C++ Overloading Conversion Operators

问题

I am trying to have a class that allows implicit casting to certain built in types, like unsigned long int and since I'm trying to do this as correct as possible (this is my first important project in C++), I have hit a strange issue regarding const correctness:

This works:

#include <iostream>

class CustomizedInt
{
private:
    int data;
public:
    CustomizedInt();
    CustomizedInt(int input);
    operator unsigned long int () const
    {
        unsigned long int output;
        output = (unsigned long int)data;
        return output;
    }
};

CustomizedInt::CustomizedInt()
{
    this->data = 0;
}

CustomizedInt::CustomizedInt(int input)
{
    this->data = input;
}

int main()
{
    CustomizedInt x;
    unsigned long int y = x;

    std::cout << y << std::endl;

    return 0;
}

But this:

#include <iostream>

class CustomizedInt
{
private:
    int data;
public:
    CustomizedInt();
    CustomizedInt(int input);
    operator unsigned long int () const;
};

CustomizedInt::CustomizedInt()
{
    this->data = 0;
}

CustomizedInt::CustomizedInt(int input)
{
    this->data = input;
}

CustomizedInt::operator unsigned long()
{
    unsigned long int output;
    output = (unsigned long int)data;
    return output;
}

int main()
{
    CustomizedInt x;
    unsigned long int y = x;

    std::cout << y << std::endl;

    return 0;
}

gives me this error in Visual Studio 2010: error C2511: 'CustomizedInt::operator unsigned long(void)' : overloaded member function not found in 'CustomizedInt'

Now, if I remove the keyword const from the operator definition, everything is OK. Is this a bug? I read that I'm supposed to use the const keyword after each (public) method / operator in order to clearly state that it does not alter the current object in any way.

Also, I know that defining such an operator may be poor practice, but I am not sure I fully understand the associated caveats. Could somebody please outline them? Would it be better practice to just define a public method called ToUnsignedLongInt?

回答1:

The function signature does not match the function definition.

operator unsigned long int () const;

and

CustomizedInt::operator unsigned long()    { ... }
                                       ^^^
                                   const missing

In this case you should mark the conversion operator as const since it doesn't affect the internal state of the object.

Also, use constructor initialization lists to initialize your member variables.

CustomizedInt::CustomizedInt()
: data()
{
}

CustomizedInt::CustomizedInt(int input)
: data(input)
{
}

回答2:

You could remove the const from the declaration, but what you almost certainly want to do is add it to the definition:

CustomizedInt::operator unsigned long() const
{
    unsigned long int output;
    output = (unsigned long int)data;
    return output;
}

回答3:

Yes, if your member function doesn't affect the logical state of the object, then you should indeed postfix it with const, so that the compiler will enforce that.

But in that case, you also need to add const when you define the function body!

回答4:

You just need to copy the same function prototype into the implementation. ie.

CustomizedInt::operator unsigned long int() const

来源：https://stackoverflow.com/questions/10281334/c-overloading-conversion-operators

标签

c++

operator-overloading

const-correctness