c++ - std::tuple，通过继承类型获取项目

Question

在 c++11 中，我有非常简洁且有效的代码用于按类型提取 std::tuple 项(据我所知，此功能甚至是 placed 到 c++14 STL)

现在我面临着通过继承类规范选择项目的任务

 struct A
 {
     int a;
 };
 struct B : public A
 {
     int b;
 };

 ...
 auto tval = std::make_tuple(1, B());
 //now I would like to reference items as following:
 tuple_ref_by_inheritance<A>(tval).a = 5; //Access to B instance by parent A

以下代码是我不成功的尝试:

    template< class T, class Tuple >
    struct tuple_ref_index;

    // recursive case
    template<class T, class Head, class... Tail >
    struct tuple_ref_index<T, std::tuple<Head, Tail...> >  
    { 
        enum { value = tuple_ref_index<T, std::tuple<Tail...>>::value + 1 };
    };

    template<class T, class Head, class... Tail >
    struct tuple_ref_index<T, std::tuple<Head, Tail...> >  
    { 
        const static typename std::enable_if<
              std::is_same<T, Head>::value>::type* _= nullptr;
        enum { value = 0 };
    };

    template <class T, class Tuple>
    inline T& tuple_ref_by_inheritance(Tuple& tuple)
    {
        return std::get< tuple_ref_index<T, Tuple>::value >(tuple);
    }

Answer 1

#include <type_traits>
#include <utility>
#include <cstddef>
#include <tuple>

template <typename Base, typename Tuple, std::size_t I = 0>
struct tuple_ref_index;

template <typename Base, typename Head, typename... Tail, std::size_t I>
struct tuple_ref_index<Base, std::tuple<Head, Tail...>, I>  
    : std::conditional<std::is_base_of<Base, Head>::value
                     , std::integral_constant<std::size_t, I>
                     , tuple_ref_index<Base, std::tuple<Tail...>, I+1>
                     >::type
{
};

template <typename Base, typename Tuple>
auto tuple_ref_by_inheritance(Tuple&& tuple)
    -> decltype(std::get<tuple_ref_index<Base, typename std::decay<Tuple>::type>::value>(std::forward<Tuple>(tuple)))
{
    return std::get<tuple_ref_index<Base, typename std::decay<Tuple>::type>::value>(std::forward<Tuple>(tuple));
}

DEMO