在编译时确定类型是否为STL容器 [英] Determine if a type is an STL container at compile time

问题描述

我想编写一个模板，它将在编译时确定类型是否为stl容器。

我有以下代码： / p>

  struct is_cont {}; 
 struct not_cont {}; 
 
 template< typename T> 
 struct is_cont {typedef not_cont result_t; }; 
  

但我不知道如何为 std创建必要的专业化： ：vector< T，Alloc> ;, deque< T，Alloc> ;, set< T，Alloc，Comp> etc ...

首先，定义您的主要模板，它将有一个在默认情况下为false的成员：

  template< typename T> 
 struct is_cont {
 static const bool value = false; 
}; 
  

然后您将为您的容器类型定义具有值为true的部分特化：

  template< typename T，typename Alloc> 
 struct is_cont< std :: vector< T，Alloc> > {
 static const bool value = true; 
}; 
  

然后对于要检查的类型X，使用

  if（is_cont< X> :: value）{...} 
  

I would like to write a template that will determine if a type is an stl container at compile time.

I've got the following bit of code:

  struct is_cont{};
  struct not_cont{};

  template <typename T>
  struct is_cont { typedef not_cont result_t; };

but I'm not sure how to create the necessary specializations for std::vector<T,Alloc>, deque<T,Alloc>, set<T,Alloc,Comp> etc...

解决方案

First, you define your primary template, which will have a member which is false in the default case:

template <typename T>
struct is_cont {
  static const bool value = false;
};

Then you will define partial specializations for your container types which have a value of true instead:

template <typename T,typename Alloc>
struct is_cont<std::vector<T,Alloc> > {
  static const bool value = true;
};

Then for a type X that you want to check, use it like

if (is_cont<X>::value) { ... } 

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