解决基本模板 class 成员的歧义

Question

我有一个 classes 的层次结构，它们中的每一个都必须有一个特定的基础 class。比基础 class 提供了 post 日志记录的能力，并接收日志通道的 ctor 名称（基本上是使用日志的 class 的名称）。让我们称之为 class Logable.

为了让我的 class 继承自那个 Logable class 几次，我给了它一个模板参数，每个后代都使用它自己作为这个参数。

实际上我使用的是 boost::log 库，但是有一个非常简化的示例，该层次结构具有简单的 LogableImpl class，它正在替换 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;
}

很明显，我尝试使用 MSVC 2008 编译此源代码时出错

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 ==========

如何在 LOG_* 宏中指定正确的基成员？我觉得它可以通过一些模板魔术来完成，但就是想不通。

必须用不支持C++11x特性的MSVC2008来完成

Answer 1

我通过明确指定应该使用哪个 Logable 来使用 c++11 设法做到这一点。由于我们不知道 this 的类型，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

Answer 2

不幸的是，我找不到将 decltype 替换为 msvc 2008 可以理解的内容以使用 Petr 的答案的方法。连boost::typeof都不合适（只要我用对了）

所以我想出了一个解决方案，添加一个 using 带有宏的案例

#include <iostream>
#include <string>

#define USE_APPROPRIATE_LOGGER(classname) using Logable<classname>::_loggerObj
#define LOG_DEBUG _loggerObj.logStream("debug")
#define LOG_INFO  _loggerObj.logStream("info")
#define LOG_WARN  _loggerObj.logStream("warning")
#define LOG_ERROR _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>
{
   USE_APPROPRIATE_LOGGER(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>
{
   USE_APPROPRIATE_LOGGER(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();
   std::cout << std::endl;
   d.someAnotherMethod();

   return 0;
}

使用继承来提供对 class 的可登录性的想法很丑陋并且扼杀了想法，但似乎没有 c++11

就没有其他方法了

解决基本模板 class 成员的歧义

Resolve ambiguousness of a base template class member

c++

templates

ambiguity

boost-log