Return Type Covariance with Smart Pointers

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In C++ we can do this:

struct Base
{
   virtual Base* Clone() const { ... }
   virtual ~Base(){}
};

struct Derived : Base
{
   virtual Derived* Clone() const {...} //overrides Base::Clone
};

However, the following won't do the same trick:

struct Base
{
   virtual shared_ptr<Base> Clone() const { ... }
   virtual ~Base(){}
};

struct Derived : Base
{
   virtual shared_ptr<Derived> Clone() const {...} //hides Base::Clone
};

In this example Derived::Clone hides Base::Clone rather than overrides it, because the standard says that the return type of an overriding member may change only from reference(or pointer) to base to reference (or pointer) to derived. Is there any clever workaround for this? Of course one could argue that the Clone function should return a plain pointer anyway, but let's forget it for now - this is just an illustratory example. I am looking for a way to enable changing the return type of a virtual function from a smart pointer to Base to a smart pointer to Derived.

Thanks in advance!

Update: My second example indeed doesn't compile, thanks to Iammilind

return-type virtual-functions smart-pointers c++
4 Answers

There is an improvement on a great answer by @ymett using CRTP technique. That way you needn't worry about forgetting to add a non-virtual function in the Derived.

struct Base
{
private:
   virtual Base* doClone() const { ... }

public:
   shared_ptr<Base> Clone() const { return shared_ptr<Base>(doClone()); }

   virtual ~Base(){}
};

template<class T>
struct CRTP_Base : Base
{
public:
   shared_ptr<T> Clone() const { return shared_ptr<T>(doClone()); }
};

struct Derived : public CRTP_Base<Derived>
{
private:
   virtual Derived* doClone() const { ... }
};
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