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What will be the output of the following code?123456789101112#include <stdio.h>void negate(int num) { *num = -(*num);}int main() { int value = 8; negate(&value); printf("%d ", value); return 0;}

Question

What will be the output of the following code?123456789101112#include <stdio.h>void negate(int num) { *num = -(*num);}int main() { int value = 8; negate(&value); printf("%d ", value); return 0;}

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Solution

The code will not compile successfully. The reason is that the function negate is trying to dereference an integer, not a pointer. The correct way to write this function is to pass a pointer to the integer. Here is the corrected code:

#include <stdio.h>

void negate(int *num) {
    *num = -(*num);
}

int main() {
    int value = 8;
    negate(&value);
    printf("%d ", value);
    return 0;
}

In this corrected code, the negate function takes a pointer to an integer as its argument. Inside the function, it dereferences the pointer to get the integer that the pointer points to, negates it, and then stores the result back in the memory location that the pointer points to.

When this corrected code is run, it will print -8 to the console. This is because the negate function changes the value of value from 8 to -8, and then this new value is printed out.

This problem has been solved

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