Question

我希望能够编写generate_tuple_type<int, 3>内部有一个类型别名type，在这种情况下会std::tuple<int, int, int>。

一些示例用法：

int main()
{
    using gen_tuple_t = generate_tuple_type<int, 3>::type;
    using hand_tuple_t = std::tuple<int, int, int>;
    static_assert( std::is_same<gen_tuple_t, hand_tuple_t>::value, "different types" );
}

我该如何做到这一点？

Answer 1

相当直接的递归表达式：

template<typename T, unsigned N, typename... REST>
struct generate_tuple_type
{
 typedef typename generate_tuple_type<T, N-1, T, REST...>::type type;
};

template<typename T, typename... REST>
struct generate_tuple_type<T, 0, REST...>
{
  typedef std::tuple<REST...> type;
};

Live example

[更新]

好的，所以我只考虑N的适度值。以下公式更复杂，但也更快，更少编译器破坏大型参数。

#include <tuple>

template<typename /*LEFT_TUPLE*/, typename /*RIGHT_TUPLE*/>
struct join_tuples
{
};

template<typename... LEFT, typename... RIGHT>
struct join_tuples<std::tuple<LEFT...>, std::tuple<RIGHT...>>
{
  typedef std::tuple<LEFT..., RIGHT...> type;
};

template<typename T, unsigned N>
struct generate_tuple_type
{
  typedef typename generate_tuple_type<T, N/2>::type left;
  typedef typename generate_tuple_type<T, N/2 + N%2>::type right;
  typedef typename join_tuples<left, right>::type type;
};

template<typename T>
struct generate_tuple_type<T, 1>
{
  typedef std::tuple<T> type;
};

template<typename T>
struct generate_tuple_type<T, 0>
{
  typedef std::tuple<> type;
};

int main()
{
  using gen_tuple_t = generate_tuple_type<int, 30000>::type;
  static_assert( std::tuple_size<gen_tuple_t>::value == 30000, "wrong size" );
}

Live example

此版本最多执行2 * log（N）+1个模板实例化，假设您的编译器会记住它们。证明留给读者练习。

Answer 2

您可以使用std::make_index_sequence为您提供足够长的包装，然后将其包装为您需要的类型。不需要递归：

template <typename T, size_t N>
class generate_tuple_type {
    template <typename = std::make_index_sequence<N>>
    struct impl;

    template <size_t... Is>
    struct impl<std::index_sequence<Is...>> {
        template <size_t >
        using wrap = T;

        using type = std::tuple<wrap<Is>...>;
    };

public:
    using type = typename impl<>::type;
};

Answer 3

查看此链接的底部以获取示例：

http://en.cppreference.com/w/cpp/utility/integer_sequence

您需要做一些工作来将结果元组封装为类型别名，但这里的关键构造是std::integer_sequence和朋友。

如何生成N型T的元组？

3 个答案: