operator overloading in c++ ( with and without friend )

Question

Hey, I would like to know the difference between these 2 operator definitions:

1:

class Rational{
//...
public:
//...
Rational operator -() const{ return Rational(-t,b);}
//...
};

2:

class Rational{
//...
public:
//...
friend Rational operator -(const Rational& v) {return Rational(-t,b);}
//...
};

as far as i understand, for the usage of:

Rational s = -r 

r.operator-()   // should happen

would like some explenation for the difference, thanks !

Answer 1

For the most part, they are the same.

First of all, I don't think you have either written right. They should be:

 // Member function.      "-r" calls r.operator-() 
 Rational Rational::operator -() const{ return Rational(-t,b);} 

 // (technically a) global function.   "-r"  calls ::operator-(r) 
 friend Rational operator -(const Rational& v) {return Rational(-v.t,v.b);} 

The major difference is that if you have another type (say MyRational) which is convertible to a Rational object, then:

  MyRational mr = MyRational();
  Rational r = -mr;

will work with the second definition, but not for the first.