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:

class Base
{
    virtual Base* clone() const = 0;
};

class Derived : public Base
{
    virtual Derived* clone() const
};

The compiler detects that clone() returns an pointer to the type of the object, and allows Derived to override it to return a pointer to derived.

It would desirable to have clone() return a smart pointer that implies transfer of ownership semantics, like the following:

class Base
{
   virtual std::auto_ptr<Base> clone() const = 0;
};

class Derived : public Base
{
    virtual std::auto_ptr<Derived> clone() const;
};

Unfortunately, the relaxation of the conventions does not apply to templated smart pointers, and the compiler will not allow the override.

So, it seems I am left with two options:

1. Have clone() return a "dumb" pointer, and document that clients are responsible for disposing of it.
2. Have clone() return a smart base pointer, and have clients use dynamic_cast to save them to a Derived pointer if they need it.

Is one of these approaches preferred? Or is there a way for me to eat my transfer of ownership semantics and have my strong type safety too?

Answer 1

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); }
};