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# C | Pointer Basics | Question 3

Output of following program?

 #include   int main(){    int *ptr;    int x;      ptr = &x;    *ptr = 0;      printf(" x = %d\n", x);    printf(" *ptr = %d\n", *ptr);      *ptr += 5;    printf(" x  = %d\n", x);    printf(" *ptr = %d\n", *ptr);      (*ptr)++;    printf(" x = %d\n", x);    printf(" *ptr = %d\n", *ptr);      return 0;}

(A) x = 0
*ptr = 0
x = 5
*ptr = 5
x = 6
*ptr = 6
(B) x = garbage value
*ptr = 0
x = garbage value
*ptr = 5
x = garbage value
*ptr = 6
(C) x = 0
*ptr = 0
x = 5
*ptr = 5
x = garbage value
*ptr = garbage value
(D) x = 0
*ptr = 0
x = 0
*ptr = 0
x = 0
*ptr = 0

Explanation: See the comments below for explanation.

int *ptr;  /* Note: the use of * here is not for dereferencing,
it is for data type int */
int x;

ptr = &x;   /* ptr now points to x (or ptr is equal to address of x) */
*ptr = 0;   /* set value ate ptr to 0 or set x to zero */

printf(" x = %d\n", x);   /* prints x =  0 */
printf(" *ptr = %d\n", *ptr);  /* prints *ptr =  0 */

*ptr += 5;        /* increment the value at ptr by 5 */
printf(" x  = %d\n", x);  /* prints x = 5 */
printf(" *ptr = %d\n", *ptr); /* prints *ptr =  5 */

(*ptr)++;         /* increment the value at ptr by 1 */
printf(" x  = %d\n", x);  /* prints x = 6 */
printf(" *ptr = %d\n", *ptr);  /* prints *ptr =  6 */
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