c++ - 如何在C++模板元编程中返回不止一种类型？

Question

我有一个简单的模板元程序，可以将许多std::ratio转换为尽可能少的ratio。如果没有溢出(并且确实如此)，则必须返回1 ratio；如果有溢出，则必须返回2或更多。有可能达到目标吗？

template<class R, class ...Rs> struct unite_ratio;

template<intmax_t Num, intmax_t Den> 
struct unite_ratio<std::ratio<Num, Den>>
{
    using type = std::ratio<Num, Den>;
};

template<
    intmax_t Num1, intmax_t Den1, 
    intmax_t Num2, intmax_t Den2,
    class ...Rs
> struct unite_ratio<std::ratio<Num1, Den1>, std::ratio<Num2, Den2>, Rs...>
{
    // here might be an overflow
    static constexpr intmax_t num = 
        std::__static_sign<Den1>::value * std::__static_sign<Den2>::value *
        (Num1 / std::__static_gcd<Num1, Den2>::value) * 
        (Num2 / std::__static_gcd<Num2, Den1>::value);

    // Here might be an overflow as well
    static constexpr intmax_t den =
        (std::__static_abs<Den1>::value / std::__static_gcd<Num2, Den1>::value) *
        (std::__static_abs<Den2>::value / std::__static_gcd<Num1, Den2>::value);

    // returns only one type
    using type = unite_ratio<std::ratio<num, den>, Rs...>;
};

Answer 1

How to return more than one type in C++ template meta-programming?

带有类型列表。它只是一个模板类，其可变参数的模板参数列表为类型列表:

template<class...T>
struct types{};