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

Tr1::shared_ptr<> can be casted like it were a raw pointer.

I think have clone() return a shared_ptr pointer is a pretty clean solution. You can cast the pointer to shared_ptr by means of tr1::static_pointer_cast or tr1::dynamic_pointer_cast in case it is not possible to determine the kind of cloned object at compile time.

To ensure the kind of object is predictible you can use a polymorphic cast for shared_ptr like this one:

template 
inline std::tr1::shared_ptr polymorphic_pointer_downcast(T &p)
{
    assert( std::tr1::dynamic_pointer_cast(p) );
    return std::tr1::static_pointer_cast(p);
}

The overhead added by the assert will be thrown away in the release version.