatures) will uniquely identify a function.
In real life we see suppose, in class there are two guys with same name, but they
can be easily identified by their signatures. The same concept is applied here.
ex:
class person
{
public:
char getsex();
void setsex(char);
void setsex(int);
};
In the above example we see that there is a function setsex() with same name but
with different signature.
93
Quantitative
Aptitude
Quantitative Aptitude
Exercise 1
Solve the following and check with the answers given at the end.
1.
It was calculated that 75 men could complete a piece of work in 20 days. When
work was scheduled to commence, it was found necessary to send 25 men to
another project. How much longer will it take to complete the work?
2.
3.
A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a
false weight of 950gm. for a kg. His gain is %.
4.
A software engineer has the capability of thinking 100 lines of code in five
minutes and can type 100 lines of code in 10 minutes. He takes a break for five
minutes after every ten minutes. How many lines of codes will he complete typing
after an hour?
5.
A man was engaged on a job for 30 days on the condition that he would get a
wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for each
day of his absence. If he gets Rs. 216 at the end, he was absent for work for ...
days.
6.
7.
(d)
10
(d)
100
8.
A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at
20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the
cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was
Rs._______ for the horse and Rs.________ for the cart.
9.
A tennis marker is trying to put together a team of four players for a tennis
tournament out of seven available. males - a, b and c; females m, n, o and p. All
players are of equal ability and there must be at least two males in the team. For a
team of four, all players must be able to play with each other under the following
restrictions:
94
5
1
4
22
6
3
10.
11.
The number on the faces adjacent to the face marked 5 are _______ .
12.
Which of the following pairs does not correctly give the numbers on the opposite
faces.
(1)
6,5
(2)
4,1
(3)
1,3
(4)
4,2
13.
Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards.
Last year, each of them discovered that every tree in their own orchard bore
exactly the same number of apples. Further, if the third farmer gives one apple to
the first, and the fifth gives three to each of the second and the fourth, they would
all have exactly the same number of apples. What were the yields per tree in the
orchards of the third and fourth farmers?
14.
Five boys were climbing a hill. J was following H. R was just ahead of G. K was
between G & H. They were climbing up in a column. Who was the second?
15-18 John is undecided which of the four novels to buy. He is considering a spy
thriller, a Murder mystery, a Gothic romance and a science fiction novel. The
books are written by Rothko, Gorky, Burchfield and Hopper, not necessary in that
order, and published by Heron, Piegon, Blueja and sparrow, not necessary in that
order.
1 (1) The book by Rothko is published by Sparrow.
2 (2) The Spy thriller is published by Heron.
(3) The science fiction novel is by Burchfield and is not published by Blueja.
3 (4)The Gothic romance is by Hopper.
4
15.
Pigeon publishes ____________.
16.
95
17.
John purchases books by the authors whose names come first and third in
alphabetical order. He does not buy the books ______.
18.
On the basis of the first paragraph and statement (2), (3) and (4) only, it is
possible to deduce that
1. Rothko wrote the murder mystery or the spy thriller
2. Sparrow published the murder mystery or the spy thriller
3. The book by Burchfield is published by Sparrow.
19.
If a light flashes every 6 seconds, how many times will it flash in of an hour?
20.
If point P is on line segment AB, then which of the following is always true?
(1) AP = PB (2) AP > PB (3) PB > AP (4) AB > AP (5) AB > AP + PB
21.
All men are vertebrates. Some mammals are vertebrates. Which of the following
conclusions drawn from the above statement is correct.
All men are mammals
All mammals are men
Some vertebrates are mammals.
None
22.
Which of the following statements drawn from the given statements are correct?
Given:
All watches sold in that shop are of high standard. Some of the HMT watches are
sold in that shop.
a) All watches of high standard were manufactured by HMT.
b) Some of the HMT watches are of high standard.
c) None of the HMT watches is of high standard.
d) Some of the HMT watches of high standard are sold in that shop.
23-27.
1.
2.
3.
4.
5.
6.
23.
24.
25.
Which of the following towns must be situated both south and west of at least one
other town?
A. Ashland only
B. Ashland and Fredericktown
C. Dover and Fredericktown
96
Which of the following statements, if true, would make the information in the
numbered statements more specific?
(a) Coshocton is north of Dover.
(b) East Liverpool is north of Dover
(c) Ashland is east of Bowling green.
(d) Coshocton is east of Fredericktown
(e) Bowling green is north of Fredericktown
27.
Which of the numbered statements gives information that can be deduced from
one or more of the other statements?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
28.
Eight friends Harsha, Fakis, Balaji, Eswar, Dhinesh, Chandra, Geetha, and Ahmed
are sitting in a circle facing the center. Balaji is sitting between Geetha and
Dhinesh. Harsha is third to the left of Balaji and second to the right of Ahmed.
Chandra is sitting between Ahmed and Geetha and Balaji and Eshwar are not
sitting opposite to each other. Who is third to the left of Dhinesh?
29.
30.
The length of the side of a square is represented by x+2. The length of the side of
an equilateral triangle is 2x. If the square and the equilateral triangle have equal
perimeter, then the value of x is _______.
31.
32.
(2)
39/50 (3)
7/25
(4)
3/10
(5)
59/100
33.
34.
There are 3 persons Sudhir, Arvind, and Gauri. Sudhir lent cars to Arvind and
Gauri as many as they had already. After some time Arvind gave as many cars to
Sudhir and Gauri as many as they have. After sometime Gauri did the same thing.
At the end of this transaction each one of them had 24. Find the cars each
originally had.
35.
A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at
20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the
cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was
Rs._______ for the horse and Rs.________ for the cart.
97
Answers:
1.
Answer:
30 days.
Explanation:
Before:
One day work
One mans one day work
Now:
No. Of workers
One day work
=
=
1 / 20
1 / ( 20 * 75)
=
=
50
50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 = 30
2.
Answer:
0%
Explanation:
Since 3x / 2 = x / (2 / 3)
3.
Answer:
5.3 %
Explanation:
He sells 950 grams of pulses and gains 50 grams.
If he sells 100 grams of pulses then he will gain (50 / 950) *100 = 5.26
4.
Answer:
250 lines of codes
5.
Answer:
7 days
Explanation:
The equation portraying the given problem is:
10 * x 2 * (30 x) = 216 where x is the number of working days.
Solving this we get x = 23
Number of days he was absent was 7 (30-23) days.
6.
Answer:
150 men.
Explanation:
One days work
One hours work
One mans work
=
=
=
2 / (7 * 90)
2 / (7 * 90 * 8)
2 / (7 * 90 * 8 * 75)
The remaining work (5/7) has to be completed within 60 days, because the
total number of days allotted for the project is 150 days.
So we get the equation
(2 * 10 * x * 60) / (7 * 90 * 8 * 75) = 5/7 where x is the number of men
working after the 90th day.
98
We get x = 225
Answer:
(c) 1
Explanation:
a percent of b : (a/100) * b
b percent of a : (b/100) * a
a percent of b divided by b percent of a : ((a / 100 )*b) / (b/100) * a )) = 1
8.
Answer:
Cost price of horse = Rs. 400 & the cost price of cart = 200.
Explanation:Let x be the cost price of the horse and y be the cost price of the cart.
In the first sale there is no loss or profit. (i.e.) The loss obtained is equal to the
gain.
Therefore
(10/100) * x
= (20/100) * y
X
= 2 * y -----------------(1)
In the second sale, he lost Rs. 10. (i.e.) The loss is greater than the profit by Rs.
10.
Therefore
(5 / 100) * x = (5 / 100) * y + 10 -------(2)
Substituting (1) in (2) we get
(10 / 100) * y = (5 / 100) * y + 10
(5 / 100) * y = 10
y = 200
From (1) 2 * 200 = x = 400
9.
Answer:
3.
Explanation:
Since inclusion of any male player will reject a female from the team.
Since there should be four member in the team and only three males are available,
the girl, n should included in the team always irrespective of others selection.
10.
Answer:
5
11.
Answer:
1,2,3 & 4
12.
Answer:
B
13.
Answer:
11 & 9 apples per tree.
Explanation:
Let a, b, c, d & e be the total number of apples bored per year in A, B, C,
D & E s orchard. Given that a + 1 = b + 3 = c 1 = d + 3 = e 6
99
But the question is to find the number of apples bored per tree in C and D s
orchard. If is enough to consider c 1 = d + 3.
Since the number of trees in Cs orchard is 11 and that of Ds orchard is
13. Let x and y be the number of apples bored per tree in C & d s orchard
respectively.
Therefore 11 x 1 = 13 y + 3
By trial and error method, we get the value for x and y as 11 and 9
14.
Answer:
G.
Explanation:
The order in which they are climbing is R G K H J
15 18
Answer:
Novel Name
Spy thriller
Murder mystery
Gothic romance
Science fiction
Author
Rathko
Gorky
Burchfield
Hopper
Publisher
Heron
Piegon
Blueja
Sparrow
Explanation:
Given
Novel Name
Spy thriller
Murder mystery
Gothic romance
Science fiction
Author
Rathko
Gorky
Burchfield
Hopper
Publisher
Heron
Piegon
Blueja
Sparrow
Since Blueja doesnt publish the novel by Burchfield and Heron publishes
the novel spy thriller, Piegon publishes the novel by Burchfield.
Since Hopper writes Gothic romance and Heron publishes the novel spy
thriller, Blueja publishes the novel by Hopper.
Since Heron publishes the novel spy thriller and Heron publishes the novel
by Gorky, Gorky writes Spy thriller and Rathko writes Murder mystery.
19.
Answer:
451 times.
Explanation:
There are 60 minutes in an hour.
In of an hour there are (60 * ) minutes = 45 minutes.
In of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will flashes 451 times in of
an hour.
20.
Answer:
(4)
Explanation:
100
A
B
Since p is a point on the line segment AB, AB > AP
21.
Answer: (c)
22.
Ahmed
23 - 27.Answer:
Fakis
28.
Answer: Fakis
Explanation:
Chandra
Harsha
Geetha
Eswar
Balaji
Dhinesh
29.
Answer:
(5).
Explanation:
Since every alternative letter starting from B of the English alphabet is
written in small letter, the letters written in small letter are b, d, f...
In the first two answers the letter E is written in both small & capital
letters, so they are not the correct answers. But in third and fourth answers the
letter is written in small letter instead capital letter, so they are not the answers.
30.
Answer:
x=4
Explanation:
Since the side of the square is x + 2, its perimeter = 4 (x + 2) = 4x + 8
Since the side of the equilateral triangle is 2x, its perimeter = 3 * 2x = 6x
Also, the perimeters of both are equal.
(i.e.) 4x + 8 = 6x
(i.e.) 2x = 8 x = 4.
31.
Answer:
5
(y 2) / y.
Explanation:
To type a manuscript karthik took y hours.
Therefore his speed in typing = 1/y.
He was called away after 2 hours of typing.
Therefore the work completed = 1/y * 2.
Therefore the remaining work to be completed = 1 2/y.
(i.e.) work to be completed = (y-2)/y
101
32.
Answer:
(2)
33.
Answer:
1
Explanation:
One is the only number exists without reciprocal because the reciprocal of
one is one itself.
34.
Answer:
Sudhir had 39 cars, Arvind had 21 cars and Gauri had 12 cars.
Explanation:
Sudhir
Arvind
Finally
24
Before Gauris transaction 12
Before Arvinds transaction 6
Before Sudhir s transaction 39
35.
24
12
42
21
Gauri
24
48
24
12
Answer:
Cost price of horse: Rs. 400 &
Cost price of cart:
Rs. 200
Explanation:
Let x be the cost of horse & y be the cost of the cart.
10 % of loss in selling horse = 20 % of gain in selling the cart
Therefore
(10 / 100) * x = (20 * 100) * y
x = 2y -----------(1)
5 % of loss in selling the horse is 10 more than the 5 % gain in selling the
cart.
Therefore
(5 / 100) * x - 10 = (5 / 100) * y
5x - 1000
=
5y
Substituting (1)
10y - 1000 = 5y
5y = 1000
y = 200
x = 400
from (1)
Exercise 2.1
For the following, find the next term in the series
1. 6, 24, 60,120, 210
a) 336
b) 366
c) 330
d) 660
Answer : a) 336
Explanation : The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, .....
2. 1, 5, 13, 25
Answer
102
: 41
Explanation : The series is of the form 0^2+1^2, 1^2+2^2,...
3. 0, 5, 8, 17
Answer : 24
Explanation : 1^2-1, 2^2+1, 3^2-1, 4^2+1, 5^2-1
4. 1, 8, 9, 64, 25
Answer : 216
Explanation : 1^2, 2^3, 3^2, 4^3, 5^2, 6^3
5. 8,24,12,36,18,54
Answer : 27
6. 71,76,69,74,67,72
Answer : 67
7. 5,9,16,29,54
Answer : 103
Explanation : 5*2-1=9; 9*2-2=16; 16*2-3=29; 29*2-4=54; 54*2-5=103
8. 1,2,4,10,16,40,64 (Successive terms are related)
Answer : 200
Explanation : The series is powers of 2 (2^0,2^1,..).
All digits are less than 8. Every second number is in octal number system.
128 should follow 64. 128 base 10 = 200 base 8.
Exercise 2.2
Find the odd man out.
1. 3,5,7,12,13,17,19
Answer : 12
Explanation : All but 12 are odd numbers
2. 2,5,10,17,26,37,50,64
Answer : 64
Explanation : 2+3=5; 5+5=10; 10+7=17; 17+9=26; 26+11=37; 37+13=50; 50+15=65;
3. 105,85,60,30,0,-45,-90
Answer : 0
Explanation : 105-20=85; 85-25=60; 60-30=30; 30-35=-5; -5-40=-45; -45-45=-90;
Exercise 3
Solve the following.
1. What is the number of zeros at the end of the product of the numbers from 1 to 100?
Answer : 127
103
2. A fast typist can type some matter in 2 hours and a slow typist can type the same in 3
hours. If both type combinely, in how much time will they finish?
Answer : 1 hr 12 min
Explanation : The fast typist's work done in 1 hr = 1/2
The slow typist's work done in 1 hr = 1/3
If they work combinely, work done in 1 hr = 1/2+1/3 = 5/6
So, the work will be completed in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min
3. Gavaskar's average in his first 50 innings was 50. After the 51st innings, his average
was 51. How many runs did he score in his 51st innings. (supposing that he lost his
wicket in his 51st innings)
Answer : 101
Explanation : Total score after 50 innings = 50*50 = 2500
Total score after 51 innings = 51*51 = 2601
So, runs made in the 51st innings = 2601-2500 = 101
If he had not lost his wicket in his 51st innings, he would have scored an
unbeaten 50 in his 51st innings.
4. Out of 80 coins, one is counterfeit. What is the minimum number of weighings needed
to find out the counterfeit coin?
Answer : 4
5. What can you conclude from the statement : All green are blue, all blue are red. ?
(i)
some blue are green
(ii)
some red are green
(iii)
some green are not red
(iv)
all red are blue
(a) i or ii but not both
(b) i & ii only
(c) iii or iv but not both
(d) iii & iv
Answer : (b)
6. A rectangular plate with length 8 inches, breadth 11 inches and thickness 2 inches is
available. What is the length of the circular rod with diameter 8 inches and equal to the
volume of the rectangular plate?
Answer : 3.5 inches
Explanation : Volume of the circular rod (cylinder) = Volume of the rectangular
plate
(22/7)*4*4*h = 8*11*2
h = 7/2 = 3.5
7. What is the sum of all numbers between 100 and 1000 which are divisible by 14 ?
Answer : 35392
Explanation : The number closest to 100 which is greater than 100 and divisible
by 14 is 112, which is the first term of the series which has to be summed.
The number closest to 1000 which is less than 1000 and divisible by 14 is
994, which is the last term of the series.
112 + 126 + .... + 994 = 14(8+9+ ... + 71) = 35392
104
8. If s(a) denotes square root of a, find the value of s(12+s(12+s(12+ ...... upto infinity.
Answer : 4
Explanation : Let x = s(12+s(12+s(12+.....
We can write x = s(12+x). i.e., x^2 = 12 + x. Solving this quadratic equation, we
get x = -3 or x=4. Sum cannot be -ve and hence sum = 4.
9. A cylindrical container has a radius of eight inches with a height of three inches.
Compute how many inches should be added to either the radius or height to give the same
increase in volume?
Answer : 16/3 inches
Explanation : Let x be the amount of increase. The volume will increase by the
same amount if the radius increased or the height is increased.
So, the effect on increasing height is equal to the effect on increasing the radius.
i.e., (22/7)*8*8*(3+x) = (22/7)*(8+x)*(8+x)*3
Solving the quadratic equation we get the x = 0 or 16/3. The possible increase
would be by 16/3 inches.
10. With just six weights and a balance scale, you can weigh any unit number of kgs from
1 to 364. What could be the six weights?
Answer : 1, 3, 9, 27, 81, 243 (All powers of 3)
11. Diophantus passed one sixth of his life in childhood, one twelfth in youth, and one
seventh more as a bachelor; five years after his marriage a son was born who died four
years before his father at half his final age. How old is Diophantus?
Answer : 84 years
Explanation : x/6 + x/12 + x/7 + 5 + x/2 + 4 = x
12 . If time at this moment is 9 P.M., what will be the time 23999999992 hours later?
Answer : 1 P.M.
Explanation : 24 billion hours later, it would be 9 P.M. and 8 hours before that it
would be 1 P.M.
13. How big will an angle of one and a half degree look through a glass that magnifies
things three times?
Answer : 1 1/2 degrees
Explanation : The magnifying glass cannot increase the magnitude of an angle.
14. Divide 45 into four parts such that when 2 is added to the first part, 2 is subtracted
from the second part, 2 is multiplied by the third part and the fourth part is divided by
two, all result in the same number.
Answer: 8, 12, 5, 20
Explanation: a + b + c + d =45;
a+2 = b-2 = 2c = d/2; a=b-4; c = (b-2)/2; d =
2(b-2); b-4 + b + (b-2)/2 + 2(b-2) = 45;
15. I drove 60 km at 30 kmph and then an additional 60 km at 50 kmph. Compute my
average speed over my 120 km.
Answer : 37 1/2
Explanation : Time reqd for the first 60 km = 120 min.; Time reqd for the second
60 km = 72 min.; Total time reqd = 192 min
Avg speed = (60*120)/192 = 37 1/2
105
106
above 25 years, of which 7 are males; 12 males are above 25 years of age; and 15
males are married. How many bachelor girls are there and how many of these are
above 25?
2. A man sailed off from the North Pole. After covering 2,000 miles in one direction
he turned West, sailed 2,000 miles, turned North and sailed ahead another 2,000
miles till he met his friend. How far was he from the North Pole and in what
direction?
3. Here is a series of comments on the ages of three persons J, R, S by themselves.
S : The difference between R's age and mine is three years.
J : R is the youngest.
R : Either I am 24 years old or J 25 or S 26.
J : All are above 24 years of age.
S : I am the eldest if and only if R is not the youngest.
R : S is elder to me.
J : I am the eldest.
R : S is not 27 years old.
S : The sum of my age and J's is two more than twice R's age.
One of the three had been telling a lie throughout whereas others had spoken the
truth. Determine the ages of S,J,R.
4. In a group of five people, what is the probability of finding two persons with the
same month of birth?
5. A father and his son go out for a 'walk-and-run' every morning around a track
formed by an equilateral triangle. The father's walking speed is 2 mph and his
running speed is 5 mph. The son's walking and running speeds are twice that of
his father. Both start together from one apex of the triangle, the son going
clockwise and the father anti-clockwise. Initially the father runs and the son walks
for a certain period of time. Thereafter, as soon as the father starts walking, the
son starts running. Both complete the course in 45 minutes. For how long does the
father run? Where do the two cross each other?
6. The Director of Medical Services was on his annual visit to the ENT Hospital.
While going through the out patients' records he came across the following data
for a particular day : " Ear consultations 45; Nose 50; Throat 70; Ear and Nose
30; Nose and Throat 20; Ear and Throat 30; Ear, Nose and Throat 10; Total
patients 100." Then he came to the conclusion that the records were bogus. Was
he right?
7. Amongst Ram, Sham and Gobind are a doctor, a lawyer and a police officer. They
are married to Radha, Gita and Sita (not in order). Each of the wives have a
profession. Gobind's wife is an artist. Ram is not married to Gita. The lawyer's
wife is a teacher. Radha is married to the police officer. Sita is an expert cook.
Who's who?
8. What should come next?
1, 2, 4, 10, 16, 40, 64,
107
Three adults Roberto, Sarah and Vicky will be traveling in a van with five
children Freddy, Hillary, Jonathan, Lupe, and Marta. The van has a drivers seat
and one passenger seat in the front, and two benches behind the front seats, one
beach behind the other. Each bench has room for exactly three people. Everyone
must sit in a seat or on a bench, and seating is subject to the following restrictions:
An adult must sit on each bench.
Either Roberto or Sarah must sit in the drivers seat.
Jonathan must sit immediately beside Marta.
9. Of the following, who can sit in the front passenger seat ?
(a) Jonathan (b) Lupe
(c) Roberto (d) Sarah
(e) Vicky
10. Which of the following groups of three can sit together on a bench?
(a) Freddy, Jonathan and Marta
(b) Freddy, Jonathan and Vicky
(c) Freddy, Sarah and Vicky
(d) Hillary, Lupe and Sarah
(e) Lupe, Marta and Roberto
11. If Freddy sits immediately beside Vicky, which of the following cannot be true ?
a. Jonathan sits immediately beside Sarah
b. Lupe sits immediately beside Vicky
c. Hillary sits in the front passenger seat
d. Freddy sits on the same bench as Hillary
e. Hillary sits on the same bench as Roberto
12. If Sarah sits on a bench that is behind where Jonathan is sitting, which of the
following must be true ?
a. Hillary sits in a seat or on a bench that is in front of where Marta is sitting
b. Lupe sits in a seat or on a bench that is in front of where Freddy is sitting
c. Freddy sits on the same bench as Hillary
d. Lupe sits on the same bench as Sarah
e. Marta sits on the same bench as Vicky
13. Make six squares of the same size using twelve match-sticks. (Hint : You will
need an adhesive to arrange the required figure)
14. A farmer has two rectangular fields. The larger field has twice the length and 4
times the width of the smaller field. If the smaller field has area K, then the are of
the larger field is greater than the area of the smaller field by what amount?
(a) 6K (b) 8K (c) 12K
(d) 7K
15. Nine equal circles are enclosed in a square whose area is 36sq units. Find the area
of each circle.
16. There are 9 cards. Arrange them in a 3*3 matrix. Cards are of 4 colors. They are
red, yellow, blue, green. Conditions for arrangement: one red card must be in first
row or second row. 2 green cards should be in 3 rd column. Yellow cards must be in
the 3 corners only. Two blue cards must be in the 2nd row. At least one green card
in each row.
17. Is z less than w? z and w are real numbers.
(I) z2 = 25
108
(II) w = 9
To answer the question,
a) Either I or II is sufficient
b) Both I and II are sufficient but neither of them is alone sufficient
c) I & II are sufficient
d) Both are not sufficient
18. A speaks truth 70% of the time; B speaks truth 80% of the time. What is the
probability that both are contradicting each other?
19. In a family 7 children don't eat spinach, 6 don't eat carrot, 5 don't eat beans, 4
don't eat spinach & carrots, 3 don't eat carrot & beans, 2 don't eat beans &
spinach. One doesn't eat all 3. Find the no. of children.
20. Anna, Bena, Catherina and Diana are at their monthly business meeting. Their
occupations are author, biologist, chemist and doctor, but not necessarily in that
order. Diana just told the neighbour, who is a biologist that Catherina was on her
way with doughnuts. Anna is sitting across from the doctor and next to the
chemist. The doctor was thinking that Bena was a good name for parent's to
choose, but didn't say anything. What is each person's occupation?
109
UNIX Concepts
UNIX Concepts
SECTION - I
FILE MANAGEMENT IN UNIX
1. How are devices represented in UNIX?
All devices are represented by files called special files that are located
in/dev directory. Thus, device files and other files are named and accessed in the same
way. A 'regular file' is just an ordinary data file in the disk. A 'block special file' represents
a device with characteristics similar to a disk (data transfer in terms of blocks). A
'character special file' represents a device with characteristics similar to a keyboard (data
transfer is by stream of bits in sequential order).
2. What is 'inode'?
All UNIX files have its description stored in a structure called 'inode'. The inode
contains info about the file-size, its location, time of last access, time of last modification,
permission and so on. Directories are also represented as files and have an associated
inode. In addition to descriptions about the file, the inode contains pointers to the data
blocks of the file. If the file is large, inode has indirect pointer to a block of pointers to
additional data blocks (this further aggregates for larger files). A block is typically 8k.
Inode consists of the following fields:
File owner identifier
File type
File access permissions
File access times
Number of links
File size
Location of the file data
3. Brief about the directory representation in UNIX
A Unix directory is a file containing a correspondence between filenames and
inodes. A directory is a special file that the kernel maintains. Only kernel modifies
directories, but processes can read directories. The contents of a directory are a list of
filename and inode number pairs. When new directories are created, kernel makes two
entries named '.' (refers to the directory itself) and '..' (refers to parent directory).
System call for creating directory is mkdir (pathname, mode).
4. What are the Unix system calls for I/O?
open(pathname,flag,mode) - open file
creat(pathname,mode) - create file
close(filedes) - close an open file
110read(filedes,buffer,bytes) - read data from an open file
The system call mknod creates special files in the following sequence.
1. kernel assigns new inode,
2. sets the file type to indicate that the file is a pipe, directory or special file,
3. If it is a device file, it makes the other entries like major, minor device numbers.
For example:
If the device is a disk, major device number refers to the disk controller and minor
device number is the disk.
9. Discuss the mount and unmount system calls
The privileged mount system call is used to attach a file system to a directory of
another file system; the unmount system call detaches a file system. When you mount
another file system on to your directory, you are essentially splicing one directory tree
onto a branch in another directory tree. The first argument to mount call is the mount
point, that is , a directory in the current file naming system. The second argument is the
file system to mount to that point. When you insert a cdrom to your unix system's drive,
the file system in the cdrom automatically mounts to /dev/cdrom in your system.
10. How does the inode map to data block of a file?
Inode has 13 block addresses. The first 10 are direct block addresses of the first
10 data blocks in the file. The 11th address points to a one-level index block. The 12th
address points to a two-level (double in-direction) index block. The 13th address points to
a three-level(triple in-direction)index block. This provides a very large maximum file size
with efficient access to large files, but also small files are accessed directly in one disk
read.
11. What is a shell?
A shell is an interactive user interface to an operating system services that allows an user
to enter commands as character strings or through a graphical user interface. The shell
converts them to system calls to the OS or forks off a process to execute the command.
System call results and other information from the OS are presented to the user through
an interactive interface. Commonly used shells are sh,csh,ks etc.
SECTION - II
PROCESS MODEL and IPC
1. Brief about the initial process sequence while the system boots up.
While booting, special process called the 'swapper' or 'scheduler' is created with
Process-ID 0. The swapper manages memory allocation for processes and influences
CPU allocation. The swapper inturn creates 3 children:
the process dispatcher,
vhand and
dbflush
with IDs 1,2 and 3 respectively.
This is done by executing the file /etc/init. Process dispatcher gives birth to the
shell. Unix keeps track of all the processes in an internal data structure called the Process
Table (listing command is ps -el).
2. What are various IDs associated with a process?
Unix identifies each process with a unique integer called ProcessID. The process
that
executes
the request for creation of a process is called the 'parent process' whose PID
112
is 'Parent Process ID'. Every process is associated with a particular user called the 'owner'
who has privileges over the process. The identification for the user is 'UserID'. Owner is
the user who executes the process. Process also has 'Effective User ID' which determines
the access privileges for accessing resources like files.
getpid() -process id
getppid() -parent process id
getuid() -user id
geteuid() -effective user id
3. Explain fork() system call.
The `fork()' used to create a new process from an existing process. The new
process is called the child process, and the existing process is called the parent. We can
tell which is which by checking the return value from `fork()'. The parent gets the child's
pid returned to him, but the child gets 0 returned to him.
4. Predict the output of the following program code
main()
{
fork();
printf("Hello World!");
}
Answer:
Hello World!Hello World!
Explanation:
The fork creates a child that is a duplicate of the parent process. The child begins
from the fork().All the statements after the call to fork() will be executed twice.(once by
the parent process and other by child). The statement before fork() is executed only by
the parent process.
5. Predict the output of the following program code
main()
{
fork(); fork(); fork();
printf("Hello World!");
}
Answer:
"Hello World" will be printed 8 times.
Explanation:
2^n times where n is the number of calls to fork()
6. List the system calls used for process management:
System calls
Description
fork()
To create a new process
exec()
To execute a new program in a process
wait()
To wait until a created process completes its execution
exit()
To exit from a process execution
getpid()
To get a process identifier of the current process
getppid()
To get parent process identifier
nice()
To bias the existing priority of a process
brk()
To increase/decrease the data segment size of a process
113
Message Queues :
116
Address
Units
10,000
Nice value is the value that controls {increments or decrements} the priority of
the process. This value that is returned by the nice () system call. The equation for using
nice value is:
Priority = (recent CPU usage/constant) + (base- priority) + (nice value)
Only the administrator can supply the nice value. The nice () system call works
for the running process only. Nice value of one process cannot affect the nice value of the
other process.
19. What are conditions on which deadlock can occur while swapping the processes?
All processes in the main memory are asleep.
All ready-to-run processes are swapped out.
There is no space in the swap device for the new incoming process that are
swapped out of the main memory.
There is no space in the main memory for the new incoming process.
20. What are conditions for a machine to support Demand Paging?
Memory architecture must based on Pages,
The machine must support the restartable instructions.
21. What is the principle of locality?
Its the nature of the processes that they refer only to the small subset of the total
data space of the process. i.e. the process frequently calls the same subroutines or
executes the loop instructions.
22. What is the working set of a process?
The set of pages that are referred by the process in the last n, references, where
n is called the window of the working set of the process.
23. What is the window of the working set of a process?
The window of the working set of a process is the total number in which the
process had referred the set of pages in the working set of the process.
24. What is called a page fault?
Page fault is referred to the situation when the process addresses a page in the
working set of the process but the process fails to locate the page in the working set. And
on a page fault the kernel updates the working set by reading the page from the secondary
device.
25. What are data structures that are used for Demand Paging?
Kernel contains 4 data structures for Demand paging. They are,
Page table entries,
Disk block descriptors,
Page frame data table (pfdata),
Swap-use table.
26. What are the bits that support the demand paging?
Valid, Reference, Modify, Copy on write, Age. These bits are the part of the page
table entry, which includes physical address of the page and protection bits.
119
Page address
Modify Reference
Valid
Protection
27. How the Kernel handles the fork() system call in traditional Unix and in the System V
Unix, while swapping?
Kernel in traditional Unix, makes the duplicate copy of the parents address space
and attaches it to the childs process, while swapping. Kernel in System V Unix,
manipulates the region tables, page table, and pfdata table entries, by incrementing the
reference count of the region table of shared regions.
28. Difference between the fork() and vfork() system call?
During the fork() system call the Kernel makes a copy of the parent processs
address space and attaches it to the child process.
But the vfork() system call do not makes any copy of the parents address space,
so it is faster than the fork() system call. The child process as a result of the vfork()
system call executes exec() system call. The child process from vfork() system call
executes in the parents address space (this can overwrite the parents data and stack )
which suspends the parent process until the child process exits.
29. What is BSS(Block Started by Symbol)?
A data representation at the machine level, that has initial values when a program
starts and tells about how much space the kernel allocates for the un-initialized data.
Kernel initializes it to zero at run-time.
30. What is Page-Stealer process?
This is the Kernel process that makes rooms for the incoming pages, by swapping
the memory pages that are not the part of the working set of a process. Page-Stealer is
created by the Kernel at the system initialization and invokes it throughout the lifetime of
the system. Kernel locks a region when a process faults on a page in the region, so that
page stealer cannot steal the page, which is being faulted in.
31. Name two paging states for a page in memory?
The two paging states are:
The page is aging and is not yet eligible for swapping,
The page is eligible for swapping but not yet eligible for reassignment to other virtual
address space.
32. What are the phases of swapping a page from the memory?
Page stealer finds the page eligible for swapping and places the page number
in the list of pages to be swapped.
Kernel copies the page to a swap device when necessary and clears the valid
bit in the page table entry, decrements the pfdata reference count, and places
the pfdata table entry at the end of the free list if its reference count is 0.
33. What is page fault? Its types?
Page fault refers to the situation of not having a page in the main memory when
any process references it.
There are two types of page fault :
Validity fault,
120Protection fault.
34. In what way the Fault Handlers and the Interrupt handlers are different?
Fault handlers are also an interrupt handler with an exception that the interrupt
handlers cannot sleep. Fault handlers sleep in the context of the process that caused the
memory fault. The fault refers to the running process and no arbitrary processes are put to
sleep.
35. What is validity fault?
If a process referring a page in the main memory whose valid bit is not set, it
results in validity fault.
The valid bit is not set for those pages:
that are outside the virtual address space of a process,
that are the part of the virtual address space of the process but no physical address is
assigned to it.
36. What does the swapping system do if it identifies the illegal page for swapping?
If the disk block descriptor does not contain any record of the faulted page, then
this causes the attempted memory reference is invalid and the kernel sends a
Segmentation violation signal to the offending process. This happens when the
swapping system identifies any invalid memory reference.
37. What are states that the page can be in, after causing a page fault?
On a swap device and not in memory,
On the free page list in the main memory,
In an executable file,
Marked demand zero,
Marked demand fill.
38. In what way the validity fault handler concludes?
It sets the valid bit of the page by clearing the modify bit.
It recalculates the process priority.
39. At what mode the fault handler executes?
At the Kernel Mode.
40. What do you mean by the protection fault?
Protection fault refers to the process accessing the pages, which do not have the
access permission. A process also incur the protection fault when it attempts to write a
page whose copy on write bit was set during the fork() system call.
41. How the Kernel handles the copy on write bit of a page, when the bit is set?
In situations like, where the copy on write bit of a page is set and that page is
shared by more than one process, the Kernel allocates new page and copies the content to
the new page and the other processes retain their references to the old page. After copying
the Kernel updates the page table entry with the new page number. Then Kernel
decrements the reference count of the old pfdata table entry.
In cases like, where the copy on write bit is set and no processes are sharing the
page, the Kernel allows the physical page to be reused by the processes. By doing so, it
clears the copy on write bit and disassociates the page from its disk copy (if one exists),
121
because
other process may share the disk copy. Then it removes the pfdata table entry
from the page-queue as the new copy of the virtual page is not on the swap device. It
decrements the swap-use count for the page and if count drops to 0, frees the swap space.
42. For which kind of fault the page is checked first?
The page is first checked for the validity fault, as soon as it is found that the page
is invalid (valid bit is clear), the validity fault handler returns immediately, and the
process incur the validity page fault. Kernel handles the validity fault and the process will
incur the protection fault if any one is present.
43. In what way the protection fault handler concludes?
After finishing the execution of the fault handler, it sets the modify and protection
bits and clears the copy on write bit. It recalculates the process-priority and checks for
signals.
44. How the Kernel handles both the page stealer and the fault handler?
The page stealer and the fault handler thrash because of the shortage of the
memory. If the sum of the working sets of all processes is greater that the physical
memory then the fault handler will usually sleep because it cannot allocate pages for a
process. This results in the reduction of the system throughput because Kernel spends too
much time in overhead, rearranging the memory in the frantic pace.
122
RDBMS Concepts
RDBMS Concepts
1. What is database?
A database is a logically coherent collection of data with some inherent meaning,
representing some aspect of real world and which is designed, built and populated with
data for a specific purpose.
2. What is DBMS?
It is a collection of programs that enables user to create and maintain a database.
In other words it is general-purpose software that provides the users with the processes of
defining, constructing and manipulating the database for various applications.
3. What is a Database system?
The database and DBMS software together is called as Database system.
4.
Advantages of DBMS?
Redundancy is controlled.
Unauthorised access is restricted.
Providing multiple user interfaces.
Enforcing integrity constraints.
Providing backup and recovery.
5.
This data model is based on real world that consists of basic objects called entities
and of relationship among these objects. Entities are described in a database by a set of
attributes.
15. What is Object Oriented model?
This model is based on collection of objects. An object contains values stored in
instance variables with in the object. An object also contains bodies of code that operate
on the object. These bodies of code are called methods. Objects that contain same types
of values and the same methods are grouped together into classes.
16. What is an Entity?
It is a 'thing' in the real world with an independent existence.
17. What is an Entity type?
It is a collection (set) of entities that have same attributes.
18. What is an Entity set?
It is a collection of all entities of particular entity type in the database.
19. What is an Extension of entity type?
The collections of entities of a particular entity type are grouped together into an
entity set.
20. What is Weak Entity set?
An entity set may not have sufficient attributes to form a primary key, and its
primary key compromises of its partial key and primary key of its parent entity, then it is
said to be Weak Entity set.
21. What is an attribute?
It is a particular property, which describes the entity.
22. What is a Relation Schema and a Relation?
A relation Schema denoted by R(A1, A2, , An) is made up of the relation name
R and the list of attributes Ai that it contains. A relation is defined as a set of tuples. Let r
be the relation which contains set tuples (t1, t2, t3, ..., tn). Each tuple is an ordered list of
n-values t=(v1,v2, ..., vn).
23. What is degree of a Relation?
It is the number of attribute of its relation schema.
24. What is Relationship?
It is an association among two or more entities.
25. What is Relationship set?
The collection (or set) of similar relationships.
26. What is Relationship type?
Relationship type defines a set of associations or a relationship set among a given
set of entity types.
125
When one of the data elements stored within a construct is utilized as the
primary key, then it is called the natural key.
53. What is indexing and what are the different kinds of indexing?
Indexing is a technique for determining how quickly specific data can be found.
Types:
Binary search style indexing
B-Tree indexing
Inverted list indexing
Memory resident table
Table indexing
54. What is system catalog or catalog relation? How is better known as?
A RDBMS maintains a description of all the data that it contains, information
about every relation and index that it contains. This information is stored in a collection
of relations maintained by the system called metadata. It is also called data dictionary.
55. What is meant by query optimization?
The phase that identifies an efficient execution plan for evaluating a query that
has the least estimated cost is referred to as query optimization.
56. What is join dependency and inclusion dependency?
Join Dependency:
A Join dependency is generalization of Multivalued dependency.A
JD {R1, R2, ..., Rn} is said to hold over a relation R if R1, R2, R3, ..., Rn is a losslessjoin decomposition of R . There is no set of sound and complete inference rules for JD.
Inclusion Dependency:
An Inclusion Dependency is a statement of the form that some columns of
a relation are contained in other columns. A foreign key constraint is an example of
inclusion dependency.
57. What is durability in DBMS?
Once the DBMS informs the user that a transaction has successfully completed,
its effects should persist even if the system crashes before all its changes are reflected on
disk. This property is called durability.
58. What do you mean by atomicity and aggregation?
Atomicity:
Either all actions are carried out or none are. Users should not have to
worry about the effect of incomplete transactions. DBMS ensures this by undoing the
actions of incomplete transactions.
Aggregation:
A concept which is used to model a relationship between a collection of
entities and relationships. It is used when we need to express a relationship among
relationships.
59. What is a Phantom Deadlock?
In distributed deadlock detection, the delay in propagating local information
might cause the deadlock detection algorithms to identify deadlocks that do not really
exist.
129 Such situations are called phantom deadlocks and they lead to unnecessary aborts.
70. Are the resulting relations of PRODUCT and JOIN operation the same?
No.
PRODUCT: Concatenation of every row in one relation with every row in
another.
JOIN: Concatenation of rows from one relation and related rows from another.
71. What is RDBMS KERNEL?
Two important pieces of RDBMS architecture are the kernel, which is the
software, and the data dictionary, which consists of the system-level data structures used
by the kernel to manage the database
You might think of an RDBMS as an operating system (or set of subsystems),
designed specifically for controlling data access; its primary functions are storing,
retrieving, and securing data. An RDBMS maintains its own list of authorized users and
their associated privileges; manages memory caches and paging; controls locking for
concurrent resource usage; dispatches and schedules user requests; and manages space
usage within its table-space structures
.
72. Name the sub-systems of a RDBMS
I/O, Security, Language Processing, Process Control, Storage Management,
Logging and Recovery, Distribution Control, Transaction Control, Memory Management,
Lock Management
73. Which part of the RDBMS takes care of the data dictionary? How
Data dictionary is a set of tables and database objects that is stored in a special
area of the database and maintained exclusively by the kernel.
74. What is the job of the information stored in data-dictionary?
The information in the data dictionary validates the existence of the objects,
provides access to them, and maps the actual physical storage location.
75. Not only RDBMS takes care of locating data it also
determines an optimal access path to store or retrieve the data
76. How do you communicate with an RDBMS?
You communicate with an RDBMS using Structured Query Language (SQL)
77. Define SQL and state the differences between SQL and other conventional
programming Languages
SQL is a nonprocedural language that is designed specifically for data access
operations on normalized relational database structures. The primary difference between
SQL and other conventional programming languages is that SQL statements specify what
data operations should be performed rather than how to perform them.
78. Name the three major set of files on disk that compose a database in Oracle
There are three major sets of files on disk that compose a database. All the files
are binary. These are
Database files
Control files
Redo logs
The most important of these are the database files where the actual data resides.
The
control
files and the redo logs support the functioning of the architecture itself.
131
All three sets of files must be present, open, and available to Oracle for any data
on the database to be useable. Without these files, you cannot access the database, and the
database administrator might have to recover some or all of the database using a backup, if
there is one.
79. What is an Oracle Instance?
The Oracle system processes, also known as Oracle background processes,
provide functions for the user processesfunctions that would otherwise be done by the
user processes themselves
Oracle database-wide system memory is known as the SGA, the system global
area or shared global area. The data and control structures in the SGA are shareable, and
all the Oracle background processes and user processes can use them.
The combination of the SGA and the Oracle background processes is known as an
Oracle instance
80. What are the four Oracle system processes that must always be up and running for
the database to be useable
The four Oracle system processes that must always be up and running for the
database to be useable include DBWR (Database Writer), LGWR (Log Writer), SMON
(System Monitor), and PMON (Process Monitor).
81. What are database files, control files and log files. How many of these files should a
database have at least? Why?
Database Files
The database files hold the actual data and are typically the largest in size.
Depending on their sizes, the tables (and other objects) for all the user accounts can go in
one database filebut that's not an ideal situation because it does not make the database
structure very flexible for controlling access to storage for different users, putting the
database on different disk drives, or backing up and restoring just part of the database.
You must have at least one database file but usually, more than one files
are used. In terms of accessing and using the data in the tables and other objects, the
number (or location) of the files is immaterial.
The database files are fixed in size and never grow bigger than the size at
which they were created
Control Files
The control files and redo logs support the rest of the architecture. Any
database must have at least one control file, although you typically have more than one to
guard against loss. The control file records the name of the database, the date and time it
was created, the location of the database and redo logs, and the synchronization
information to ensure that all three sets of files are always in step. Every time you add a
new database or redo log file to the database, the information is recorded in the control
files.
Redo Logs
Any database must have at least two redo logs. These are the journals for
the database; the redo logs record all changes to the user objects or system objects. If any
type of failure occurs, the changes recorded in the redo logs can be used to bring the
database to a consistent state without losing any committed transactions. In the case of
non-data loss failure, Oracle can apply the information in the redo logs automatically
without intervention from the DBA.
The redo log files are fixed in size and never grow dynamically from the
132
size at which they were created.
87. How are exceptions handled in PL/SQL? Give some of the internal exceptions' name
PL/SQL exception handling is a mechanism for dealing with run-time errors
encountered during procedure execution. Use of this mechanism enables execution to
continue if the error is not severe enough to cause procedure termination.
The exception handler must be defined within a subprogram specification. Errors
cause the program to raise an exception with a transfer of control to the exception-handler
block. After the exception handler executes, control returns to the block in which the
handler was defined. If there are no more executable statements in the block, control
returns to the caller.
User-Defined Exceptions
PL/SQL enables the user to define exception handlers in the declarations
area of subprogram specifications. User accomplishes this by naming an exception as in
the following example:
ot_failure EXCEPTION;
In this case, the exception name is ot_failure. Code associated with this handler is written
in the EXCEPTION specification area as follows:
EXCEPTION
when OT_FAILURE then
out_status_code := g_out_status_code;
out_msg
:= g_out_msg;
The following is an example of a subprogram exception:
EXCEPTION
when NO_DATA_FOUND then
g_out_status_code := 'FAIL';
RAISE ot_failure;
Within this exception is the RAISE statement that transfers control back to the ot_failure
exception handler. This technique of raising the exception is used to invoke all userdefined exceptions.
System-Defined Exceptions
Exceptions internal to PL/SQL are raised automatically upon error.
NO_DATA_FOUND is a system-defined exception. Table below gives a complete list of
internal exceptions.
PL/SQL internal exceptions.
Exception Name
CURSOR_ALREADY_OPEN
DUP_VAL_ON_INDEX
INVALID_CURSOR
INVALID_NUMBER
LOGIN_DENIED
NO_DATA_FOUND
NOT_LOGGED_ON
PROGRAM_ERROR
STORAGE_ERROR
TIMEOUT_ON_RESOURCE
TOO_MANY_ROWS
TRANSACTION_BACKED_OUT
VALUE_ERROR
134
ZERO_DIVIDE
Oracle Error
ORA-06511
ORA-00001
ORA-01001
ORA-01722
ORA-01017
ORA-01403
ORA-01012
ORA-06501
ORA-06500
ORA-00051
ORA-01422
ORA-00061
ORA-06502
ORA-01476
101.
What are Armstrong rules? How do we say that they are complete and/or sound
The well-known inference rules for FDs
Reflexive rule :
If Y is subset or equal to X then X
Y.
Augmentation rule:
If X
Y then XZ
YZ.
Transitive rule:
If {X Y, Y
Z} then X
Z.
Decomposition rule :
If X
YZ then X
Y.
Union or Additive rule:
If {X
Y, X
Z} then X
YZ.
Pseudo Transitive rule :
If {X
Y, WY
Z} then WX
Z.
Of these the first three are known as Amstrong Rules. They are sound because it is
enough if a set of FDs satisfy these three. They are called complete because using these
three rules we can generate the rest all inference rules.
137
104.
Minimal key is one which can identify each tuple of the given relation schema
uniquely. For finding the minimal key it is required to find the closure that is the set of all
attributes that are dependent on any given set of attributes under the given set of
functional dependency.
Algo. I Determining X+, closure for X, given set of FDs F
1. Set X+ = X
2. Set Old X+ = X+
3. For each FD Y
Z in F and if Y belongs to X + then add Z
to X+
4. Repeat steps 2 and 3 until Old X+ = X+
Algo.II Determining minimal K for relation schema R, given set of FDs F
1. Set K to R that is make K a set of all attributes in R
2. For each attribute A in K
a. Compute (K A)+ with respect to F
b. If (K A)+ = R then set K = (K A)+
105.
138
SQL
SQL
1. Which is the subset of SQL commands used to manipulate Oracle Database
structures, including tables?
Data Definition Language (DDL)
2. What operator performs pattern matching?
LIKE operator
3. What operator tests column for the absence of data?
IS NULL operator
4. Which command executes the contents of a specified file?
START <filename> or @<filename>
5. What is the parameter substitution symbol used with INSERT INTO command?
&
6. Which command displays the SQL command in the SQL buffer, and then executes it?
RUN
7. What are the wildcards used for pattern matching?
_ for single character substitution and % for multi-character substitution
8. State true or false. EXISTS, SOME, ANY are operators in SQL.
True
9. State true or false. !=, <>, ^= all denote the same operation.
True
10. What are the privileges that can be granted on a table by a user to others?
Insert, update, delete, select, references, index, execute, alter, all
11. What command is used to get back the privileges offered by the GRANT command?
REVOKE
12. Which system tables contain information on privileges granted and privileges
obtained?
USER_TAB_PRIVS_MADE, USER_TAB_PRIVS_RECD
13. Which system table contains information on constraints on all the tables created?
139
USER_CONSTRAINTS
14.
The privilege receiver can further grant the privileges he/she has obtained from
the owner to any other user.
23. What is the use of the DROP option in the ALTER TABLE command?
It is used to drop constraints specified on the table.
24. What is the value of comm and sal after executing the following query if the initial
value of sal is 10000?
UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1;
sal = 11000, comm = 1000
25. What is the use of DESC in SQL?
Answer :
DESC has two purposes. It is used to describe a schema as well as to retrieve
rows from table in descending order.
Explanation :
The query SELECT * FROM EMP ORDER BY ENAME DESC will display the
output sorted on ENAME in descending order.
26. What is the use of CASCADE CONSTRAINTS?
When this clause is used with the DROP command, a parent table can be dropped
even when a child table exists.
27. Which function is used to find the largest integer less than or equal to a specific
value?
FLOOR
28. What is the output of the following query?
SELECT TRUNC(1234.5678,-2) FROM DUAL;
1200
SQL QUERIES
I. SCHEMAS
Table 1 : STUDIES
PNAME (VARCHAR), SPLACE (VARCHAR), COURSE (VARCHAR), CCOST
(NUMBER)
Table 2 : SOFTWARE
PNAME (VARCHAR), TITLE (VARCHAR), DEVIN
(NUMBER), DCOST (NUMBER), SOLD (NUMBER)
141
Table 3 : PROGRAMMER
(VARCHAR),
SCOST
KEYS:
1. SELECT AVG(SCOST) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
2. SELECT
PNAME,TRUNC(MONTHS_BETWEEN(SYSDATE,DOB)/12)
"AGE", TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) "EXPERIENCE"
FROM PROGRAMMER;
3. SELECT PNAME FROM STUDIES WHERE COURSE = 'PGDCA';
4. SELECT MAX(SOLD) FROM SOFTWARE;
5. SELECT PNAME, DOB FROM PROGRAMMER WHERE DOB LIKE '%APR
%';
6. SELECT MIN(CCOST) FROM STUDIES;
7. SELECT COUNT(*) FROM STUDIES WHERE COURSE = 'DCA';
8. SELECT SUM(SCOST*SOLD-DCOST) FROM SOFTWARE GROUP BY
DEVIN HAVING DEVIN = 'C';
9. SELECT * FROM SOFTWARE WHERE PNAME = 'RAKESH';
10. SELECT * FROM STUDIES WHERE SPLACE = 'PENTAFOUR';
11. SELECT * FROM SOFTWARE WHERE SCOST*SOLD-DCOST > 5000;
12. SELECT CEIL(DCOST/SCOST) FROM SOFTWARE;
13. SELECT * FROM SOFTWARE WHERE SCOST*SOLD >= DCOST;
14. SELECT MAX(SCOST) FROM SOFTWARE GROUP BY DEVIN HAVING
DEVIN = 'VB';
15. SELECT COUNT(*) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
16. SELECT COUNT(*) FROM STUDIES WHERE SPLACE = 'PRAGATHI';
17. SELECT COUNT(*) FROM STUDIES WHERE CCOST BETWEEN 10000
AND 15000;
18. SELECT AVG(CCOST) FROM STUDIES;
19. SELECT * FROM PROGRAMMER WHERE PROF1 = 'C' OR PROF2 = 'C';
20. SELECT * FROM PROGRAMMER WHERE PROF1 IN ('C','PASCAL') OR
PROF2 IN ('C','PASCAL');
21. SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C++') AND
PROF2 NOT IN ('C','C++');
22. SELECT TRUNC(MAX(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM
PROGRAMMER WHERE SEX = 'M';
23. SELECT TRUNC(AVG(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM
PROGRAMMER WHERE SEX = 'F';
24. SELECT PNAME, TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) FROM
PROGRAMMER ORDER BY PNAME DESC;
25. SELECT PNAME FROM PROGRAMMER WHERE TO_CHAR(DOB,'MON') =
TO_CHAR(SYSDATE,'MON');
26. SELECT COUNT(*) FROM PROGRAMMER WHERE SEX = 'F';
27. SELECT DISTINCT(PROF1) FROM PROGRAMMER WHERE SEX = 'M';
28. SELECT AVG(SAL) FROM PROGRAMMER;
29. SELECT COUNT(*) FROM PROGRAMMER WHERE SAL BETWEEN 5000
AND 7500;
30. SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C+
+','PASCAL') AND PROF2 NOT IN ('C','C++','PASCAL');
31. SELECT PNAME,TITLE,SCOST FROM SOFTWARE WHERE SCOST IN
(SELECT MAX(SCOST) FROM SOFTWARE GROUP BY PNAME);
32.SELECT
'Mr.'
||
PNAME
||
'
has
'
||
TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12)
||
'
years
of
experience'
143
Programmer FROM PROGRAMMER WHERE SEX = 'M' UNION SELECT
II . SCHEMA :
Table 1 : DEPT
DEPTNO (NOT NULL , NUMBER(2)), DNAME (VARCHAR2(14)),
LOC (VARCHAR2(13)
Table 2 : EMP
EMPNO (NOT NULL , NUMBER(4)), ENAME (VARCHAR2(10)),
JOB (VARCHAR2(9)), MGR (NUMBER(4)), HIREDATE (DATE),
SAL (NUMBER(7,2)), COMM (NUMBER(7,2)), DEPTNO (NUMBER(2))
MGR is the empno of the employee whom the employee reports to. DEPTNO is a foreign
key.
QUERIES
1. List all the employees who have at least one person reporting to them.
2. List the employee details if and only if more than 10 employees are present in
department no 10.
3. List the name of the employees with their immediate higher authority.
4. List all the employees who do not manage any one.
5. List the employee details whose salary is greater than the lowest salary of an
employee belonging to deptno 20.
6. List the details of the employee earning more than the highest paid manager.
7. List the highest salary paid for each job.
8. Find the most recently hired employee in each department.
9. In which year did most people join the company? Display the year and the number of
employees.
10. Which department has the highest annual remuneration bill?
11. Write a query to display a * against the row of the most recently hired employee.
12. Write a correlated sub-query to list out the employees who earn more than the
average salary of their department.
13. Find the nth maximum salary.
14. Select the duplicate records (Records, which are inserted, that already exist) in the
EMP table.
15. Write a query to list the length of service of the employees (of the form n years and m
months).
KEYS:
1. SELECT DISTINCT(A.ENAME) FROM EMP A, EMP B WHERE A.EMPNO =
B.MGR; or SELECT ENAME FROM EMP WHERE EMPNO IN (SELECT MGR
FROM
EMP);
144
145
Computer
Networks
Computer Networks
1. What are the two types of transmission technology available?
(i) Broadcast and
(ii) point-to-point
2. What is subnet?
A generic term for section of a large networks usually separated by a bridge or
router.
3. Difference between the communication and transmission.
Transmission is a physical movement of information and concern issues like bit
polarity, synchronisation, clock etc.
Communication means the meaning full exchange of information between two
communication media.
4. What are the possible ways of data exchange?
(i) Simplex (ii) Half-duplex
(iii) Full-duplex.
5. What is SAP?
Series of interface points that allow other computers to communicate with the
other layers of network protocol stack.
6. What do you meant by "triple X" in Networks?
The function of PAD (Packet Assembler Disassembler) is described in a document
known as X.3. The standard protocol has been defined between the terminal and the PAD,
called X.28; another standard protocol exists between hte PAD and the network, called
X.29. Together, these three recommendations are often called "triple X"
7. What is frame relay, in which layer it comes?
Frame relay is a packet switching technology. It will operate in the data link layer.
8. What is terminal emulation, in which layer it comes?
Telnet is also called as terminal emulation. It belongs to application layer.
9. What is Beaconing?
The process that allows a network to self-repair networks problems. The stations
on the network notify the other stations on the ring when they are not receiving the
transmissions. Beaconing is used in Token ring and FDDI networks.
10. What is redirector?
Redirector is software that intercepts file or prints I/O requests and translates them
146
into
network requests. This comes under presentation layer.
a) Guided Media:
These are those that provide a conduit from one device to another that
include twisted-pair, coaxial cable and fiber-optic cable. A signal traveling along any of
these media is directed and is contained by the physical limits of the medium. Twistedpair and coaxial cable use metallic that accept and transport signals in the form of
electrical current. Optical fiber is a glass or plastic cable that accepts and transports
signals in the form of light.
b) Unguided Media:
This is the wireless media that transport electromagnetic waves without
using a physical conductor. Signals are broadcast either through air. This is done through
radio communication, satellite communication and cellular telephony.
23. What is Project 802?
It is a project started by IEEE to set standards to enable intercommunication
between equipment from a variety of manufacturers. It is a way for specifying functions
of the physical layer, the data link layer and to some extent the network layer to allow for
interconnectivity of major LAN
protocols.
It consists of the following:
802.1 is an internetworking standard for compatibility of different LANs and MANs
across protocols.
802.2 Logical link control (LLC) is the upper sublayer of the data link layer which is
non-architecture-specific, that is remains the same for all IEEE-defined LANs.
Media access control (MAC) is the lower sublayer of the data link layer that contains
some distinct modules each carrying proprietary information specific to the LAN
product being used. The modules are Ethernet LAN (802.3), Token ring LAN (802.4),
Token bus LAN (802.5).
802.6 is distributed queue dual bus (DQDB) designed to be used in MANs.
24. What is Protocol Data Unit?
The data unit in the LLC level is called the protocol data unit (PDU). The PDU
contains of four fields a destination service access point (DSAP), a source service access
point (SSAP), a control field and an information field. DSAP, SSAP are addresses used
by the LLC to identify the protocol stacks on the receiving and sending machines that are
generating and using the data. The control field specifies whether the PDU frame is a
information frame (I - frame) or a supervisory frame (S - frame) or a unnumbered frame
(U - frame).
25. What are the different type of networking / internetworking devices?
Repeater:
Also called a regenerator, it is an electronic device that operates only at
physical layer. It receives the signal in the network before it becomes weak, regenerates
the original bit pattern and puts the refreshed copy back in to the link.
Bridges:
These operate both in the physical and data link layers of LANs of same
type. They divide a larger network in to smaller segments. They contain logic that allow
them to keep the traffic for each segment separate and thus are repeaters that relay a
frame only the side of the segment containing the intended recipent and control
congestion.
Routers:
148
39. Why should you care about the OSI Reference Model?
It provides a framework for discussing network operations and design.
40. What is logical link control?
One of two sublayers of the data link layer of OSI reference model, as defined by
the IEEE 802 standard. This sublayer is responsible for maintaining the link between
computers when they are sending data across the physical network connection.
41. What is virtual channel?
Virtual channel is normally a connection from one source to one destination,
although multicast connections are also permitted. The other name for virtual channel is
virtual circuit.
42. What is virtual path?
Along any transmission path from a given source to a given destination, a group
of virtual circuits can be grouped together into what is called path.
43. What is packet filter?
Packet filter is a standard router equipped with some extra functionality. The extra
functionality allows every incoming or outgoing packet to be inspected. Packets meeting
some criterion are forwarded normally. Those that fail the test are dropped.
44. What is traffic shaping?
One of the main causes of congestion is that traffic is often busy. If hosts could be
made to transmit at a uniform rate, congestion would be less common. Another open loop
method to help manage congestion is forcing the packet to be transmitted at a more
predictable rate. This is called traffic shaping.
45. What is multicast routing?
Sending a message to a group is called multicasting, and its routing algorithm is
called multicast routing.
46. What is region?
When hierarchical routing is used, the routers are divided into what we will call
regions, with each router knowing all the details about how to route packets to
destinations within its own region, but knowing nothing about the internal structure of
other regions.
47. What is silly window syndrome?
It is a problem that can ruin TCP performance. This problem occurs when data are
passed to the sending TCP entity in large blocks, but an interactive application on the
receiving side reads 1 byte at a time.
48. What are Digrams and Trigrams?
The most common two letter combinations are called as digrams. e.g. th, in, er, re
and an. The most common three letter combinations are called as trigrams. e.g. the, ing,
and, and ion.
49.
Expand IDEA.
151
IDEA stands for International Data Encryption Algorithm.
153
Operating Systems
Operating Systems
Following are a few basic questions that cover the essentials of OS:
1. Explain the concept of Reentrancy.
It is a useful, memory-saving technique for multiprogrammed timesharing
systems. A Reentrant Procedure is one in which multiple users can share a single copy of
a program during the same period. Reentrancy has 2 key aspects: The program code
cannot modify itself, and the local data for each user process must be stored separately.
Thus, the permanent part is the code, and the temporary part is the pointer back to the
calling program and local variables used by that program. Each execution instance is
called activation. It executes the code in the permanent part, but has its own copy of local
variables/parameters. The temporary part associated with each activation is the activation
record. Generally, the activation record is kept on the stack.
Note: A reentrant procedure can be interrupted and called by an interrupting
program, and still execute correctly on returning to the procedure.
2. Explain Belady's Anomaly.
Also called FIFO anomaly. Usually, on increasing the number of frames allocated
to a process' virtual memory, the process execution is faster, because fewer page faults
occur. Sometimes, the reverse happens, i.e., the execution time increases even when more
frames are allocated to the process. This is Belady's Anomaly. This is true for certain page
reference patterns.
3. What is a binary semaphore? What is its use?
A binary semaphore is one, which takes only 0 and 1 as values. They are used to
implement mutual exclusion and synchronize concurrent processes.
4. What is thrashing?
It is a phenomenon in virtual memory schemes when the processor spends most of
its time swapping pages, rather than executing instructions. This is due to an inordinate
number of page faults.
5.
Long term scheduler determines which programs are admitted to the system for
processing. It controls the degree of multiprogramming. Once admitted, a job becomes a
process.
Medium term scheduling is part of the swapping function. This relates to
processes that are in a blocked or suspended state. They are swapped out of real-memory
until they are ready to execute. The swapping-in decision is based on memorymanagement criteria.
Short term scheduler, also know as a dispatcher executes most frequently, and
makes the finest-grained decision of which process should execute next. This scheduler is
invoked whenever an event occurs. It may lead to interruption of one process by
preemption.
7. What are turnaround time and response time?
Turnaround time is the interval between the submission of a job and its
completion. Response time is the interval between submission of a request, and the first
response to that request.
8. What are the typical elements of a process image?
User data: Modifiable part of user space. May include program data, user stack area,
and programs that may be modified.
User program: The instructions to be executed.
System Stack: Each process has one or more LIFO stacks associated with it. Used to
store parameters and calling addresses for procedure and system calls.
Process control Block (PCB): Info needed by the OS to control processes.
9. What is the Translation Lookaside Buffer (TLB)?
In a cached system, the base addresses of the last few referenced pages is
maintained in registers called the TLB that aids in faster lookup. TLB contains those
page-table entries that have been most recently used. Normally, each virtual memory
reference causes 2 physical memory accesses-- one to fetch appropriate page-table entry,
and one to fetch the desired data. Using TLB in-between, this is reduced to just one
physical memory access in cases of TLB-hit.
10. What is the resident set and working set of a process?
Resident set is that portion of the process image that is actually in real-memory at
a particular instant. Working set is that subset of resident set that is actually needed for
execution. (Relate this to the variable-window size method for swapping techniques.)
11. When is a system in safe state?
The set of dispatchable processes is in a safe state if there exists at least one
temporal order in which all processes can be run to completion without resulting in a
deadlock.
12. What is cycle stealing?
We encounter cycle stealing in the context of Direct Memory Access (DMA).
Either the DMA controller can use the data bus when the CPU does not need it, or it may
force the CPU to temporarily suspend operation. The latter technique is called cycle
stealing. Note that cycle stealing can be done only at specific break points in an
instruction cycle.
155
If one or a few processes have a high access rate to data on one track of a storage
disk, then they may monopolize the device by repeated requests to that track. This
generally happens with most common device scheduling algorithms (LIFO, SSTF, CSCAN, etc). High-density multisurface disks are more likely to be affected by this than
low density ones.
14. What are the stipulations of C2 level security?
C2 level security provides for:
Discretionary Access Control
Identification and Authentication
Auditing
Resource reuse
15. What is busy waiting?
The repeated execution of a loop of code while waiting for an event to occur is
called busy-waiting. The CPU is not engaged in any real productive activity during this
period, and the process does not progress toward completion.
16. Explain the popular multiprocessor thread-scheduling strategies.
Load Sharing: Processes are not assigned to a particular processor. A global queue of
threads is maintained. Each processor, when idle, selects a thread from this queue.
Note that load balancing refers to a scheme where work is allocated to processors on
a more permanent basis.
Gang Scheduling: A set of related threads is scheduled to run on a set of processors at
the same time, on a 1-to-1 basis. Closely related threads / processes may be scheduled
this way to reduce synchronization blocking, and minimize process switching. Group
scheduling predated this strategy.
Dedicated processor assignment: Provides implicit scheduling defined by assignment
of threads to processors. For the duration of program execution, each program is
allocated a set of processors equal in number to the number of threads in the program.
Processors are chosen from the available pool.
Dynamic scheduling: The number of thread in a program can be altered during the
course of execution.
17. When does the condition 'rendezvous' arise?
In message passing, it is the condition in which, both, the sender and receiver are
blocked until the message is delivered.
18. What is a trap and trapdoor?
Trapdoor is a secret undocumented entry point into a program used to grant access
without normal methods of access authentication. A trap is a software interrupt, usually
the result of an error condition.
19. What are local and global page replacements?
Local replacement means that an incoming page is brought in only to the relevant
process' address space. Global replacement policy allows any page frame from any
process to be replaced. The latter is applicable to variable partitions model only.
20. Define latency, transfer and seek time with respect to disk I/O.
156
Seek time is the time required to move the disk arm to the required track.
Rotational delay or latency is the time it takes for the beginning of the required sector to
reach the head. Sum of seek time (if any) and latency is the access time. Time taken to
actually transfer a span of data is transfer time.
21. Describe the Buddy system of memory allocation.
Free memory is maintained in linked lists, each of equal sized blocks. Any such
block is of size 2^k. When some memory is required by a process, the block size of next
higher order is chosen, and broken into two. Note that the two such pieces differ in
address only in their kth bit. Such pieces are called buddies. When any used block is
freed, the OS checks to see if its buddy is also free. If so, it is rejoined, and put into the
original free-block linked-list.
22. What is time-stamping?
It is a technique proposed by Lamport, used to order events in a distributed
system without the use of clocks. This scheme is intended to order events consisting of
the transmission of messages. Each system 'i' in the network maintains a counter Ci.
Every time a system transmits a message, it increments its counter by 1 and attaches the
time-stamp Ti to the message. When a message is received, the receiving system 'j' sets
its counter Cj to 1 more than the maximum of its current value and the incoming timestamp Ti. At each site, the ordering of messages is determined by the following rules: For
messages x from site i and y from site j, x precedes y if one of the following conditions
holds....(a) if Ti<Tj or (b) if Ti=Tj and i<j.
23. How are the wait/signal operations for monitor different from those for semaphores?
If a process in a monitor signal and no task is waiting on the condition variable,
the signal is lost. So this allows easier program design. Whereas in semaphores, every
operation affects the value of the semaphore, so the wait and signal operations should be
perfectly balanced in the program.
24. In the context of memory management, what are placement and replacement
algorithms?
Placement algorithms determine where in available real-memory to load a
program. Common methods are first-fit, next-fit, best-fit. Replacement algorithms are
used when memory is full, and one process (or part of a process) needs to be swapped out
to accommodate a new program. The replacement algorithm determines which are the
partitions to be swapped out.
25. In loading programs into memory, what is the difference between load-time dynamic
linking and run-time dynamic linking?
For load-time dynamic linking: Load module to be loaded is read into memory.
Any reference to a target external module causes that module to be loaded and the
references are updated to a relative address from the start base address of the application
module.
With run-time dynamic loading: Some of the linking is postponed until actual
reference during execution. Then the correct module is loaded and linked.
26. What are demand- and pre-paging?
With demand paging, a page is brought into memory only when a location on that
157
page is actually referenced during execution. With pre-paging, pages other than the one
demanded by a page fault are brought in. The selection of such pages is done based on
common access patterns, especially for secondary memory devices.
27. Paging a memory management function, while multiprogramming a processor
management function, are the two interdependent?
Yes.
28. What is page cannibalizing?
Page swapping or page replacements are called page cannibalizing.
29. What has triggered the need for multitasking in PCs?
Increased speed and memory capacity of microprocessors together with the support
fir virtual memory and
Growth of client server computing
30. What are the four layers that Windows NT have in order to achieve independence?
Hardware abstraction layer
Kernel
Subsystems
System Services.
31. What is SMP?
To achieve maximum efficiency and reliability a mode of operation known as
symmetric multiprocessing is used. In essence, with SMP any process or threads can be
assigned to any processor.
32. What are the key object oriented concepts used by Windows NT?
Encapsulation
Object class and instance
33. Is Windows NT a full blown object oriented operating system? Give reasons.
No Windows NT is not so, because its not implemented in object oriented
language and the data structures reside within one executive component and are not
represented as objects and it does not support object oriented capabilities .
34. What is a drawback of MVT?
It does not have the features like
ability to support multiple processors
virtual storage
source level debugging
35. What is process spawning?
When the OS at the explicit request of another process creates a process, this
action is called process spawning.
36. How many jobs can be run concurrently on MVT?
15 jobs
37. List out some reasons for process termination.
158
Normal completion
160