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What will be the output of the following C program?

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#include <stdio.h>

int change(int (**temp)(int (*)(void)));

int main() {
    int change(int (*)(int (*)(void)));
    int (*ptr)(int (*)(void)) = change;
    printf("%d", (*ptr)(&ptr));
    return 0;
}

int change(int (**temp)(int (*)(void))) {
    static int m;
    if (++m < 10)
        (**temp)(temp);
    return (m);
}

Show answer & explanation

Answer: A. 10

The program exhibits type mismatch issues that cause undefined behavior leading to output 10. The function change expects int (**temp)(int (*)(void)) (pointer to function pointer), but &ptr of type int (*(*)(int (*)(void))) is passed. The static variable m increments from 1 to 10 across recursive calls before the recursion terminates when m == 10, and the final return value is 10.

Step-by-step Derivation:
Execution trace:

main() declares ptr as function pointer to change
Calls (*ptr)(&ptr) → change(&ptr) with type mismatch
Inside change(): static int m = 0 (initialized once)
Recursion loop: ++m increments m, check m < 10:
- m=1: 1<10 → call (**temp)(temp) → (*ptr)(&ptr) (recursive)
- m=2: 2<10 → recurse
- ...
- m=9: 9<10 → recurse
- m=10: 10<10 is FALSE → exit recursion
Return m=10 from innermost call; propagates up
printf("%d", 10) outputs: 10

What will be the output of the following C program?

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