What's the best signature for clone() in C++?

Question

As Scott Myers wrote, you can take advantage of a relaxation in C++\'s type-system to declare clone() to return a pointer to the actual type being declared:

Answer 1

I think the function semantics are so clear in this case that there is little space for confusion. So I think you can use the covariant version (the one returning a dumb pointer to the real type) with an easy conscience, and your callers will know that they are getting a new object whose property is transferred to them.

Answer 2

Use the Public non-virtual / Private virtual pattern :

class Base {
    public:
    std::auto_ptr<Base> clone () { return doClone(); }
    private:
    virtual Base* doClone() { return new (*this); }
};
class Derived : public Base {
    public:
    std::auto_ptr<Derived> clone () { return doClone(); }
    private:
    virtual Derived* doClone() { return new (*this); }
};