Thread: Unobvious output

Eh, that program results in undefined behaviour, methinks. You made p point to b in func(), but then reduced p by 2. This means that p no longer points to b, so your (*p)+= 10; line changes something other than b.

You probably wanted to write:

Code:

#include <stdio.h>

void func(int a, int b)
{
    int * p;
    p = &b;
    *p -= 2;
    *p += 10;
}

int main()
{
    int num;
    num = 0;
    func(1, 2);
    num = 12;
    printf(" Num is now %d \n", num);
    return 0;
}

With this revised program, I get the output you expected.

Changing the value that is 2 integer sizes before b in the memory is pretty undefined - that could be ANYTHING. Someone who wrote this code obviously meant for it to modify the "num" value, but there's no guarantee whatsoever that there isn't "stuff" between num and b that is there because the compiler decided to put other stuff there.... Such as the program counter, frame pointer, locally stored registers, etc, etc.

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Mats