Double inheritance of enable_shared_from_this

问题

I have an object (Z) which derives from two other objects (A and B).

A and B both derive from enable_shared_from_this<>, respectively enable_shared_from_this<A> and enable_shared_from_this<B>.

Of course I call shared_from_this() on Z. And of course the compiler reports this as ambiguous.

My questions are :

is it safe to inherit twice from enable_shared_from_this<> or will it create two separated reference counts (bad !)
If not safe, how do I solve this ?

Note : I've found this other question bad weak pointer when base and derived class both inherit from boost::enable_shared_from_this but it doesn't really answer. Should I use the virtual trick too ?

回答1:

Yes, as per bad weak pointer when base and derived class both inherit from boost::enable_shared_from_this the solution is to use virtual inheritance. Here's an implementation for the C++11 standard shared_ptr (not Boost):

#include <memory>

struct virtual_enable_shared_from_this_base:
   std::enable_shared_from_this<virtual_enable_shared_from_this_base> {
   virtual ~virtual_enable_shared_from_this_base() {}
};
template<typename T>
struct virtual_enable_shared_from_this:
virtual virtual_enable_shared_from_this_base {
   std::shared_ptr<T> shared_from_this() {
      return std::dynamic_pointer_cast<T>(
         virtual_enable_shared_from_this_base::shared_from_this());
   }
};

struct A: virtual_enable_shared_from_this<A> {};
struct B: virtual_enable_shared_from_this<B> {};
struct Z: A, B { };
int main() {
   std::shared_ptr<Z> z = std::make_shared<Z>();
   std::shared_ptr<B> b = z->B::shared_from_this();
}

This isn't part of the default implementation, probably because of the overhead of virtual inheritance.

回答2:

Yep, your class will be derived from two distinct classes enable_shared_from_this<A> and enable_shared_from_this<B>, and have two different weak ref's

Trick from that answer allows to have one base class, because of virtual inheritance

回答3:

Using the shared_ptr aliasing constructor, a variation of ecatmur's answer can be derived:

#include <memory>

struct virtual_enable_shared_from_this_base:
   std::enable_shared_from_this<virtual_enable_shared_from_this_base> {
   virtual ~virtual_enable_shared_from_this_base() {}
};
template<typename T>
struct virtual_enable_shared_from_this:
virtual virtual_enable_shared_from_this_base {
   std::shared_ptr<T> shared_from_this() {
      return std::shared_ptr<T>(
         virtual_enable_shared_from_this_base::shared_from_this(),
         static_cast<T*>(this));
   }
   std::shared_ptr<const T> shared_from_this() const {
      return std::shared_ptr<const T>(
         virtual_enable_shared_from_this_base::shared_from_this(),
         static_cast<const T*>(this));
   }
};

struct A: virtual_enable_shared_from_this<A> {};
struct B: virtual_enable_shared_from_this<B> {};
struct Z: A, B { };
int main() {
   std::shared_ptr<Z> z = std::make_shared<Z>();
   std::shared_ptr<B> b = z->B::shared_from_this();
}

I expect this version to be faster in many circumstances, since it avoids a costly dynamic cast. However, as usual, only becnhmarks have the final word. Also, I have added the const variation.

来源：https://stackoverflow.com/questions/15549722/double-inheritance-of-enable-shared-from-this

标签

c++

shared-ptr

multiple-inheritance

enable-shared-from-this