问题
I've got a type which inherits from enable_shared_from_this<type>, and another type that inherits from this type. Now I can't use the shared_from_this method because it returns the base type and in a specific derived class method I need the derived type. Is it valid to just construct a shared_ptr from this directly?
Edit:
In a related question, how can I move from an rvalue of type shared_ptr<base> to a type of shared_ptr<derived>? I used dynamic_cast to verify that it really was the correct type, but now I can't seem to accomplish the actual move.
回答1:
Once you obtain the shared_ptr<Base>, you can use static_pointer_cast to convert it to a shared_ptr<Derived>.
You can't just create a shared_ptr directly from this; that would be equivalent to:
shared_ptr<T> x(new T());
shared_ptr<T> y(x.get()); // now you have two separate reference counts
来源：https://stackoverflow.com/questions/4491420/enable-shared-from-this-and-inheritance