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我目前正在尝试按照Dr.Dobb概述的方法为自定义模板化容器类实现迭代器。在我尝试为嵌套的迭代器定义迭代器成员函数之前，它工作正常

template <typename T>
class Container
{
  template <bool isConst = false>
  class Iterator
  {
    Iterator& operator++();
    Iterator  operator++(int);  
  };
};

像这样的课外：

template <typename T>
template <bool isConst>
//Container<T>::Iterator<isConst>& // g++ 4.8.1 is fine with that
typename Container<T>::Iterator<isConst>& // VS 2012 is fine with that
//typename Container<T>::template Iterator<isConst>& // again fine with g++
  Container<T>::Iterator<isConst>::operator++()
{
  ++index_;

  return *this;
}

template <typename T>
template <bool isConst>
//Container<T>::Iterator<isConst> // g++ 4.8.1 is fine with that
typename Container<T>::Iterator<isConst> // VS 2012 is fine with that
//typename Container<T>::template Iterator<isConst> // again fine with g++
  Container<T>::Iterator<isConst>::operator++(int)
{
  auto tmp(*this);

  this->operator++();

  return tmp;
}

到目前为止，我已经能够识别出三种不同的变体（下文详述），它们适用于 g++ 4.8.1（使用 C++0x）或 Visual Studio 2012（更新 4），但不适用于两者同时。

这样做的正确方法是什么？

从 2009 年开始似乎有一个相关的问题，但据我所知，它不包含Variant 4，我认为编译器错误现在已经修复。;)

还有这个问题，让我觉得Variant 3应该是那个，g++接受Variant 1是一种礼貌。

我在底部包含了完整的精简测试应用程序；它应该按照提供的方式编译（注释适当的行）。

变体 1：没有关键字

Container<T>::Iterator<isConst> // g++ 4.8.1 is fine with that

对于这个变体，VS 抱怨依赖名称不是类型并建议使用typename关键字：

2>..\demo\demo.cpp(70): warning C4346: 'Container<T>::?$Iterator@$Q_N(*(_CAAB@))' : dependent name is not a type  
2>          prefix with 'typename' to indicate a type  
2>..\demo\demo.cpp(70): error C2143: syntax error : missing ';' before '&'  
2>..\demo\demo.cpp(70): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int  
2>..\demo\demo.cpp(70): error C2936: 'Container<T>::Iterator<isConst>' : template-class-id redefined as a global data variable  
2>..\demo\demo.cpp(70): fatal error C1903: unable to recover from previous error(s); stopping compilation

变体 2：只有typename关键字

typename Container<T>::Iterator<isConst>& // VS 2012 is fine with that

如果只使用typename关键字（正如 VS 对 Variant 1 的建议），g++ 对此不太满意：

main.cpp:71:28: error: non-template 'Iterator' used as template                                                      
     typename Container<T>::Iterator<isConst>& // VS 2012 is fine with that  
                            ^                           
main.cpp:71:28: note: use 'Container<T>::template Iterator' to indicate that it is a template                        
main.cpp:71:28: error: expected unqualified-id at end of input

变体 3：两个关键字

typename Container<T>::template Iterator<isConst>& // again fine with g++

如果使用了两个关键字（如 g++ 对变体 2 的建议），VS 再次抱怨：

2>..\demo\demo.cpp(78): error C2244: 'Container<T>::Iterator<isConst>::operator ++' : unable to match function definition to an existing declaration  
2>          definition  
2>          'Container<T>::Iterator<isConst>  &Container<T>::Iterator<isConst>::operator ++(void)'  
2>          existing declarations  
2>          'Container<T>::Iterator<isConst>  Container<T>::Iterator<isConst>::operator ++(int)'  
2>          'Container<T>::Iterator<isConst> &Container<T>::Iterator<isConst>::operator ++(void)'  
2>..\demo\demo.cpp(92): error C2244: 'Container<T>::Iterator<isConst>::operator ++' : unable to match function definition to an existing declaration  
2>          definition  
2>          'Container<T>::Iterator<isConst> Container<T>::Iterator<isConst>::operator ++(int)'  
2>          existing declarations  
2>          'Container<T>::Iterator<isConst> Container<T>::Iterator<isConst>::operator ++(int)'  
2>          'Container<T>::Iterator<isConst> &Container<T>::Iterator<isConst>::operator ++(void)'

完整的测试应用程序

由于auto关键字，需要-std=c++0x和 g++。

template <bool isConst, typename IsTrue, typename IsFalse>
struct Choose;

template <typename IsTrue, typename IsFalse>
struct Choose<true, IsTrue, IsFalse>
{
  typedef IsTrue type;
};

template <typename IsTrue, typename IsFalse>
struct Choose<false, IsTrue, IsFalse>
{
  typedef IsFalse type;
};

template <typename T>
class Container
{
public:
  template <bool isConst = false>
  class Iterator
  {
  public:
    typedef typename Choose<isConst, T const&, T&>::type reference;
    typedef typename Choose<isConst, T const*, T*>::type pointer;

    typedef typename Choose<isConst,
      Container<T> const*,
      Container<T>*>::type ObjectPointer;

    friend class Iterator<true>;

    Iterator(
      ObjectPointer object           = 0,
      unsigned long long const index = 0);

    Iterator(Iterator<false> const& other);

    Iterator& operator++();
    Iterator  operator++(int);  

  private:
    ObjectPointer object_;

    unsigned long long index_;
  };

  typedef Iterator<>           iterator;
  typedef Iterator<true> const_iterator;
};

template <typename T>
template <bool isConst>
Container<T>::Iterator<isConst>::Iterator(
  ObjectPointer object,
  unsigned long long const index)
  : object_(object)
  , index_ (index )
{}

template <typename T>
template <bool isConst>
Container<T>::Iterator<isConst>::Iterator(Iterator<false> const& other)
  : object_(other.object_)
  , index_ (other.index_ )
{}

template <typename T>
template <bool isConst>
//Container<T>::Iterator<isConst>& // g++ 4.8.1 is fine with that
typename Container<T>::Iterator<isConst>& // VS 2012 is fine with that
//typename Container<T>::template Iterator<isConst>& // again fine with g++
  Container<T>::Iterator<isConst>::operator++()
{
  ++index_;

  return *this;
}

template <typename T>
template <bool isConst>
//Container<T>::Iterator<isConst> // g++ 4.8.1 is fine with that
typename Container<T>::Iterator<isConst> // VS 2012 is fine with that
//typename Container<T>::template Iterator<isConst> // again fine with g++
  Container<T>::Iterator<isConst>::operator++(int)
{
  auto tmp(*this);

  this->operator++();

  return tmp;
}

int main(int, char*[])
{
  Container<int> c;

  Container<int>::iterator it(&c, 0);

  Container<int>::const_iterator cIt = it;

  Container<int>::const_iterator cIt0 = ++cIt;
  Container<int>::const_iterator cIt1 = cIt++;

  return 0;
}
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