Making operator

by

The problem with this setup is that the operator<< you defined above is a free function, which can’t be virtual (it has no receiver object). In order to make the function virtual, it must be defined as a member of some class, which is problematic here because if you define operator<< as a member of a class then the operands will be in the wrong order:

class MyClass {
public:
    virtual ostream& operator<< (ostream& out) const;
};

means that

MyClass myObject;
cout << myObject;

will not compile, but

MyClass myObject;
myObject << cout;

will be legal.

To fix this, you can apply the Fundamental Theorem of Software Engineering – any problem can be solved by adding another layer of indirection. Rather than making operator<< virtual, consider adding a new virtual function to the class that looks like this:

class MyClass {
public:
    virtual void print(ostream& where) const;
};

Then, define operator<< as

ostream& operator<< (ostream& out, const MyClass& mc) {
    mc.print(out);
    return out;
}

This way, the operator<< free function has the right parameter order, but the behavior of operator<< can be customized in subclasses.

Hope this helps!

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