GATE CS 2022 Q.21: Pointers, Arrays and Strings

What is printed by the following ANSI C program?

#include<stdio.h>

int main(int argc, char *argv[])
{
       int x = 1, z[2] = {10, 11};
       int *p = NULL;
       p = &x;
       *p = 10;
       p = &z[1];
       *(&z[0] + 1) += 3;
       printf("%d, %d, %d\n", x, z[0], z[1]);
       return 0;
}

1, 10, 11

1, 10, 14

10, 14, 11

10, 10, 14

Official answer and solution

Official answer: D
From: Official GATE 2022 answer key

Solution

Trace pointer assignments and array modifications step by step.

Step-by-step program trace:

int x = 1, z[2] = {10, 11}; sets $x = 1$ , $z[0] = 10$ , and $z[1] = 11$ .
int *p = NULL; p = &x; makes $p$ point to variable $x$ .
*p = 10; assigns the value 10 to $x$ . Now $x = 10$ .
p = &z[1]; updates $p$ to point to $z[1]$ .
*(&z[0] + 1) += 3;: Here &z[0] is a pointer to the first element of array $z$ . Adding 1 moves the pointer to &z[1]. Dereferencing yields the lvalue $z[1]$ . Adding 3 gives $z[1] = 11 + 3 = 14$ .
Variable $z[0]$ is not modified, so $z[0] = 10$ .
printf("%d, %d, %d\n", x, z[0], z[1]); prints 10, 10, 14.

Final answer

Option (D)

Common trap

Overlooking that *(&z[0] + 1) modifies z[1], not z[0].

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2022 Computer Science and Information Technology (CS)
Question: 21 of 65
Section: Computer Science and Information Technology

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