Consider the following ANSI-C program.
#include <stdio.h>
int main(){
int *ptr, a, b, c;
a=5; b=11; c=20;
ptr=&a; *ptr=c; ptr=&c;
a=*(&b); c=*ptr-a;
printf("%d",c);
return(0);
}The output of this program is ____________. (answer in integer)
Note: Assume that the program compiles and runs successfully.
ANSWER
The program prints 9.
Official answer: 9Source: Official answer key, GATE 2026 CS2 (IIT Guwahati)
Initialize: a=5, b=11, c=20.
ptr=&a; sets ptr to point to a.
*ptr=c; writes c's current value (20) into a, so a=20; c is unchanged at 20.
ptr=&c; retargets ptr to point to c (it no longer points to a).
a=*(&b); — &b dereferenced by * is simply b, so a=11 (overwriting the earlier a=20).
c=*ptr-a; — since ptr points to c, *ptr yields c's current value, 20; so c = 20 - 11 = 9.
printf("%d",c); outputs 9.
KEY IDEA
Track pointer targets and values separately: an assignment through *ptr changes whatever object ptr currently points to, not necessarily the same variable it pointed to earlier or later.
WATCH OUT FOR
Assuming *ptr always refers to a throughout the program (since ptr was first set to &a), and missing that ptr=&c; re-targets it to c before the final c=*ptr-a; line — this leads students to wrongly read *ptr as a's value there instead of c's.
From the GATE 2026 CS2 paper. Every question of this paper · Download the full paper · All 11 Pointers & arrays questions, with notes
