了解别名模板 [英] Understanding Alias Templates

问题描述

我问了一个问题，其中有多个代码引用：

I asked a question that has several references to the code:

template <typename...>
using void_t = void;

我相信我有一个一般误解别名模板：

I believe I have a generally misunderstand alias templates:

您是否在 enable_if_t 或 conditional_t 语句中传递别名模板？

Why wouldn't you just evaluate whatever template parameter you're passing into an alias template in an enable_if_t or conditional_t statement?

上面的代码只是一次对多个模板参数执行 enable_if_t ？

Is the code above just about doing an enable_if_t on multiple template parameters at once?

其次，我相信我对 void_t 的作用有一个具体的误解。 此评论说明C ++ 17标准定义 void_t 。这里是我不能得到的：

Secondly, I believe that I have a specific misunderstanding of the role of void_t. This comment states that the C++17 standard defines void_t. Here's what I don't get:

不是 void_t 只是一个任意名称？如果我还需要定义 template< typename ...>使用void_t = void; 无论我计划使用 void_t 什么是标准化任意名称的意义？

Isn't void_t just an arbitrary name? If I still have to define template <typename...> using void_t = void; wherever I plan to use void_t what's the point of standardizing an arbitrary name?

推荐答案

我不认为所示的例子真的显示了 void_t 显示一个用例，但是当您查看

I don't think the shown example really shows what void_t is good for as it only shows one use case, but when you look at

template<typename T>
struct has_to_string<T, 
    void_t<decltype(std::to_string(std::declval<T>()))>
    > 
: std::true_type { };

与

template<typename T>
struct has_to_string<T, 
    decltype(std::to_string(std::declval<T>()), void())
    > 
: std::true_type { };

对于此语句：

前一个版本更容易阅读， void_t 不需要 decltype 工作。

我认为可读性的优势很小，第二部分没有意义，当 decltype 不起作用，SFINAE会按预期插入。

I think the advantage in readability is quite small and the second part makes no sense, when decltype doesn't work, SFINAE kicks in as expected.

其中 void_t 有用的是提案中的一个：

One example where void_t is more useful is the one from the proposal:

// primary template handles types that have no nested ::type member
template< class, class = void_t<> >
struct has_type_member
: std::false_type { };

// specialization recognizes types that do have a nested ::type member
template< class T >
struct has_type_member<T, void_t<typename T::type>>
: std::true_type { }

正如你所看到的， void_t 以增加可读性，因为它现在匹配专业化。这不是绝对必要的，但我喜欢它。真正的力量来自于你想到的替代品。没有 void_t ，专业化现在更复杂：

As you can see, even the primary template uses void_t to increase the readability as it now matches the specialization. That is not strictly necessary, but I like it. The real power comes when you think about the alternatives. Without void_t, the specialization is now more complicated:

template< class T >
struct has_type_member<T, decltype(typename T::type, void())>
: std::true_type { }

不能作为 T :: type 命名一个类型，而不是一个表达式。因此，您需要

wouldn't work as T::type names a type, not an expression. You therefore need

template< class T >
struct has_type_member<T, decltype(std::declval<typename T::type>(), void())>
: std::true_type { }

整个表达式变得更长，更棘手可能会遇到你忘记处理的边缘情况。这是 void_t 真正有帮助的地方，其他用途只是一个小小的改进，它们提高了一致性。

The whole expression becomes longer, more tricky and it might suffer from edge-cases you forgot to handle. This is where void_t really helps, the other uses are then just a small improvement and they increase consistency.

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