使用尾随返回类型的条件重载可能？ [英] Conditional overloading with trailing-return-type possible?

问题描述

假设您尝试执行以下操作：

Suppose you attempt to do the following:

template</* args */>
typename std::enable_if< /*conditional*/ , /*type*/ >::type
static auto hope( /*args*/) -> decltype( /*return expr*/ )
{
}

可以结合条件包含/重载（ std :: enable_if ）与trailing-return-type（ auto ... - > decltype / code>）？

Is it possible to combine conditional inclusion/overloading (std::enable_if) with trailing-return-type (auto ... -> decltype())?

我不会在使用预处理器的解决方案有趣。我总是可以做像

I would not be interesting in solutions using the preprocessor. I can always do things like

#define RET(t) --> decltype(t) { return t; }

并扩展它以接受整个条件。相反，如果语言支持它，而不使用另一个trait为返回类型，即 ReturnType< A，B> :: type_t 或任何在函数体中使用，我感兴趣。

and extend it to take also the whole conditional. Instead I am interested if the language supports it without using another trait for the return type, i.e. ReturnType<A,B>::type_t or whatever is used in the function body.

推荐答案

trailing-return-type 与正常的return类型它在参数列表和cv- / ref-qualifiers之后指定。此外，它不一定需要 decltype ，正常类型也很好：

The trailing-return-type isn't much different from the normal return type, except that it's specified after the parameter list and cv-/ref-qualifiers. Also, it doesn't necessarily need decltype, a normal type is fine too:

auto answer() -> int{ return 42; }

现在你应该看到你的问题的答案是：

So by now you should see what the answer to your question is:

template<class T>
using Apply = typename T::type; // I don't like to spell this out

template</* args */>
static auto hope( /*args*/)
    -> Apply<std::enable_if</* condition */, decltype( /*return expr*/ )>>
{
}

虽然我个人更喜欢使用 decltype 和表达式SFINAE，只要条件可以表达为一个表达式（例如，你可以调用某个类型的对象上的函数）：

Though I personally prefer using just decltype and expression SFINAE, as long as the condition can be expressed as an expression (e.g., can you invoke a function on an object of a certain type):

template<class T>
static auto hope(T const& arg)
  -> decltype(arg.foo(), void())
{
  // ...
}

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