如何返回映射到可变参数位置的类型

Question

我很喜欢学习元编程，并开始了解一些功能，但我无法调用 at Works，它需要一个索引来返回映射到可变位置的类型。

#include <iostream>
#include <type_traits>
#include <utility>

template <typename...>
struct type_list {};

template<typename T ,typename... Ns>
auto pop_front(type_list<T, Ns...> t) {
    return type_list<Ns...>{};
}

template<typename T ,typename... Ns>
auto front(type_list<T, Ns...> t) 
{
    return type_list<T>{};
}

template<typename T ,typename... Ns>
auto at(type_list<T, Ns...> t, size_t i)
{
    if (i) return at(pop_front(t), i-1);
    return front(t);
}

int main()
{
    type_list<int, bool, float> x;
    at(x,2);
}

Answer 1

如果我们将索引移动到模板参数中，使其成为编译时常量，那么我们可以利用std::tuple_element提供的索引来获取类型，例如

template<size_t Idx, typename... Ts>
auto at(type_list<Ts...>)
{
    return std::tuple_element_t<Idx, std::tuple<Ts...>>{};
}

然后我们可以这样称呼它

at<2>(x);

It should be noted that this approach requires all types in the list to be default constructible.

If you are only ever going to use at in a context like decltype(at<2>(x)) then you actually don't need to define the function and can instead use what I call a "meta function" like

template<size_t Idx, typename... Ts>
auto at(type_list<Ts...>t) -> std::tuple_element_t<Idx, std::tuple<Ts...>>;

and now you can't run this function, but you can use it an any unevaluated context, like in decltype(at<2>(x)), and the types in the list do not need to be default constructible.