为什么type_traits实现与专门的模板结构而不是constexpr？ [英] Why are type_traits implemented with specialized template structs instead of constexpr?

问题描述

有什么原因为什么标准将它们指定为模板 struct s而不是简单布尔 constexpr ？

在另一个问题可能会回答一个好的回答主要问题，如何 enable_if stuff非结构化版本？

解决方案

一个原因是 constexpr 函数不能提供嵌套的类型成员，这在某些元编程情况下非常有用。

很清楚，我不是只谈到产生类型的显然不能做<$ c $转换特性（例如 make_unsigned c> constexpr 函数。 所有类型traits 提供这样的嵌套类型成员，甚至一元类型特性 em>。例如 is_void< int> :: type 是 false_type 。

当然，这可以用 std :: integral_constant< bool，the_constexpr_function_version_of_some_trait< T>（）> 来解决，但是它不会那么实用。

在任何情况下，如果你真的想要类似函数的语法，这是可能的。您只需使用traits构造函数，并利用 intl_constant ：

  static_assert（std :: is_void< void>（），void is void;谁会有thunk？）; 
 

对于转换traits ，您可以使用模板别名来获取该语法：

  template< bool Condition，typename T = void> 
使用enable_if = typename std :: enable_if< Condition，T> :: type; 
 //用法：
 // template< typename T> enable_if< is_void< T>（），int> F（）; 
 // 
 // make_unsigned< T> X; 
  




Is there any reason why the standard specifies them as template structs instead of simple boolean constexpr?

In an additional question that will probably be answered in a good answer to the main question, how would one do enable_if stuff with the non-struct versions?

解决方案

One reason is that constexpr functions can't provide a nested type member, which is useful in some meta-programming situations.

To make it clear, I'm not talking only of transformation traits (like make_unsigned) that produce types and obviously can't be made constexpr functions. All type traits provide such a nested type member, even unary type traits and binary type traits. For example is_void<int>::type is false_type.

Of course, this could be worked around with std::integral_constant<bool, the_constexpr_function_version_of_some_trait<T>()>, but it wouldn't be as practical.

In any case, if you really want function-like syntax, that is already possible. You can just use the traits constructor and take advantage of the constexpr implicit conversion in integral_constant:

static_assert(std::is_void<void>(), "void is void; who would have thunk?");

For transformation traits you can use a template alias to obtain something close to that syntax:

template <bool Condition, typename T = void>
using enable_if = typename std::enable_if<Condition, T>::type;
// usage:
// template <typename T> enable_if<is_void<T>(), int> f();
//
// make_unsigned<T> x;

这篇关于为什么type_traits实现与专门的模板结构而不是constexpr？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！