在编译时确定类型是否为 STL 容器

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

问题描述

我想编写一个模板，用于在编译时确定类型是否为 stl 容器.  

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

我有以下代码:

struct is_cont{};
struct not_cont{};

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

但我不确定如何为 std::vector、deque、set 等创建必要的特化...

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

推荐答案

首先，你定义你的主模板，它有一个默认情况下为 false 的成员:

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;
};

然后您将为您的容器类型定义部分特化，其值为 true:

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;
};

然后对于要检查的类型 X，像这样使用它

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

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

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