c++ - 压缩 `std::tuple` 和可变参数

Question

我有以下类(class):

template<typename... Tkeys>
class C
{
public:
    std::tuple<std::unordered_map<Tkeys, int>... > maps;

    // Not real function:
    void foo(Tkeys... keys) {
        maps[keys] = 1;
    }
};

我将如何实现 foo 以便它分配给 maps 中的每个 std::map 并使用它匹配的键调用？

例如，如果我有

C<int, int, float, std::string> c;

我打电话

c.foo(1, 2, 3.3, "qwerty")

那么c.maps应该等同于

m1 = std::map<int, int>()
m1[1] = 1;
m2 = std::map<int, int>()
m2[2] = 1;
m3 = std::map<float, int>()
m3[3.3] = 1;
m4 = std::map<std::string, int>()
m4["qwerty"] = 1;
c.maps = std::make_tuple(m1, m2, m3, m4);

Answer 1

#include <unordered_map>
#include <utility>
#include <tuple>
#include <cstddef>

template <typename... Tkeys>
class C
{
public:
    std::tuple<std::unordered_map<Tkeys, int>... > maps;

    template <typename... Args>
    void foo(Args&&... keys)
    {
        foo_impl(std::make_index_sequence<sizeof...(Args)>{}, std::forward<Args>(keys)...);
    }

private:
    template <typename... Args, std::size_t... Is>
    void foo_impl(std::index_sequence<Is...>, Args&&... keys)
    {
        using expand = int[];
        static_cast<void>(expand{ 0, (
            std::get<Is>(maps)[std::forward<Args>(keys)] = 1
        , void(), 0)... });
    }
};

DEMO