具有专用功能的分体功能类型 [英] Pull Apart Function Type With Specialized Function

问题描述

此问题的答案使用类模板来区分函数类型:

The answer to this question picks apart a function type using a class template:

template <typename T>
struct function_args {};

template <typename R, typename... Args>
struct function_args<R(Args...)> {
    using type = tuple<Args...>;
};

template <typename T>
using decltypeargs = typename function_args<T>::type;

当我研究这里所做的事情时，我试图重写function_args.我尝试使用一个函数来执行此操作，以消除对decltypeargs模板的需要.但是发现自己陷入了语法错误:

As I studied what was being done here I tried to rewrite function_args. I attempted to do this using a function so as to eliminate the need for the decltypeargs template. But found myself mired in improper syntax:

template <typename T>
tuple<> myTry();

template <typename Ret, typename... Args>
tuple<Args...> myTry<Ret(Args...)>();

我的希望是调用decltype(myTry<decltype(foo)>())来获取tuple类型，而不必调用decltypeargs<decltype(foo)>.有没有办法通过函数声明来做到这一点?

My hope had been to call decltype(myTry<decltype(foo)>()) to get the tuple type instead of having to call decltypeargs<decltype(foo)>. Is there a way to do this with a function declaration?

推荐答案

//------------------------ Machinery:
#include <tuple>

template< class Ret, class... Args >
std::tuple<Args...> m( Ret(Args...) );

//------------------------- Example:
#include <iostream>
#include <typeinfo>

void foo( double );

using namespace std;
auto main()
    -> int
{
    using Args_tuple = decltype( m( foo ) );
    cout << typeid( Args_tuple ).name() << endl;
}

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