可变模板 [英] dependent types with variadic templates

问题描述

您可以看到此函数声明有什么问题吗？

 模板< typename ... Containers> 
 std :: tuple< typename Containers :: value_type ...> 
 foo（const Containers& ... args）; 
  

当我尝试调用它，像这样：

  foo（std :: list< int>（），std :: vector< float>（））; 
  

MSVC2013说错误C2027：使用未定义类型'std :: tuple&容器:: value_type> 。

我尝试使用late return语法重写函数声明， >

有什么方法可以实现这个代码要做什么？

解决方案

p>您赢得了填写微软错误报告的权利连接 ...代码在clang和gcc上确定。

VS2013和gcc 4.7的解决方法：

  < typename T> 
使用ValueType = typename T :: value_type; 
 
 template< typename ... Containers> 
 std :: tuple< ValueType< Containers> ...> 
 foo（const Containers& ... args）{return {}; } 
  

Can you see anything wrong with this function declaration?

template<typename... Containers>
std::tuple<typename Containers::value_type...>
foo(const Containers &...args);

When I try to call it, like this:

foo(std::list<int>(), std::vector<float>());

MSVC2013 says error C2027: use of undefined type 'std::tuple<Containers::value_type>.

I tried rewriting the function declaration with the "late return" syntax and it made no difference.

Is there any way I can achieve what this code is trying to do?

解决方案

You won the right to fill a bug report on microsoft connect… The code is ok on clang and gcc.

A workaround on VS2013 and maybe gcc 4.7 :

template <typename T>
using ValueType = typename T::value_type;

template<typename... Containers>
std::tuple<ValueType<Containers>...>
foo( const Containers &...args ) { return {}; }

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