如何从另一个仅静态选择满足某种类型规则的索引的元组实例创建一个元组实例？

Question

假设我有一个基本类型和一些派生类型：

struct base {};
struct d1 : base {};
struct d2 : base {};

假设我创建了一个元组，例如：

std::tuple<double, d1, d2, int> t(1.0, d1(), d2(), 1);

是否可以使用元/运行时编程来生成：

template<typename T>
using my_predicate = std::is_base_of<base, T>;

std::tuple<d1, d2> t_derived = filter(t, my_predicate);

我已经看过有关堆栈交换的答案，该答案可以让我定义新类型，但是我还没有弄清楚如何从原始元组中选择运行时组件。 堆栈交换帖子在这里：

https://codereview.stackexchange.com/questions/115740/filtering-variadic-template-arguments

Answer 1

您可以使用以下内容：

template <typename TupleOfIntegralConstant>
struct as_sequence;

template <typename ... Ts>
struct as_sequence<std::tuple<Ts...>> {
    using type = std::index_sequence<Ts::value...>;
};

template <template <typename> class Pred, typename Tuple, typename Seq>
struct make_filtered_sequence;

template <template <typename> class Pred, typename Tuple, std::size_t ... Is>
struct make_filtered_sequence<Pred, Tuple, std::index_sequence<Is...>>
{
    using type = typename as_sequence<decltype(std::tuple_cat(
           std::conditional_t<
             Pred<std::tuple_element_t<Is, Tuple>>::value,
             std::tuple<std::integral_constant<std::size_t, Is>>,
             std::tuple<>>{}...))>::type;
};

template <typename Tuple, std::size_t ... Is>
auto filter_impl(const Tuple& t, std::index_sequence<Is...>)
-> std::tuple<std::tuple_element_t<Is, Tuple>...>
{
    return {std::get<Is>(t)...};
}

template <template <typename> class Pred, typename Tuple>
auto filter(const Tuple& t)
{
    using filtered_seq = typename make_filtered_sequence<Pred, Tuple, std::make_index_sequence<std::tuple_size<Tuple>::value>>::type;

    return filter_impl(t, filtered_seq());
}

想法是创建一个过滤后的序列，然后对该序列使用std::tuple_element_t和std::get 。