【发布时间】:2016-04-15 17:22:01
【问题描述】:
我只是想知道为什么我的代码无法编译。下面的好吗?我正在尝试使用 Category1 和 Category2 typedef-s 声明一个简单的类。
Category1 typedef 可以正常编译,但 Category2 one 不能。
似乎无法编译 Category2 typedef,因为尽管未实例化周围的类 X,但仍实例化了类 iterator_traits... 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) { }
};
注意,我不需要实例化X类得到以下错误(以上是完整的编译单元)。
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。而且我需要确定返回的容器是否具有随机访问迭代器。但是我遇到了这个编译错误。谢谢你的解释!
【问题讨论】:
-
containerGetterFun()替换为std::declval<typename std::result_of<GetterFunType()>::type>()? -
@Yakk:没有变化...令人惊讶的是代码:
typedef typename std::iterator_traits< decltype(std::vector<int>().begin()) >::iterator_category Category2;不编译... - 即使非参数化类型用于iterator_traits。 -
不足为奇:2012 年对 decltype 的支持介于差和蹩脚之间。所以有一些查询成员存在的内置内在函数,也许有一个用于成员返回值?
&const_X::begin可能是另一种值得考虑的方法。 -
试试 MSVS 2015,它可以工作
标签: c++ templates visual-studio-2012 compiler-errors iterator-traits