为什么即使保证返回，我也会收到警告“控制到达非空函数的结尾”

Question

I understand that, given the code never enters the loop, would give me the error. However, given that r is equal to 1 , it should return it, ie , it would always enter it.

int arroz() {

    int r = 1;
    while (r) {
        return r;
    }
}

So, why does this code give me the warning control reaches end of non-void function?

Answer 1

The compiler correctly assumes that r is a variable value, it does not process it as a constant, that being the case it cannot be sure what the value is going to be and therefore if the body of the cycle will or will not be executed.

If you use a constant the warning goes away because the compiler will assemble the code using a constant value.

You can check this behavior here .

Answer 2

IMO it is a bug in the compiler. The analyzer should see that there is no execution path avoiding the return statement. The variable is not volatile .

The code is compiled to:

arroz:                                  # @arroz
        mov     eax, 1
        ret

This shows that compiler know that it will always return 1

You should report the bug.