Question

Here is the code,

char *foo()
{
    static char s[10] = "abcde";
    return s;
}

char *bar()
{
    char *c = foo();
    return c;
}

int main()
{
    printf("%s\n", bar()); 
}

Typically, it is wrong to return a local pointer as I did in bar, but now c points to a static var returned by foo, would it be correct to return a local var c in bar?

I tried, it printf the right value, but I don’t understand how it works. I thought, when bar() finishes, the var c should vanish, which should make printf print undefined stuff, right?

Follow Up

char *c is a local var, if char *c = "abcde";, I assume this: c is a local var which resides in the function’s stack, while "abcde" is a constant var which resides in the constants-area (part of the heap?), so when bar() finishes, c vanishes, but "abcde" still remains in the heap, right?

Answer 1

Variable c is only a pointer. It is not wrong to return a local pointer from a function, you do it all the time. For example, when you store a result of malloc in a pointer, the pointer is local, but the storage it points to is not. It is, however, wrong to return a pointer to a local storage. Since in your example c never points to a locally allocated data, your code works correctly as written.

EDIT (in response to the Follow Up)

“abcde” is a constant var which resides in the constants-area (part of the heap?)

The constants area is not usually part of the heap, it is either a separate arrea, usually adjacent to the area where the machine code of your program is stored.

c vanishes, but “abcde” still remains in the heap, right?

“abcde” remains in the constants area, not in the heap, but the concept is correct: the pointer to that constant remains valid throughout the entire run-time of your program.