Question

Hi I have a code like this, I think both the friend overloaded operator and conversion operator have the similar function. However, why does the friend overloaded operator is called in this case? What’s the rules?

Thanks so much!

class A{

    double i;
public:
    A(int i):i(i) {}
    operator double () const { cout<<"conversion operator"<<endl;return i;}                            // a conversion operator
    friend bool operator>(int i, A  a);                            // a friend funcion of operator >
};

bool operator>(int i, A  a ){
    cout<<"Friend"<<endl;
    return i>a.i;
}
int main()
{
    A  aa(1);
     if (0 > aa){
         return 1;
      }
}

Answer 1

No conversion is necessary for the overloaded operator> to be called. In order for the built-in operator> to be called, one conversion is necessary (the user-defined conversion operator. Overload resolution prefers options with fewer required conversions, so the overloaded operator> is used.

Note that if you were to change the definition of your overloaded operator> to be, for example:

friend bool operator>(double i, A  a);

you would get a compilation error because both the overloaded operator> and the built-in operator> would require one conversion, and the compiler would not be able to resolve the ambiguity.