Question

I understand that in C++ you can overload an operator like a function. And as common with functions in C++, you have to specify a return value:

struct A {

    int operator +();

};

Here I’ve overloaded operator+ as a function that returns an int. But I find that when I overload a type while giving it a return value, I get error: return type specified for 'operator int'.

struct A {

    void operator int() {} // error

};

But if I take away the return value it works fine.

struct A {

    operator int() {} // pass

};

Does the error mean that by my use of int before the function parameters, that I’m creating a function that returns an int. Or is this some mistake? What if I want the function to not return a value? Can someone please explain why I’m getting this error? Thanks.

Answer 1

By definition, operator int() returns an int. It will be called in contexts where an object of type A needs to be converted to int. That’s quite different from the first code snippet, where the operator is operator+(); you can define it to return pretty much anything you like.