Question

我只是想知道为什么我的代码无法编译。以下是好吗？我正在尝试使用Category1和Category2 typedef-s声明一个简单的类。

Category1 typedef编译正常，但Category2编译不正确。

似乎无法编译Category2 typedef，因为类iterator_traits＆lt;＆gt;虽然周围的类没有实例化，但是实例化了...对我来说似乎很困惑。

#include <iterator>

template <class GetterFunType>
struct X {
  GetterFunType containerGetterFun;

  // works
  typedef typename std::iterator_traits<typename GetterFunType::iterator>::iterator_category Category1;

  // compile error - std::iterator_traits<> is instantiated with type 'unknown'
  typedef typename std::iterator_traits<
    decltype(containerGetterFun().begin())>::iterator_category Category2;

  X(GetterFunType _containerGetterFun) : containerGetterFun(_containerGetterFun) { }
};

请注意，我不需要实例化类X来获得以下错误（上面是完整的编译单元）。

在Visual Studio 2012中我得到了这个：

1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C2146: syntax error : missing ';' before identifier 'iterator_category'
1>          c:\data\fsl\apif_src_review\apif_src\systemns.cpp(11) : see reference to class template instantiation 'std::iterator_traits<_Iter>' being compiled
1>          with
1>          [
1>              _Iter=unknown
1>          ]
1>          c:\data\fsl\apif_src_review\apif_src\systemns.cpp(14) : see reference to class template instantiation 'X<GetterFunType>' being compiled
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C3254: 'std::iterator_traits<_Iter>' : class contains explicit override 'iterator_category' but does not derive from an interface that contains the function declaration
1>          with
1>          [
1>              _Iter=unknown
1>          ]
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C2838: 'iterator_category' : illegal qualified name in member declaration
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C2602: 'std::iterator_traits<_Iter>::iterator_category' is not a member of a base class of 'std::iterator_traits<_Iter>'
1>          with
1>          [
1>              _Iter=unknown
1>          ]
1>          c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364) : see declaration of 'std::iterator_traits<_Iter>::iterator_category'
1>          with
1>          [
1>              _Iter=unknown
1>          ]
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xutility(364): error C2868: 'std::iterator_traits<_Iter>::iterator_category' : illegal syntax for using-declaration; expected qualified-name
1>          with
1>          [
1>              _Iter=unknown
1>          ]

在xutility（364）中有：

template<class _Iter>
    struct iterator_traits
    {   // get traits from iterator _Iter
    typedef typename _Iter::iterator_category iterator_category;
    typedef typename _Iter::value_type value_type;
    typedef typename _Iter::difference_type difference_type;
    typedef difference_type distance_type;  // retained
    typedef typename _Iter::pointer pointer;
    typedef typename _Iter::reference reference;
    };

我的情况是我想声明一个在构造函数中获取lambda的类。 lambda应该返回对容器的引用。我需要确定返回的容器是否有随机访问迭代器。但是我遇到了这个编译错误。谢谢你的解释！

Answer 1

我能够使用gcc 5.3.1编译相同的代码而没有任何错误，使用-std = c ++ 11

您的编译器是一个相对较旧的编译器，不支持当前的C ++ 1x标准。如果您需要使用现代C ++功能，切换到另一个编译器是我在这里可以看到的唯一选项。

Answer 2

我看到代码没问题，但有一些Visual Studio 2012限制不允许编译。至少在MSVS 2015中，它的工作原理与gcc 5.3.1相同。谢谢，朋友们！

意外的模板实例化导致编译错误

2 个答案: