给定这些类型：

template<Ratio r, Symbol s>
struct base_unit {
    using ratio = r;
    using symbol = s;
};

template <BaseUnit... baseUnits>
struct derived_unit {
    using units = std::tuple<baseUnits...>;
};

例如，如果我有一个如下类型：

template <typename T>
struct Computations {
    using collection_ratios = /* */
    static constexpr ratios_product = /* */
}

我想知道我该如何：

• 将给定derived_unit的所有比率存储在容器中，如std：：array〈T，N〉，并使用折叠表达式（collection_ratios）
• 存储derived_unit的units成员中的每个元素相乘的结果，并将其存储在具有折叠表达式（ratios_product）的变量中

编辑：
其中T是derived_unit的特殊化，例如：

struct MetersPerSecond :
   public speed,
   public derived_unit<         
        base_unit<Kilo, m>,
        base_unit<Root, s>
   >
{};

编辑二：

template <typename T>
concept RatioV = (std::is_integral_v<T> || std::is_floating_point_v<T>)
    && !std::is_same_v<T, char>;

consteval double getFactor(double base, double exponent);

template <RatioV T = short, T Base = 10, T Exponent = 0>
struct ratio {
    static constexpr T base = Base;
    static constexpr T exponent = Exponent;
    static constexpr T value = getFactor(base, exponent);
};

consteval double getFactor(double base, double exponent) {
    double result = 1;
    for (int i = 0; i < exponent; i++)
        result *= base;
    return result;
}

using Yocto = ratio<short, 10, -24>;
using Zepto = ratio<short, 10, -21>;
using Atto = ratio<short, 10, -18>;
using Femto = ratio<short, 10, -15>;
using Pico = ratio<short, 10, -12>;
using Nano = ratio<short, 10, -9>;
using Micro = ratio<short, 10, -6>;
using Milli = ratio<short, 10, -3>;
using Centi = ratio<short, 10, -2>;
using Deci = ratio<short, 10, -1>;
using Root = ratio<short, 10, 0>;
using Deca = ratio<short, 10, 1>;
using Hecto = ratio<short, 10, 2>;
using Kilo = ratio<short, 10, 3>;
using Mega = ratio<short, 10, 6>;
using Giga = ratio<short, 10, 9>;
using Tera = ratio<short, 10, 12>;
using Peta = ratio<short, 10, 15>;
using Exa = ratio<short, 10, 18>;
using Zetta = ratio<short, 10, 21>;
using Yotta = ratio<short, 10, 24>;