enable_shared_from_this和继承 [英] enable_shared_from_this and inheritance

问题描述

我已经得到了从 enable_shared_from_this＆LT继承的类型;类型＆GT; ，和另一种类型的，从这个类型继承。现在，因为它返回的基本类型和具体的派生类的方法，我需要派生类型，我不能用shared_from_this方法。它是有效的，只是从这个构造一个shared_ptr直接？

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?

编辑：
在一个相关的问题，我怎么能在类型的右值移动的shared_ptr＆LT;基地＆GT; 来一个类型的的shared_ptr＆LT;衍生＆GT; ？我用的dynamic_cast以验证它确实是正确的类型，但现在我似乎无法完成的实际行动。

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.

推荐答案

一旦你获得的shared_ptr＆LT;基地以及GT; ，你可以使用<一个href=\"http://www.boost.org/doc/libs/1_45_0/libs/smart_ptr/shared_ptr.htm#static_pointer_cast\"><$c$c>static_pointer_cast将其转换为一个的shared_ptr＆LT;衍生方式＆gt;

Once you obtain the shared_ptr<Base>, you can use static_pointer_cast to convert it to a shared_ptr<Derived>.

您不能只是创建一个的shared_ptr 直接这个;这将是等同于：

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

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