Output of following program?
#include <stdio.h>
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
Answer: (A)
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 */