没有价值的回报

Question

If I have an int function, with a return;

int f(int val, int *hasResult) {
  if (val) {
    *hasResult = 0;
    return;
  }
  *hasResult = 1;
  return 2;
}

and use it as

int hasResult;
int unwrapped = f(val, &hasResult);
if (hasResult) {
  printf("%d", unwrapped);
}

Is this valid C89? I know it's not valid C99+, but can I do this, exactly as it is, in C89? If not, how would I have to do this?

( f doesn't have side-effects)

Answer 1

In C89 draft paragraph 3.6.6.4 we find:

If a return statement without an expression is executed, and the value of the function call is used by the caller, the behavior is undefined . Reaching the } that terminates a function is equivalent to executing a return statement without an expression.

In short, to be avoided, whether "legal"or not.