Type No return, in function returning non-void

Question

My C++ code looks like this:

int f(int i){
    if (i > 0) return 1;
    if (i == 0) return 0;
    if (i < 0) return -1;
}

It's working but I still get:

Warning: No return, in function returning non-void

Even though it is obvious that all cases are covered. Is there any way to handle this in "proper" way?

Answer 1

The compiler doesn't grasp that the if conditions cover all possible conditions. Therefore, it thinks the execution flow can still fall through past all ifs.

Because either of these conditions assume the others to be false, you can write it like this:

int f(int i) {
    if (i > 0) return 1;
    else if (i == 0) return 0;
    else return -1;
}

And because a return statement finally terminates a function, we can shorten it to this:

int f(int i) {
    if (i > 0) return 1;
    if (i == 0) return 0;
    return -1;
}

Note the lack of the two elses.

Answer 2

Is there any way to handle this in "proper" way?

A simple fix is to get rid of the last if. Since the first two are either called or not the third case must be called if you get to it

int f(int i){
    if (i > 0) return 1;
    if (i == 0) return 0;
    return -1;
}

The reason we have to do this is that the compiler cannot guarantee that your if statements will be called in every case. Since it reaches the end of the function and it might not have executed any of the if statements it issues the warning.

Answer 3

Just help the compiler to understand your code. Rewrite the function the following way

int f(int i){
    if (i > 0) return 1;
    else if (i == 0) return 0;
    else return -1;
}

you could also write the function for example like

int f( int i )
{
    return i == 0 ? 0 : ( i < 0 ? -1 : 1 );
}

Another way to write the function in one line is the following

int f( int i )
{
    return ( i > 0 ) - ( i < 0 );
}

Answer 4

The compiler doesn't know that all options are covered, because in terms of your function's syntax there's nothing to suggest it.

A simplified example:

                           int f(int i)

        if                      if                   if

(int > int), return    (int == int), return   (int < int), return

A clearer structure like if/else with a return in each yields an Abstract-Syntax-Tree which clearly shows there's a return in each case. Yours, however, is dependent on the evaluation of nodes in the AST, which isn't covered in the syntax check (by which you're being issued a warning).

Beyond pure syntax, if the compiler were to rely on "possible" evaluations as well in trying to figure out the behavior of your program, it would ultimately need to entangle itself and probably hitting the halting problem. Even if it managed to cover some cases, this would probably spawn more questions from users than it would answer, and also risk an entirely new level of bugs.