键入返回所选类型元组的函数

Question

我已经实现了一个类型函数Tuple,可以将My_enum值列表转换std::tuple为相应的类型:

#include <tuple>

enum My_enum{ t_int, t_double };

// Bind_type is a type function that given a My_enum returns the corresponding type
template<My_enum E> struct Bind_type;
template<> struct Bind_type<t_int>{ using type = int; };
template<> struct Bind_type<t_double>{ using type = double; };

// Tuple is a type function that given a template value parameter pack of My_enums returns a std::tuple of correspondig types
template<My_enum First, My_enum... Others>
struct Tuple {
    using type = decltype(std::tuple_cat(
                typename Tuple<First>::type{},
                typename Tuple<Others...>::type{}
            ));
};

template<>
struct Tuple<t_int> {
    using type = std::tuple<Bind_type<t_int>::type>;
};
template<>
struct Tuple<t_double> {
    using type = std::tuple<Bind_type<t_double>::type>;
};

我希望能够Tuple一次性声明递归基本情况,因为Tuple只要我添加或删除值My_enum,我就不想手动管理特化,因为它容易出错(而且很无聊).我试过了:

template<My_enum E>
struct Tuple {
    using type = std::tuple<Bind_type<E>::type>;
};

但这不是可变版本的有效专业化.

我的问题是:Tuple当它只有一个模板值参数时,是否有一种智能的方法来声明特化？

Answer 1

只需将参数包直接扩展为std::tuple:

template<My_enum... Enums>
struct Tuple {
    using type = std::tuple<typename Bind_type<Enums>::type...>;
};

要更直接地回答您的问题,您可以声明一个可变参数主模板,然后编写两个特化:当至少有两个参数时,以及何时只有一个:

//primary template, takes any number of My_enums
template <My_enum... Enums>
struct Tuple {
    //this case will be chosen if we instantiate a Tuple with no args
    using type = std::tuple<>;
}

//specialization for when there are at least two arguments
template<My_enum First, My_enum Second, My_enum... Others>
struct Tuple<First,Second,Others...> {
    using type = decltype(std::tuple_cat(
                typename Tuple<First>::type{},
                typename Tuple<Second,Others...>::type{}
            ));
};

//base case, only one argument left
template<My_enum Last>
struct Tuple<Last> {
    using type = std::tuple<typename Bind_type<Last>::type>;
};