Unexpected C behaviour with pointers

Question

Suppose I have this block of code:

#include <stdio.h>
#include <unistd.h>

int *local_pointer(void)
{
    int x = 6;
    return &x;
}

void add(void)
{
    int a;
    a = 4;
    a = a + 1;
}

int main()
{
    int *result;
    result = local_pointer();
    printf("int is %d\n", *result);
    add();
    printf("int is %d\n", *result);
    return 0;
}

The output of this code is

int is 6
int is 5

I am confused as to why this is?

If we go through main, the result variable is returned a pointer to x and then when it the pointer is dereferenced the its value is 6 however why does calling the add function make the next printf statement print 5?

@GyaptiJain but why is this the case? everytime i run the program after compiling with gcc i get the same output — bigfocalchord, Aug 27 '19 at 05:51
@bigfocalchord Undefined behavior is undefined. Anything can happen and this is what happens to happen for no particular reason. — David Schwartz, Aug 27 '19 at 05:51
You might need to switch to python.. :D or understand where local variables are kept — Pynchia, Aug 27 '19 at 05:54
Works does it? When running your code [elsewhere it results in a segmentation fault](https://wandbox.org/permlink/DqstykBVdh6P52oJ). Sadly "working" is one way undefined behavior manifests. And it's the worst manifestation possible, since it fools you into thinking your code is okay and can be relied on. — StoryTeller - Unslander Monica, Aug 27 '19 at 05:55
@bigfocalchord Similar answer: https://stackoverflow.com/questions/27963547/memory-allocation-stack/27963631#27963631 — Gyapti Jain, Aug 27 '19 at 05:58
@GyaptiJain good catch. Strange coincidence. Or a classical homework ? — Guillaume Petitjean, Aug 27 '19 at 09:14

score 3 · Accepted Answer · answered Aug 27 '19 at 05:51

Your code has a bug. Your local_pointer function returns a pointer to its x variable. But as soon as that function returns, x no longer exists since it was local. So when you dereference the returned pointer, the results are undefined.

If I had to guess, I'd say that a happens to occupy the memory that x used to occupy. But that's just a guess. The behavior might be due to something else. It might change with compiler options. It might change with a different compiler. Who knows? Fix the bug and the mystery will go away.

With optimization, the `add()` function and the call to it might be optimized away since they have no visible effect on the computation. Then you might get 6 twice, or might get yet another result. — Jonathan Leffler, Aug 27 '19 at 06:53

score 0 · Answer 2 · answered Aug 27 '19 at 09:09

The function local_pointer probably (cf. other comments on undefined behaviour) returns a pointer to the stack. So I think you are basically printing the content of the stack. The call to add writes the value 5 to the stack, that's what your second printf is outputing 5.

Unexpected C behaviour with pointers

2 Answers2