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GATE 2025 CS1 Q.34: solution and answer

GATE 2025 · CS1Q.34 · 1 mark · NAT
#include <stdio.h>
void foo(int *p, int x){
    *p=x;
}
int main(){
    int *z;
    int a = 20, b = 25;
    z = &a;
    foo(z,b);
    printf("%d",a);
    return 0;
}

The output of the given C program is __________. (Answer in integer)

See the question in the paper
GATE 2025 CS1 question 34Tap to open full size

Answer and solution

ANSWER

The output of the program is 25.

Official answer: 25Source: Official answer key, GATE 2025 CS1 (IIT Roorkee)

  1. Declarations: int *z; and int a = 20, b = 25; set up an uninitialized pointer z and two ints, a = 20 and b = 25.

  2. z = &a; makes z hold the address of a.

  3. foo(z, b) is called: the parameter p receives the value of z (i.e., the address of a), and the parameter x receives the value of b (25) — a copy, not a reference.

  4. Inside foo, *p = x; dereferences p (which points to a) and stores the value of x (25) into that memory location, i.e., into a.

  5. Because p and z both hold the address of a, this write is visible in main: a is now 25 (its original value 20 is overwritten). b remains unchanged at 25 but plays no further role.

  6. foo returns; printf("%d", a); prints the current value of a, which is 25.

  7. Output: 25.

KEY IDEA

foo receives a pointer p and a value x. Since p is a copy of z (i.e., p also holds the address of a), the statement *p = x writes through the pointer directly into a in main, overwriting it with the value of x (a copy of b).

WATCH OUT FOR

Confusing pass-by-value with pass-by-reference: assuming that because b (value 25) is passed by value, the assignment inside foo cannot affect the caller's variables — missing that it is the pointer z (holding &a) that is passed, and dereferencing it writes directly into a.

From the GATE 2025 CS1 paper. Every question of this paper · Download the full paper · All 11 Pointers & arrays questions, with notes

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