Is copy elision in the form of named return value optimization permitted in C?

Question

Is the following C program guaranteed to exit with 0 or is the compiler allowed to identify the objects s and t with one another as is permitted in C++ as the so-called named return value optimization (NRVO) form of copy elision?

typedef struct {
    int i, j;
    double a, b;
} S;

int result;

S test(S *q) {
    S s = {0, 0, 0, 0};
    result = &s == q;
    return s;
}

int main(void)
{
  S t = test(&t);
  return result;
}

Clang exits with 1 against my expectations, see https://godbolt.org/z/ME8sPGn3n.

(In C++, both 0 and 1 are valid outcomes due to explicit permission to perform NRVO.)

Answer 1

It is guaranteed to exit with 0. t and s are different objects, pointers to them can't compare equal.

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About the code that you removed:

What about the following variation in which relevant lifetime considerations may be different?

From https://port70.net/~nsz/c/c11/n1570.html#6.2.4p2 :

The value of a pointer becomes indeterminate when the object it points to (or just past) reaches the end of its lifetime.

It may return 1 or 0 or perform a trap.