Question

我正在尝试编写这样的函数

template<
        bool b, RT = std::conditional_t<b,
               std::tuple<int, int, int, int>,
               std::tuple<int, int, int, int, double, double, double, double>
        >
RT function()
{
    int i1, i2, i3, i4;

    if constexpr(b)
    {
        double i5, i6, i7, i8;
        return { i1, i2, i3, i4, i5, i6, i7, i8 };
    }
    else
    {
        return { i1, i2, i3, i4 };
    }
}

是否可以为元组创建模板化的typedef，以便简化上述功能

template<typename T, int N>
using tuple_t = std::tuple<T, T, ... N1 times>

template<typename T1, int N1, typename T2, int N2>
using tuple_t = std::tuple<T1, T1, ... N1 times, T2, T2, ... N2 times>

Answer 1

您可以使用返回类型推导，并用对make_tuple的调用来代替聚合初始化：

template<bool b>
auto function()
{
    int i1, i2, i3, i4;

    if constexpr(b)
    {
        double i5, i6, i7, i8;
        return std::make_tuple(i1, i2, i3, i4, i5, i6, i7, i8);
    }
    else
    {
        return std::make_tuple(i1, i2, i3, i4);
    }
}

如果您仍然需要返回类型，则只需做一个特征：

template <bool b>
using return_t = decltype(function<b>());

Answer 2

玩太晚了吗？

回答您的一般性问题

是否可以为元组创建模板化的typedef，以便简化上述功能

template<typename T, int N>
using tuple_t = std::tuple<T, T, ... N1 times>

template<typename T1, int N1, typename T2, int N2>
using tuple_t = std::tuple<T1, T1, ... N1 times, T2, T2, ... N2 times>

我提出了以下完整且可编译的C ++ 14示例

#include <tuple>
#include <utility>

template <typename T, std::size_t>
using get_type = T;

template <typename T, std::size_t ... Is>
constexpr std::tuple<get_type<T, Is>...>
   get_tuple_t (std::index_sequence<Is...>);

template <typename T, std::size_t N>
using tuple_t_1 = decltype(get_tuple_t<T>(std::make_index_sequence<N>{}));

template <typename T1, std::size_t N1, typename T2, std::size_t N2>
using tuple_t_2 = decltype(std::tuple_cat(
      std::declval<tuple_t_1<T1, N1>>(),
      std::declval<tuple_t_1<T2, N2>>()));

int main ()
 {
   using t1a = tuple_t_1<int, 4u>;
   using t1b = std::tuple<int, int, int, int>;
   using t2a = tuple_t_2<int, 4u, double, 4u>;
   using t2b = std::tuple<int, int, int, int, double, double, double, double>;

   static_assert( std::is_same<t1a, t1b>::value, "!" );
   static_assert( std::is_same<t2a, t2b>::value, "!" );
 }

构造模板化的元组类型

2 个答案: