问题

我试图使用类型特性与函数对象，以下是代码:

struct fun {
  template <class T>
  constexpr auto operator()(T a) const  {
    return 0;
  }
};

template<class T> 
struct types{};

我好奇types<fun(int)>的官方名称是什么，可能是获取函数调用运算符的类型吗？

英文:

I'm trying to use type trait with functor, below is the code:

struct fun {
  template &lt;class T&gt;
  constexpr auto operator()(T a) const  {
    return 0;
  }
};

template&lt;class T&gt; 
struct types{};

I'm curious about what's the official name of the usage types<fun(int)>, probably getting the type of the function call operator?

答案1

得分: 1

fun(int) 是一个接受 int 并返回 fun 类型的函数类型。

它与 fun::operator() 无关。

英文:

fun(int) is the type of a function that takes an int and returns a fun.

It is unrelated to fun::operator().

得分: 1

"types<fun(int)>"被称为模板标识：

模板标识由两个部分组成：

英文:

> what's the official name of the usage types<fun(int)>

It (types<fun(int)>) is called a template-id:

> <h4>template-id</h4>
>
> > template-name <parameter-list > >

As you can see the template-id consists of 2 parts:

template-name: which is types in your example.
parameter-list: which is fun(int) which itself is a function type in your example.

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