Say I have a Complex number class and operator+ is overloaded twice, both as a member function and as a global function, for example:
class Complex {
public:
Complex operator+(const Complex& c);
};
Complex operator+(const Complex& a, const Complex& b);
And in the main function I will call the operator+ like follows:
Complex a, b;
Complex c = a + b;
Which operator+ function will be called? Thanks!
Members are not preferred over non-members in general, nor vice versa. C++'s overload resolution rules are applied to select one or the other.
A member function, for the purpose of overload resolution, is considered to have an implied object parameter (§13.3.1/2). So
Complex Complex::operator+(const Complex& c);
is treated as though it takes two arguments: the original const Complex& c, and another Complex& which refers to the object used to call the member function (in effect, *this).
Suppose we have two Complex variables:
Complex c1, c2;
Both c1 and c2 are non-const, so in order to call
c1.operator+(c2)
the parameter c, which is a const reference, has to bind to the non-const argument c2.
On the other hand, to call
operator+(c1, c2)
both parameters a and b, which are const references, have to bind to non-const objects, c1 and c2.
The member operator+ is better because const Complex&, Complex& is a better match for c1, c2 than const Complex&, const Complex& because it performs less qualification conversion. (§13.3.3.2/3)
If you change the declaration of the member operator+ to
Complex Complex::operator+(const Complex& c) const;
then the overload will become ambiguous, and compilation will fail.
A little hint how you could find it out by yourself: Ùse the debugger to see which method is called.
