Implicit downcast of shared_ptr in CRTP - c++

Question

I built a class Interface to be used with CRTP for static polymorphism and a Client class having a shared_ptr to the Interface. I would like to return from the Client the shared_ptr to the Implementation, something that I can (safely?) do through static_pointer_cast within the Client. Is there a way to allow an implicit downcasting from the shared_ptr to Interface to a shared_ptr to Implementation?
template<class Implementation>
struct Interface {
int foo() { return static_cast<Implementation*>(this)->fooimpl();}
};
template<class Implementation>
struct Client {
Client(std::shared_ptr<Implementation> const & pt) : pt_{pt} {}
std::shared_ptr<Interface<Implementation>> pt_;
std::shared_ptr<Implementation> getPt() {
//Can I avoid this static_pointer_cast?<
return std::static_pointer_cast<Implementation>(pt_);
}
};
One possible solution to avoid all this mess is to keep a shared_ptr to Implementation within the Client class. In this way, however, nowhere I am saying that Implementation in Client has the Interface.
template<class Implementation>
struct Client {
Client(std::shared_ptr<Implementation> const & pt) : pt_{pt} {}
std::shared_ptr<Implementation> pt_;
std::shared_ptr<Implementation> getPt() { return pt_;}
};

Answer 1

Is there a way to allow an implicit downcasting from the shared_ptr to Interface to a shared_ptr to Implementation?
Simple answer: No! As the compiler have no idea of the "reverse" inheritance, it can give you direct support for it. CRTP is a general pattern to work around the underlying problem. Here the downcast is hand coded but hidden behind the interface of the CRTP implementation.
The CRTP is because I want Client to use foo() and at the same time independent of the implementation
As you get it as a compile time implementation, it is not really independent at the moment. You will see it latest, if you want point to something of that CRTP type.
The shared_ptr is because the Implementation may be shared among Clients
Your idea have a circular problem!
If you write as given in your example:
template
struct Client {
Client(std::shared_ptr const & pt) : pt_{pt} {}
std::shared_ptr pt_;
std::shared_ptr getPt() { return pt_;}
};
the code which calls getPt() must know the type of the returned pointer! Even if you use auto you will get the type of the returned pointer. So you simply can't hide it from your using code at all.
I ended up just putting a shared_ptr to the Implementation as class member and added a "static_assert + is_base_of " to insure this.
Seems to be also a circular problem.
If you write:
template < typename T>
class CRTP: public T
{
public:
void Check()
{
static_assert( std::is_base_of_v < T, CRTP<T> > );
}
};
class A {};
int main()
{
CRTP<A> x;
x.Check();
}
What is the assert helping here? It is only a check that you wrote 4 lines above "class CRTP: public T". For me that makes no sense.
I still have no real idea what you want to achieve more than simply using CRTP.