Return value of function when return path not specified

Question

int foo(int i)
{
    if(i != 0)
        foo(i - 1);
    else
        return i;
}

GCC warns control reaches end of non-void function [-Wreturn-type].

Since the last return statement that set eax to 0, upon return from any other path, it returns 0. Also, compiler explorer produces the exact same code whether I wrote return foo(i - 1). Could one treat this as guaranteed behavior?

Your recursion construct must have a `base case`! That will have the "fall back" return statement compiler is looking for (since lookahead is limited, compiler cannot ascertain, when control will reach the return statement of your code). — anurag, Jan 08 '21 at 07:42
If there were multiple return paths and one of the returns modified eax, this might not hold, is what I'm thinking. In this case, yes you're correct. — user2338150, Jan 08 '21 at 07:43
No you cannot treat it as guaranteed behaviour for the simple reason that the C++ standard says it is undefined behaviour. You're looking at the behaviour of your compiler not at the C++ standard. — john, Jan 08 '21 at 07:44
`foo(1)` invokes `foo(0)` which returns 0 to the call `foo(1)`. Now, the control reaches end of `foo(1)` and there is no return. Hence the warning, because the return type was `int`. So, it is upto a compiler to decide what to do when a non `void` function doesn't return. **C/C++** cannot guarantee anything! — anurag, Jan 08 '21 at 07:52
See the SECOND most upvoted answer of the duplicate I proposed. — Yunnosch, Jan 08 '21 at 08:01
Thanks for the link. It is a slightly more complex disassembly, but the same that I was expecting. And I got it, this is just the particular compiler on the particular architecture. I shouldn't expect this as guaranteed behavior. — user2338150, Jan 08 '21 at 08:16
@Lundin: but if i dropped the else, code would behave differently without the return statement. — user2338150, Jan 08 '21 at 08:18
@anurag: so, you see now why such behavior cannot be guaranteed. — user2338150, Jan 08 '21 at 08:20
Well my conclusion is based on the accepted and now deleted answer! — anurag, Jan 08 '21 at 08:24

John Kugelman · Accepted Answer · 2021-01-08T07:50:37.917

Could one treat this as guaranteed behavior?

Simple answer: No. You must have the return.

Also, compiler explorer produces the exact same code...

In C and C++ you can't rely on compiler output to figure out whether you're doing something correctly. Any time you invoke undefined behavior it's possible for the compiler to do anything at all. It might even do the thing you "expect" it to. If you infer from that that your code is fine you'll be misled. Correct behavior doesn't necessarily mean correct code.

Return value of function when return path not specified

1 Answers1