Pointers

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Find out the output for the following programs.

1) #include <stdio.h >

int main()

int *p = 10;

printf(“ %u\n”, (unsigned int)p); //print the 10 but next line it will give an error

printf(“%d\n”,*p);

2) #include <stdio.h>

int main()

int *ptr, a = 10;

ptr = &a; // giving the address to the pointer

*ptr += 1; // changing the value * means change the address

printf("%d,%d/n", *ptr, a); // output is 11,11/n because the format is /n not \n

3) #include<stdio.h> //i don’t know how the ans is came

int main()

int x = -300;

unsigned char *p;

p = &x;

printf(“%d\n”,*p++); //212
 printf(“%d\n”,*p);//254

4) #include<stdio.h>

int main()

int x = 256;

char *p = &x;

*++p = 2; // the bit is set 512 so

printf(“%d”,x); //printing the decimal value

5) #include<stdio.h>

int main()

int x = 300;

if(*(char *)&x == 44) // it will take just first byte (0010 1100)=44

printf(“Little Endian\n”); // print little endian

else

printf(“Big Endian\n”);

6) #include <stdio.h>

void main()

int x = 0;
 int *ptr = &5; // we can not assign a const add it will give lvalue error

printf("%p\n", ptr);

7) #include<stdio.h>

int main()

int const *p = 5;

int q;

p = &q;

printf(“%d”,++(*p)); //we can not modify the *p we use int const *p thats why

8) #include<stdio.h>

int main()

int x = 10;

int const * const p; //at the declaration time we have to give the address other wise it will
not accept the address also and we can not modify the data also

p = &x; //we can not assign the address because it is in read only coundition

printf(“%d\n”, *p);

9) #include <stdio.h>

int x = 0;

void main()

int *const ptr = &x;

printf("%p\n", ptr);
 ptr++; // read only coundition

printf("%p\n ", ptr);

10) #include <stdio.h>

int main()

const int ary[4] = {1, 2, 3, 4};

int *p;

p = ary + 3; //this is giving the address of 3rd position

*p = 5; //modifying the 3 to 5

printf("%d\n", ary[3]);

11) #include <stdio.h>

int main()

int ary[4] = {1, 2, 3, 4};

int *p = ary + 3; //pointing to the 3rd position

printf("%d\n", p[-2]); // printing the 2 value 2nd no of the arry

12) #include <stdio.h>

void main()

char *s= "hello";

char *p = s + 2; //pointing on first L

printf("%c\t%c", *p, s[1]); // printing first L and s[1] is e

 }

13) #include <stdio.h>

int main()

void *p;

int a[4] = {1, 2, 3, 4};

p = &a[3]; // giving address of 3rd digit

int *ptr = &a[2]; // pointing to 2nd digit

int n = (int*)p - ptr; // substracting the value 4-3=1

printf("%d\n", n); // output 1

14) #include<stdio.h>

int main()

int a[ ] = {10,20,30,40,50},i;

char *p = a;

for(i=0;i<5;i++)

printf(“%d “,*p++); //take only one byte the out put is 10 0 0 0 20

15) #include<stdio.h>

int main()

int a[]={10,20,30,40,50};

char *p;
 p=(char *)a;

printf("%d\n",*((int *)p+4)); // just print 50 because ‘a’ stores base add. And in that
explicite type conversion so int+4=last add. That’s why

16) #include <stdio.h>

int main()

double *ptr = (double *)100; // the given is address

ptr = ptr + 2; // it will give 116 double is 8 byte

printf("%u\n", ptr);// output 116

17) #include <stdio.h>

int main()

int i = 10;

void *p = &i;

printf("%d\n", (int *)*p); // we can not use such type of statement *p is wrong for printing
the void pointer

// printf("%d\n", *(int*)p); // this statement is correct bt it is in comment so it will give an error

return 0;

18) #include <stdio.h>

int main()

int a[4] = {1, 2, 3, 4};

 void *p = &a[1]; //giving the address of array a[1]=....12

void *ptr = &a[2]; //giving the address or the a[2]=....16

int n = 1;

n = ptr - p;// substracting the address 16-12=4

printf("%d\n", n);//4

19) #include <stdio.h>

int main()

int *p = (int *)2;

int *q = (int *)3;

printf("%d", p + q); // we can do only substraction no other operation is performed on

address

20) Which of the following operand can be applied to pointers p and q?

(Assuming initialization as int *a = (int *)2; int *b = (int *)3;)

a) a + b

b) a – b // only subtract is possible

c) a * b

d) a / b

Ans: b)

21) Which of following logical operation can be applied to pointers?

(Assuming initialization int *a = 2; int *b = 3;)

a) a | b

b) a ^ b
 c) a & b

d) None of the mentioned

Ans: d)

22) #include <stdio.h>

void main()

char *s = "hello";

char *n = "cjn";

char *p = s + n; //we cannot add pointer

4 printf("%c\t%c", *p, s[1]);

23) #include <stdio.h>

void m(int *p)

int i = 0;

for(i = 0;i < 5; i++)

printf("%d\t", p[i]); //6 5 3 0 0

void main()

int a[5] = {6, 5, 3}; // partially initialize array so it will give other value are zero

m(&a);//calling function

24)#include <stdio.h>
 void foo(int*);

int main()

int i = 10,j=20,*p = &i;

foo(p++);

foo(p);

void foo(int *p)

printf("%d\n", *p);

25)#include <stdio.h>

int main()

int i = 97, *p = &i;

foo(&i);// don’t define the function it will give the error

printf("%d ", *p);

void foo(int *p) //

int j = 2;

p = &j;

printf("%d ", *p);

26) #include<stdio.h>

int main()
 {

const int ary[4] = {1,2,3,4}; // we can just read only we cannot modify the array

int *p = ary+3;

*p = 5;

ary[3] = 6;

printf(“%d”,ary[3]);

27) #include<stdio.h>

int main()

char *p = “Hai friends”, *p1 = p;

while(*p!='\0');

++*p++;

printf(“%s %s\n”,p,p1);

28) #include<stdio.h>

int main()

char *x = “VECTOR”;

printf(“%s\n”,x+3); //it will print the TOR from 3 it will print

printf(“%d\n”+1,123456); // 1 print from left to right

29) #include<stdio.h>
 int main()

char a[ ] = “abcdefgh”;

int *ptr = a;

printf(“%x %x\n”,ptr[0],ptr[1]);

30) #include<stdio.h>

#include<string.h>

int main()

char *str = "hello, world\n";

char *strc = "good morning\n";

strcpy(strc, str);

printf("%s\n", strc);

return 0;

31)#include <stdio.h>

int main()

char *str = "hello world";

char strc[50] = "good morning india\n";

strcpy(strc, str); // copy the content of *str into strc

printf("%s\n", strc); //it will print hello world

return 0;

}
 32) #include <stdio.h>

int main()

char *str = "hello, world\n";

str[5] = '.';

printf("%s\n", str);

return 0;

33) #include <stdio.h>

int main()

char str[] = "hello, world";

str[5] = '.'; // it will replace the , to .

printf("%s\n", str); // the output is hello.world

return 0;

34) #include <stdio.h>

int main()

char *str = "hello world"; // it is pointer so size is 4 byte

char strary[] = "hello world"; // it is char array and the given array is string \0 is added
automatically size is 12

printf("%d %d\n", sizeof(str), sizeof(strary));

return 0;

35) #include <stdio.h>

 int main()

char *str = "hello world";

char strary[] = "hello world";

printf("%d %d\n", strlen(str), strlen(strary)); // it is printing the length so both length is 11

return 0;

36) #include<stdio.h>

int main()

int a = 5,b = 4,c = 9;

*(a>b ? &a : &b) = (a+b)>c; // a+b>c is false so it return 0 and it is stored in the a address
means a modify

printf(“%d %d\n”,a,b); // so output is 0 4

37) Find the sizeof any datatype with out using sizeof operator. (Hint : Use pointers)

38) #include<stdio.h>

int main()

int i;

double a = 5.2;

char *ptr;

ptr = (char *)&a;

for(i=0;i<=7;i++)

printf(“%d\n”,*ptr++);

return 0;
 }

39) Correct the following program.

#include<stdio.h>

int main()

void *p; // we can change void to int also

int **ptr;

int a = 129;

p = &a;

ptr = &p;

printf(“ p = %d p = %u &p = %u\n”, *p, p, &p); //other is we can change *p to (int *)p is
correct

40) #include<stdio.h>

main()

char a[20];

char *p,*q;

p=&a[0];

q=&a[10];

printf("%d %d\n",q-p,&q-&p);

41) #include<stdio.h>

main()

int a=0x12345678;
 void *ptr;

ptr=&a;

printf("0x%x\n",*(int *)&*&*(char*)ptr);

42) #include<stdio.h>

main()

int a[5]={1,2,3,4,5};

int *ptr=(int *)(&a+1);

printf("%d %d\n",*(a+1),*(ptr-1));

printf("%d %d\n",*(a+1),*(ptr));

43) #include <stdio.h>

void main()

char *s= "hello";

char *p = s;

printf("%c\t%c", 1[p], s[1]); // we can use this 1[p] it will print the 1st block of the array

44) #include<stdio.h>

main()

char a[]="abcde";

char *p=a;
 p++; // address is increment by 1 byte

p++; // again address id increment by 1byte

p[2]='z'; // here p is changing the value of e because it will change the 2nd position

printf("%s",p); // cdz ci is on 0th position

45) #include<stdio.h>

main()

{
char a[]=”ABCDEFGHIJKLMNOPQRSTUVWXYZ”;

int i,*p = a;

for(i=0;i<5;i++)
printf(“%d\t”,*p++);

46) #include<stdio.h>

main()

char a[]=”abcdef”;

char *ptr1 = a;

ptr1 = ptr1+(strlen(ptr1)-1);

printf(“%c”, --*ptr1--);

printf(“%c”,--*--ptr1);

printf(“%c”,--*(ptr1--));

printf(“%c”,--*(--ptr1));

printf(“%c”,*ptr1);

47) #include<stdio.h>
char *str1 = “Hello”;

char *str2 = “Hai”;

char *str3;

str3 = strcat(str1,str2);

printf(“%s %s\n”,str3,str1);

int main()

return 0;

48) #include<stdio.h>

int main()

char a[]=”Hello”;

char *p=”Hai”;

a=”Hai”;

p=”Hello”;

printf(“%s %s\n”,a,p);

return 0;

49) #include<stdio.h>

int main()

int i,n;

char *x=”Alice”;
 n=strlen(x);

*x=x[n];

for(i=0;i<=n;i++)

printf(“%s”,x);

x++;

printf(“%s\n”,x);

return 0;

50) #include<stdio.h>

char *str=”char *str=%c%s%c;main(){printf(str,34,str,34);}”;

int main()

printf(str,34,str,34);

return 0;

51) #include <stdio.h>

void f(char *k)

k++;

k[2] = 'm';

printf("%c\n", *k);

void main()

{
 char s[] = "hello";

f(s);

printf("%s\n",s);

52) #include<stdio.h>

void t1(char *q);

main()

char *p;

p = “abcder”;

t1(p);

void t1(char *q)

{
if(*q!='r')

putchar(*q);

t1(q++);

53) #include<stdio.h>

int main()

int i;

float a=5.2;

char *ptr;

ptr=(char *)&a;
 for(i=0;i<=3;i++)

printf("%d ",*ptr++);

return 0;

54) #include <stdio.h>

void foo( int[] );

int main()

int ary[4] = {1, 2, 3, 4};

foo(ary);

printf("%d ", ary[0]);

void foo(int p[4])

int i = 10;

p = &i;

printf("%d ", p[0]);

55) #include <stdio.h>

void main()

int k = 5;

int *p = &k;

int **m = &p;

**m = 10;

printf("%d%d%d\n", k, *p, **m);

 }

56) #include <stdio.h>

int main()

int a = 1, b = 2, c = 3;

int *ptr1 = &a, *ptr2 = &b, *ptr3 = &c;

int **sptr = &ptr1;

printf("%d ",**sptr);

*sptr = ptr2;

printf(("%d ",**sptr);

57) #include <stdio.h>

void main()

int a[3] = {1, 2, 3};

int *p = a;

int *r = &p;

printf("%d\n", (**r));

58) #include <stdio.h>

int main()

int i = 97, *p = &i;

 foo(&p);

printf("%d ", *p);

return 0;

void foo(int **p)

int j = 2;

*p = &j;

printf("%d ", **p);

59) #include <stdio.h>

void foo(int *const *p);

int main()

int i = 11;

int *p = &i;

foo(&p);

printf("%d ", *p);

void foo(int *const *p)

int j = 10;

*p = &j;

printf("%d ", **p);

60) #include <stdio.h>

 void foo(int **const p);

int main()

int i = 10;

int *p = &i;

foo(&p);

printf("%d ", *p);

void foo(int **const p)

int j = 11;

*p = &j;

printf("%d ", **p);

61) #include <stdio.h>

int *f();

int main()

int *p = f();

printf("%d\n", *p);

int *f()

int *j = (int*)malloc(sizeof(int));

*j = 10;

return j;

}
62) #include <stdio.h>

void main()

char *a[10] = {"hi", "hello", "how"};

int i = 0;

for (i = 0;i < 10; i++)

printf("%s ", *(a[i]));

63) #include <stdio.h>

void main()

char *a[10] = {"hi", "hello", "how"};

int i = 0, j = 0;

a[0] = "hey"; // modifying the “hi” to “hey”

for (i = 0;i < 10; i++)

printf("%s ", a[i]); //printing the the 10 char

64) #include <stdio.h>

void main()

char *a[10] = {"hi", "hello", "how"};

printf("%d\n", sizeof(a));

65) #include <stdio.h>

 void main()

char *a[10] = {"hi", "hello", "how"};

printf("%d\n", sizeof(a[1]));

66) #include <stdio.h>

int main()

char a[2][6] = {"hello", "hi"};

printf("%s ", *a + 1);

return 0;

67) #include <stdio.h>

int main()

char *a[2] = {"hello", "hi"};

printf("%s\n", *(a + 1));

return 0;

68) #include <stdio.h>

int main(int argc, char *argv[])

while (argc--)

printf("%s\n", argv[argc]);
 return 0;

69) #include <stdio.h>

int main(int argc, char *argv[])

while (*argv++ != NULL)

printf("%s\n", *argv);

return 0;

70) #include <stdio.h>

int main(int argc, char *argv[])

while (*argv != NULL)

printf("%s\n", *(argv++));

return 0;

71) #include<stdio.h>

int main(int sizeofargv, char *argv[])

while(sizeofargv)

printf(“%s ”,argv[--sizeofargv]);

return 0;

} if i/p is sample friday tuesday sunday

72) #include<stdio.h>

int main()

char *str[]={“Progs”,”Do”,”Not”,”Die”,”They”,”Croak!”};

printf(“%d %d”,sizeof(str),strlen(str[0]));

return 0;

73) #include<stdio.h>

int main()

{
static char *s[]={“black”,”white”,”pink”,”violet”};

char **ptr[]={s+3,s+2,s+1,s},***p;

p = ptr;

printf(“%s\n”,**p+1);

return 0;

74) #include<stdio.h>

main()

char *m[]={“jan”,”feb”,”mar”};

char d[][10] = {“sun”,”mon”,”tue”};

printf(“%s\t”,m[1]);

printf(“%s\t”,d[1]);

75) #include<stdio.h>

void fun(char **);

 int main()

char *argv[]={“ab”,”cd”,”ef”,”gh”};

fun(argv);

return 0;

void fun(char **p)

char *t;

t=(p+=sizeof(int))[-1];

printf(“%s\n”,t);

76) #include <stdio.h>

void first()

printf("first");

void second()

first();

void third()

second();

void main()

{
 void (*ptr)();

ptr = third;

ptr();

77) #include <stdio.h>

int add(int a, int b)

return a + b;

int main()

int (*fn_ptr)(int, int);

fn_ptr = add;

printf("The sum of two numbers is: %d\n", (int)fn_ptr(2, 3));

78) #include <stdio.h>

int mul(int a, int b, int c)

return a * b * c;

void main()

int (*function_pointer)(int, int, int);

function_pointer = mul;

printf("The product of three numbers is:%d",function_pointer(2, 3, 4));

}
79) #include<stdio.h>

int fun(int (*)());

int main()

fun(main);

printf(“Hi\n”);

return 0;

int fun(int (*p)())

printf(“Hello\n”);

return 0;

80) #include<stdio.h>

int main()

char *p = “Hello World”;

printf(p);

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Dear Students, if any mistakes found, Kindly inform to me.

A.Tandava Ramakrishna

Email:ramakrishna@vectorindia.org