警告C4715：“ operator ==”：并非所有控制路径都返回值

Question

I keep getting this error and I do not understand why. The compiler told me that it was in this section.

Any help would be greatly appreciated

bool operator==(const Bitmap& b1, const Bitmap& b2){
    // TODO: complete the == operator
    if ((b1.height == b2.height) && (b1.width == b2.width))
    {

        for (int r = 0; r < b1.height; r++)
        {

            for (int c = 0; c < b1.width; c++)
            {
                if (b1.get(r, c) == b2.get(r, c))
                {

                }
                else
                    return false;
            }

        }

    }
    else
        return false;
}

Answer 1

The diagnostic from the compiler is exactly what it says.

Note that if the for loop runs through to the end, without taking the if condition that returns false , if r reaches the b1.height value, the execution path will reach the end of this function without an explicit return .

Answer 2

The error message is pretty clear.

bool operator==(const Bitmap& b1, const Bitmap& b2){

    if ((b1.height == b2.height) && (b1.width == b2.width))
    {
        for (int r = 0; r < b1.height; r++)
        {
            for (int c = 0; c < b1.width; c++)
            {
                ...
            }
        }
        return ???;   // What should be returned here?
    }
    else
        return false;
}

Answer 3

The error message tells what is wrong.

bool operator==(const Bitmap& b1, const Bitmap& b2){
    // TODO: complete the == operator
    if ((b1.height == b2.height) && (b1.width == b2.width))
    {

        for (int r = 0; r < b1.height; r++)
        {

            for (int c = 0; c < b1.width; c++)
            {
                if (b1.get(r, c) == b2.get(r, c))
                {

                }
                else
                    return false;
            }

        }
        return true; // I guess you forgot this line

    }
    else
        return false;
}

Answer 4

Well, it is precisely what the error says. The compiler doesn't know if there is a possibility that the code in the nested for loop can ever be triggered. Assuming the first condition is true, there is a chance that the code never gets to the return statement. Therefore the compiler will make sure that something always is returned, regardless of whatever condition you give it.

bool operator==(const Bitmap& b1, const Bitmap& b2){
if ((b1.height == b2.height) && (b1.width == b2.width))
{
    // The compiler expects a return statement that is always reachable inside this if.

    for (int r = 0; r < b1.height; r++)
    {

        for (int c = 0; c < b1.width; c++)
        {
            if (b1.get(r, c) == b2.get(r, c))
            {

            }
            else
                return false; //This isn't always reachable.
        }

    }
}
else
    return false; // This case is covered, so nothing wrong here.
}

Answer 5

It's simple.

bool operator==(const Bitmap& b1, const Bitmap& b2){
// TODO: complete the == operator
if ((b1.height == b2.height) && (b1.width == b2.width))
{

for (int r = 0; r < b1.height; r++)
{

    for (int c = 0; c < b1.width; c++)
    {
        if (b1.get(r, c) == b2.get(r, c))
        {

        }
        else
            return false;
    }

}
// if your function runs to end of for loop, you get to here
}
else
    return false;
//then here
return false;   //<-- add this
}