I am trying to understand operator overloading of plus sign. I see 2 prototypes:
Box operator+(const Box& b) { ... }
Box operator+(const Box& left, const Box& right) { ... }
Which one is right? If the difference is only that first is member function and second is non-member, then lets say I define both ways, then which one will be called on ?
Box a, b;
Box c = a + b;
What is "right" or "better" depends on your application. The member version gives access to all private attributes and methods of Box, the non-member version doesn't unless it's declared as a friend of Box. But a non-member can be templated and made to apply to a wide range of types.
Members are not preferred by the compiler over non-members in general, nor vice versa. Instead C++ overload resolution rules are applied to select one or the other.
Box Box::operator+(const Box& b) is treated as though it takes two arguments: Box& which refers to the object used to call the member function (*this), and const Box& b.
In your example, both a and b are non-const.
In order to call Box Box::operator+(const Box& b), b needs to be converted to a const reference.
In order to call Box operator+(const Box& left, const Box& right), both a and b need to be converted to const references.
So the member operator is selected because it is a better match (requires less conversions).
Had your member operator+ been declared const, you would have gotten a compiler error because the call would become ambiguous.
