将 std::integer_sequence 作为模板参数传递给元函数

我怎样才能通过std::integer_sequence作为元函数的模板参数（即不是函数模板）？

给定例如以下用例（但不限于此）：

我想使用整数序列来删除最后一个N来自参数包的类型。我以为我可以用selector from 这个问题 https://stackoverflow.com/a/19759105/678093，但我无法将整数序列传递给这个元函数。

#include <tuple>
#include <utility>

template <typename T, std::size_t... Is>
struct selector
{
    using type = std::tuple<typename std::tuple_element<Is, T>::type...>;
};

template <std::size_t N, typename... Ts>
struct remove_last_n
{
    using Indices = std::make_index_sequence<sizeof...(Ts)-N>;  
    using type = typename selector<std::tuple<Ts...>, Indices>::type; // fails
};

int main()
{
    using X = remove_last_n<2, int, char, bool, int>::type;
    static_assert(std::is_same<X, std::tuple<int, char>>::value, "types do not match");
}

编译器错误

main.cpp:15:55: error: template argument for non-type template parameter must be an expression

using type = typename selector<std::tuple<Ts...>, Indices>::type; // fails

                                                  ^~~~~~~

main.cpp:5:38: note: template parameter is declared here

template <typename T, std::size_t... Is>

live on coliru http://coliru.stacked-crooked.com/a/d8d1efd56d529309

我将如何传递整数序列？

你需要（部分）专业化selector这样就可以推导出指数std::index_sequence:

#include <tuple>
#include <utility>
#include <type_traits>

template <typename T, typename U>
struct selector;

template <typename T, std::size_t... Is>
struct selector<T, std::index_sequence<Is...>>
{
    using type = std::tuple<typename std::tuple_element<Is, T>::type...>;
};

template <std::size_t N, typename... Ts>
struct remove_last_n
{
    using Indices = std::make_index_sequence<sizeof...(Ts)-N>;  
    using type = typename selector<std::tuple<Ts...>, Indices>::type;
};

int main()
{
    using X = remove_last_n<2, int, char, bool, int>::type;
    static_assert(std::is_same<X, std::tuple<int, char>>::value, "types do not match");
}

DEMO http://coliru.stacked-crooked.com/a/2a87c38513df2eb9