如何生成N型T的元组？

Question

我希望能够编写generate_tuple_type<int, 3>内部具有类型别名的内容type,std::tuple<int, int, int>在这种情况下.

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" );
}

我怎么能做到这一点？

Answer 1

相当简单的递归公式:

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;
};

实例

[更新]

好的,所以我只考虑适度的价值观N.以下公式更复杂,但对于大型参数而言也更快且编译器更少.

#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" );
}

实例

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