Question

I have a piece of code written in C where some pointer arithmetic is performed. I would like to know how the output comes to be this?

#include <stdio.h>  
int main()
{
    char arr[] = "gookmforgookm";
    char *ptr1 = arr;
    char *ptr2 = ptr1 + 3;
    printf ("ptr2 - ptr1 = %d\n", ptr2 - ptr1);
    printf ("(int*)ptr2 - (int*) ptr1 = %d",  (int*)ptr2 - (int*)ptr1);
    getchar();
    return 0;
}

Output is below:

ptr2 - ptr1 = 3  
(int*)ptr2 - (int*) ptr1 = 0

Answer 1

Strictly speaking, you’re invoking undefined behaviour and any result that the program produces is OK according to the C standard.

However, you’re probably on a machine where sizeof(int) == 4 (as opposed to say, 2). Since there are 4 bytes to an integer, two addresses which are 3 bytes apart are part of the same integer, so the difference between the addresses is 0 * sizeof(int). You might find a different answer if you chose ptr1 = arr + 1;, or you might not. But that’s the beauty of undefined behaviour – it would be ‘right’ either way.