c++ - 解决基本模板类成员的歧义

我有一个类的层次结构，每个类都必须有一个特定的基类。比基类提供了张贴日志记录的功能，并可以获取日志通道的名称（基本上是使用日志的类的名称）。让我们将此类称为Logable。

为了允许我的类多次从该Logable类继承，我给它提供了一个模板参数，每个子代都将自身用作该参数。

实际上，我正在使用boost :: log库，但是用简单的LogableImpl类简化了上述层次结构的示例，该类替换了boost :: log接收器。

#include <iostream>
#include <string>

// macro for logging in a boost::log style
#define LOG_DEBUG this->_loggerObj.logStream("debug")
#define LOG_INFO  this->_loggerObj.logStream("info")
#define LOG_WARN  this->_loggerObj.logStream("warning")
#define LOG_ERROR this->_loggerObj.logStream("error")


class LogableImpl
{
private:
   std::string _channelName;
public:
   LogableImpl(const std::string & channelName): _channelName(channelName) {}

   std::ostream & logStream(const std::string & severetyLevel)
   {
      std::cout << _channelName << " " << severetyLevel;
      return std::cout;
   }
};


template <class Descendant>
class Logable
{
protected:
   Logable(const std::string & channelName): _loggerObj(channelName) {}
   LogableImpl _loggerObj;
};


class Base: private Logable<Base>
{
public:
   Base()
      : Logable<Base>("Base")
   {}

   void someMethod()
   {
      LOG_INFO << "some method is called" << std::endl;
      LOG_ERROR << "an error happened" << std::endl;
   }
};


class Derived: public Base, private Logable<Derived>
{
public:
   Derived()
      : Logable<Derived>("Derived")
   {}

   void someAnotherMethod()
   {
      LOG_INFO << "another method is called" << std::endl;
      LOG_ERROR << "another error is happened" << std::endl;
   }
};


int main()
{
   Base b;
   Derived d;
   b.someMethod();
   d.someMethod();

   return 0;
}

显然，我使用MSVC 2008对该源进行编译尝试时出错

error C2385: ambiguous access of '_loggerObj'
1>        could be the '_loggerObj' in base 'Logable<Base>'
1>        or could be the '_loggerObj' in base 'Logable<Derived>'
1>d:\cpp\visualstudio\tests\workbench\test\main.cpp(55) : error C2248: 'Logable<Descendant>::_loggerObj' : cannot access inaccessible member declared in class 'Logable<Descendant>'
1>        with
1>        [
1>            Descendant=Base
1>        ]
1>        d:\cpp\visualstudio\tests\workbench\test\main.cpp(29) : see declaration of 'Logable<Descendant>::_loggerObj'
1>        with
1>        [
1>            Descendant=Base
1>        ]
1>d:\cpp\visualstudio\tests\workbench\test\main.cpp(56) : error C2385: ambiguous access of '_loggerObj'
1>        could be the '_loggerObj' in base 'Logable<Base>'
1>        or could be the '_loggerObj' in base 'Logable<Derived>'
1>d:\prog\cpp\visualstudio\tests\workbench\boost_test\main.cpp(56) : error C2248: 'Logable<Descendant>::_loggerObj' : cannot access inaccessible member declared in class 'Logable<Descendant>'
1>        with
1>        [
1>            Descendant=Base
1>        ]
1>        d:\cpp\visualstudio\tests\workbench\test\main.cpp(29) : see declaration of 'Logable<Descendant>::_loggerObj'
1>        with
1>        [
1>            Descendant=Base
1>        ]
1>Build log was saved at "file://d:\cpp\visualStudio\tests\workbench\test\Debug\BuildLog.htm"
1>boost_test - 4 error(s), 0 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

如何指定在LOG_*宏中使用的正确的基本成员？
我觉得可以通过一些模板魔术来完成，但是无法弄清楚。

必须使用不支持C ++ 11x功能的MSVC2008完成

最佳答案

我通过使用c ++ 11明确指定应使用的Logable来做到这一点。由于我们不知道this的类型，因此我使用decltype：

#define LOGABLE_TYPE typename std::remove_reference<decltype(*this)>::type
#define LOG_DEBUG this->Logable<LOGABLE_TYPE>::_loggerObj.logStream("debug")
#define LOG_INFO  this->Logable<LOGABLE_TYPE>::_loggerObj.logStream("info")
#define LOG_WARN  this->Logable<LOGABLE_TYPE>::_loggerObj.logStream("warning")
#define LOG_ERROR this->Logable<LOGABLE_TYPE>::_loggerObj.logStream("error")

在此处查看完整代码：http://ideone.com/1D5jrj