递归的默认返回值

Question

Consider the following well known code to find factorial of a number.

int fact(int num)
{
    if (num != 1)
        return num * fact(num - 1);
    else
        return num;
}

Now consider one more variant of the same

int fact(int num)
{
    if (num != 1)
        return num * fact(num - 1);
}

The first one looks clean, and if we give num as 4, it'll be 4*3*2*1 . The second one is also working fine, even though compiler may give a warning. But in the latter one, there's no return statement if the num become 1. So how the calculation happens? The code will work only when it return 1 if the num is 1. Can anybody please explain what's happening here 4*3*2*x . How x is returned as 1. Thanks.

Edit: I'm using gcc 4.8.2

Answer 1

The second variant is incorrect, there is no such thing as "default return value", that behavior is totally undefined!

Think about it, how will the compiler to know that fact(1) == 1 ??

If you let me guess, I'll say you wanted to avoid using the else stuff and, in that case, the default return value should be the last sentence.

int fact(int num)
{ // pre: num > 0
    if (num != 1)
        return num * fact(num - 1); 
    return num;
}

Answer 2

The second piece of code is incorrect, the warning that you thought is not important might be something like:

warning: control reaches end of non-void function

You must provide the return value in all paths, otherwise (when num is 1 ) it's undefined behavior, there's no default return value (with the exception of main ).

Answer 3

It is an undefined behavior, which means that the result totally depends on the specific compiler. Hence I think you should try to avoid it.