Returning a struct containing a pointer to the first element of an array-CodePudding

Running this code

#include <stdlib.h>
#include <stdio.h>

char* f() {
    char s[] = "Hello, world!";
    return s;
}

int main() {
    printf("%s", f());
    return 0;
}

prints (null). This makes sense to me: The char array is valid only in the scope of f, so the pointer returned is not valid. However, if I create a struct containing the pointer to the array

#include <stdlib.h>
#include <stdio.h>

typedef struct word {
    char* str;
} word;

word f() {
    char str[] = "Hello, world!";
    return (word) { str };
}

int main() {
    printf("%s", f().str);
    return 0;
}

this will now "work", i. e. it will print Hello, world!.

Now I am wondering why it does; shouldn't the pointer here be invalid too, because the array it points to was created in f and is thus not valid anymore? Or is this just undefined behavior and I got "lucky" that nothing else was pushed onto the stack at that location? Or is there a mistake in my reasoning (I am rather new to C).

I haven't been able to find information on this specific behavior, only for structs containing arrays of constant size. I have tested this using gcc version 9.4.0 (which uses gnu11 "standard" as default). I would appreciate if anyone could point me to the explanation of this behavior.

CodePudding user response：

Or is this just undefined behavior and I got "lucky" that nothing else was pushed onto the stack at that location?

That is correct. The line

printf("%s", f().str);

will invoke undefined behavior by dereferencing a dangling pointer. This means that anything may happen; your program may crash or it may appear to work as intended.

The compiler clang is actually able to detect this error at run-time, if you compile with -fsanitize=address. Here is an online demonstration.

For further explanation, you may want to read this answer to a similar C question, which contains an interesting analogy. Most of the information in the answer also applies to C.