Can't Overload operator<< as member function

I am trying to overload operator<< as a member function. It works if simply do this:

friend ostream& operator<<(ostream& os, const MyClass& myClass); in my header file and in my MyClass.cc file:

ostream& operator<<(ostream& os, const MyClass& myClass)
{
   return myClass.print(os);
}

However, if I try to take the friend off and make it a member function, then it complains that operator<< can only take one argument. Why?

ostream& MyClass::operator<<(ostream& os, const MyClass& myClass)
{
   return myClass.print(os);
}

I read at this question that it can't be a member function, but not sure why?

Solution 1:

When overloaded as a member function, a << b is interpreted as a.operator<<(b), so it only takes one explicit parameter (with this as a hidden parameter).

Since this requires that the overload be part of the class used as the left-hand operand, it's not useful with normal ostreams and such. It would require that your overload be part of the ostream class, not part of your class. Since you're not allowed to modify ostream, you can't do that. That leaves only the global overload as an alternative.

There is, however, a fairly widely-used pattern where you overload the operator globally, but have that call a member function:

class whatever { 
    // make this public, or the global overload a friend.
    std::ostream &write(std::ostream &dest) const { 
        // write self to dest
    }
};

std::ostream &operator<<(std::ostream &os, whatever const &w) { 
     return w.write(os);
}

This is particularly useful when/if you want polymorphic behavior. You can't make the overloaded operator polymorphic itself, but you make the member function it calls virtual, so it acts polymorphic anyway.

Edit: to (I hope) clarify the situation, you can do this a few different ways. The first and probably most obvious is to just make our write member public, and have the global operator call it. Since it is public, we don't have to do anything special to let the operator use it:

class myClass {
public:
    std::ostream &write(std::ostream &os) const { 
        // write stuff to stream
        return os;
    }   
};

std::ostream &operator<<(std::ostream &os, myClas const &m) { 
   // since `write` is public, we can call it without any problem.
   return m.write(os);
}

A second alternative is to make write private, and declare operator<< a friend to give it access:

class myClass {
    // Note this is private:
    std::ostream &write(std::ostream &os) const { 
        // write stuff to stream
        return os;
    }

    // since `write` is private, we declare `operator<<` a friend to give it access:
    friend std::ostream &operator<<(std::ostream &, myClass const &);
};

std::ostream &operator<<(std::ostream &os, myClas const &m) { 
   return m.write(os);
}

There's a third possibility that's almost like the second:

class myClass {
    // Note this is private:
    std::ostream &write(std::ostream &os) const { 
        // write stuff to stream
        return os;
    }

    // since `write` is private, we declare `operator<<` a friend to give it access.
    // We also implement it right here inside the class definition though:
    friend std::ostream &operator<<(std::ostream &os, myClas const &m) { 
        return m.write(os);
    }
};

This third case uses a rather strange (and little known) rule in C++ called "name injection". The compiler knows that a friend function can't be part of the class, so instead of defining a member function, this "injects" the name of that function into the surrounding scope (the global scope, in this case). Even though operator<< is defined inside the class definition, it's not a member function at all -- it's a global function.

Solution 2:

You can overload operator<< as a member function. But you can't write a member operator<< that takes an ostream on the left side, and your class on the right side.

When you make something a (non-static) member function, there is an implied first argument, the calling object. operator<< is binary, so it only takes 2 arguments. If you make it a member function, you can only give it one paremeter, because it already has one(the calling object). And since that calling object is always the first argument, you can't possibly write the output operator as a (non-static) member (at least, not in the standard form for that particular function), because for that case, the ostream needs to be the first argument.