Is it safe to keep a pointer to a variable out of scope?

Question

Example code:

#include <stdio.h>
int main (){
    int *p;
    {
        int v = 1;
        p = &v;
    }
    printf("%d\n", *p);
    return 0;
}

This code works fine, but I'm not sure if there's a guarantee that the address of v will be preserved.

Answer 1

There is no guarantee.

Once v goes out of scope, doing anything with it at all (even via a pointer) is considered Undefined Behavior.

As with any other undefined behavior, just because it works on one operating system, compiler, compiler version, time of day, etc, doesn't mean it will work for another.

Answer 2

To add on Merlyn's answer, one case where this would probably result in behavior you didn't intend is the following:

#include <stdio.h>
int main (){
    int *p;
    {
        int v = 1;
        p = &v;
    }
    {
        int w = 2;
        printf("%d\n", w);
    }
    printf("%d\n", *p);
    return 0;
}

The compiler may optimize this by having v and w share the same allocation on the stack. Again, the compiler might also not optimize this -- that's why the behavior of using pointers to variables after their enclosing block ends isn't defined. The program might output "2" and "1", or "2" and "2", or "2" and something completely different depending on which compiler and settings are used.