为什么不能在带有模板的派生类中使用基类的别名?

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问题描述

考虑以下C ++代码:

Consider this C++ code :

template<typename Session>
class Step
{
public:
   using Session_ptr = boost::shared_ptr<Session>;
protected:
   Session_ptr m_session;
public:
   inline Step(Session_ptr session) :
      m_session(session)
   {}

};

template<typename Socket>
class Session
{
public:
   Socket a;

   Session(Socket _a):
      a(_a)
   {}
};

template <typename Socket>
class StartSession : public Step<Session<Socket> >
{
protected:
   Session_ptr m_session; //Unknown type Session_ptr
public:
   inline StartSession(Session_ptr session) :
      Step<Session<Socket> >(session)
   {}

   void operator()(const boost::system::error_code& ec);
};

template <typename Socket>
class StartSession2 : public Step<Session<Socket> >
{
protected:
   typename Step<Session<Socket> >::Session_ptr m_session;
public:
   inline StartSession2(typename Step<Session<Socket> >::Session_ptr session) :
      Step<Session<Socket> >(session)
   {}

   void operator()(const boost::system::error_code& ec);
};

int main(int argc, char * argv[])
{
   Step<Session<int> >::Session_ptr b(new Session<int>(5)); //no problem
   StartSession<int >::Session_ptr bb(new Session<int>(5)); //gcc ok, clang refuses to remember the symbol since the class has errors
   StartSession2<int >::Session_ptr bbb(new Session<int>(5)); //no problem
   std::cout << b->a; // ok
   std::cout << bb->a; // gcc ok, clang bb not declared
   std::cout << bbb->a; // ok
   return 0;
}

如您所见，这里发生了一些奇怪的事情(至少对我来说)...

As you can see, there are some strange (to me at least) things happening here...

首先，为什么在子类中不能访问Session_ptr?我知道，因为这些是模板化的类，这会使事情变得更复杂...但是我看不出有任何强制使用typename的歧义...

First, why isn't Session_ptr accessible in the child classes ?I know because these are templated class, that make things more complicated... But I don't see any ambiguity here that makes the use of typename mandatory...

然后，为什么在主目录中Session_ptr既可以作为基类的成员又可以作为子类的成员来访问?

Then, why in the main, Session_ptr is accessible either as member of the base class either as member of a child class ?

推荐答案

不合格的查找不在类模板的依赖基类中查找.

Unqualified lookup does not look in dependent base classes in class templates.

所以在这里:

template <typename Socket>
class StartSession : public Step<Session<Socket> >
{
protected:
   Session_ptr m_session; // <== unqualified name lookup on Session_ptr
   // ...
};

Step<Session<Socket>>是StartSession<Socket>的依赖基类.为了在那里查找，您必须进行 qualified 名称查找(这是您在StartSession2中所做的事情):

Step<Session<Socket>> is a dependent base class of StartSession<Socket>. In order to lookup there, you'll have to do qualified name lookup (which is what you're doing in StartSession2):

template <typename Socket>
class StartSession : public Step<Session<Socket> >
{
protected:
   typename Step<Session<Socket>>::Session_ptr m_session;
   // ...
};

或者只是自己添加别名:

Or simply add the alias yourself:

using Session_ptr = typename Step<Session<Socket>>::Session_ptr;

这篇关于为什么不能在带有模板的派生类中使用基类的别名?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

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