如何从 struct/class 的元组中提取类型列表

Question

我想在 class 中使用静态方法来获取 std::tuple<T1, T2, T3,...> 形式的类型列表。我不想使用 std::tuple<...>，而是使用 <...>。如何实施示例 struct x 导致 Ts == <T1, T2, T3,...>

template<template <typename...> typename TL, typename... Ts>
struct x {
    static void test() {
        // expecting Ts == <int, char, float, double> (without std::tuple<>)
        std::cout << __PRETTY_FUNCTION__ << '\n';
    }
};

using types = std::tuple<int, char, float, double>;
x<types>::test();

见example on godbolt

Answer 1

在我看来，您正在寻找模板专业化。

有点像

// declaration (not definition) for a template struct x
// receiving a single template parameter 
template <typename>
struct x;

// definition for a x specialization when the template
// parameter is in the form TL<Ts...>
template<template<typename...> typename TL, typename... Ts>
struct x<TL<Ts...>> {
    static constexpr void test() {
        // you can use Ts... (and also TL, if you want) here
        std::cout << __PRETTY_FUNCTION__ << ' ' << sizeof...(Ts) << '\n';            
    }
};

如何从 struct/class 的元组中提取类型列表

How to extract type list from tuple for struct/class

c++

templates

variadic-templates

template-argument-deduction

c++20