我有以下类：
- template<typename... Tkeys>
- class C
- {
- public:
- std::tuple<std::unordered_map<Tkeys,int>... > maps;
- // Not real function:
- void foo(Tkeys... keys) {
- maps[keys] = 1;
- }
- };
我如何实现foo,以便它分配给映射中的每个std :: map,使用匹配的key来调用它？
例如,如果我有
- C<int,int,float,std::string> c;
我打来电话
- 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);
解决方法
- #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)... });
- }
- };
