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？@H_403_5@
例如,如果我有@H_403_5@
C<int,int,float,std::string> c;
我打来电话@H_403_5@
c.foo(1,2,3.3,"qwerty")
那么c.maps应该等同于@H_403_5@
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)... });
}
};
