Question

Given the following code:

int x = 0;
int y = 10;
int *ptr = &x;
*ptr = -55;
x += -52;
ptr = &y;
*ptr += 52;
printf("%d\n", x);

What is printed to the screen?

Answer 1

Explanation :

It will print -107, as *ptr contains the address of x.

first time x = 0

then ptr which contains the address of x is found value -55

after that x also become -55

now x += -52, taking the value of x to -107

C Program:

#include <stdio.h>

int main()
{
    int x = 0;
    int y = 10;
    int *ptr = &x;
    *ptr = -55;
    x += -52;
    ptr = &y;
    *ptr += 52;
    printf("%d\n", x);

    return 0;
}

Output:

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