如何限制模板中的类型名 [英] How to restrict typenames in template

问题描述

有一个可以接受4种不同类型的函数。它们的实现非常相似。

  template<类型名T> 
 void foo（std :: string&& name，T value）{
 // ... 
} 
  

typename T 必须是4种预定义类型之一。

是否可以用C ++编写？

解决方案

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< type_traits> 让你将逻辑推广到类模板。这是如何工作的我们取模板参数 T 和参数包 Ts ，我们开始检查 T 与类型列表的 Head 相同。如果找到匹配，我们继承 std :: true_type ，我们就完成了。如果没有找到匹配，我们弹出头部并递归地实例化 Ts 的尾部的相同模板。最终，如果没有找到匹配，参数包大小将下降到零，编译器将实例化继承自 std :: false_type 的基本模板类。请检查此视频，以便更好地进行，并由Walter E. Brown先生进行深入解释。

 模板< class T，class ...> struct is_any_of：std :: false_type {}; 
 template< class T，class Head，class ... Tail> 
 struct is_any_of< T，Head，Tail ...> ;: std :: conditional_t< 
 std :: is_same< T，Head> :: value，
 std :: true_type，
 is_any_of< T，Tail ...> 
 {}; 
  

现在我们可以使用 enable_if 。

  include< type_traits> 
 #include< string> 
 
 template< 
 class T，
 class = std :: enable_if_t< is_any_of< T，int，float，unsigned，double> :: value> 
> 
 void foo（std :: string&& str，T value）{} 
 
 
 int main（）
 {
 foo :: string {hello}，3）; 
 foo（std :: string {world}，'0'）; //编译时错误
} 
  

SFANIE是一种语言功能，

标准库组件std :: enable_if允许使用或滥用某些人说要实现您的要求

<用于创建替换失败，以便基于在编译时评估的条件来启用或禁用特定的重载。
FYI http://en.cppreference.com/w/cpp/language/ sfinae 。

请注意， std :: conditional_t<> std :: conditional<> :: type 和 std :: enable_if_t< code> std :: enable_if<> :: type 。你可以简单地在代码中替换这些，但应该在 enable_if 之前放置 typename

Have a function which can accept 4 different types. And their implementations are very similar.

template< typename T >
void foo( std::string &&name, T value ){
    //...
}

typename T must be one of 4 predefined types. Other typenames is not acceptable, and should be restricted at compile-time.
Is it possible to write in C++?

解决方案

<type_traits> let you generalize your logic into class template. How this works is we take template parameter T and parameter pack of Ts and we start checking if T is same to Head of the list of types. If match is found, we inherit from std::true_type and we are done. If no match is found, we pop the head and recursively instantiate same template with tail of Ts. Eventually, if no match is found at all, parameter pack size will drop to zero and the compiler will instantiate base template class that inherit from std::false_type. Please check this video for much better and in dept explanation by Mr. Walter E. Brown.

template<class T, class...> struct is_any_of: std::false_type{};
template<class T, class Head, class... Tail> 
struct is_any_of<T, Head, Tail...>: std::conditional_t<
  std::is_same<T, Head>::value, 
  std::true_type,
  is_any_of<T, Tail...>>
{};

Now we can SFINAE over, using enable_if in almost English wording.

#include <type_traits>
#include <string>

template<
  class T,
  class = std::enable_if_t<is_any_of<T, int, float, unsigned, double>::value>
>
void foo(std::string &&str, T value) {}


int main()
{
  foo(std::string{"hello"}, 3);
  foo(std::string{"world"}, '0'); //compile-time error
}

SFANIE is a language feature, a tool, that's used or abused some say to achieve what you ask for,

The standard library component std::enable_if allows for creating a substitution failure in order to enable or disable particular overloads based on a condition evaluated at compile time. FYI http://en.cppreference.com/w/cpp/language/sfinae.

Note that std::conditional_t<> and std::enable_if_t<> are shorthanded from std::conditional<>::type and std::enable_if<>::type respectively. You could simple replace these in code, but should put typename keyword before enable_if then.

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