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c++templatesc++11stdtuple

How to get N-th type from a tuple?


I want to make a template where I can input an index and it will give me the type at that index. I know I can do this with decltype(std::get<N>(tup)) but I would like to implement this myself. For example, I would like to do this,

typename get<N, std::tuple<int, bool, std::string>>::type;

...and it will give me the type at position N - 1 (because arrays indexed starting from 0). How can I do this? Thanks.


Solution

  • You can use a class template and partial specializations to do what you want. (Note that std::tuple_element does almost the same like the other answer says):

    #include <tuple>
    #include <type_traits>
    
    template <int N, typename... Ts>
    struct get;
    
    template <int N, typename T, typename... Ts>
    struct get<N, std::tuple<T, Ts...>>
    {
        using type = typename get<N - 1, std::tuple<Ts...>>::type;
    };
    
    template <typename T, typename... Ts>
    struct get<0, std::tuple<T, Ts...>>
    {
        using type = T;
    };
    
    int main()
    {
        using var = std::tuple<int, bool, std::string>;
        using type = get<2, var>::type;
    
        static_assert(std::is_same<type, std::string>::value, ""); // works
    }