20

如何通过给定的变体类型

using V = std::variant<bool, char, std::string, int, float, double, std::vector<int>>;

声明两种变体类型

using V1 = std::variant<bool, char, int, float, double>;
using V2 = std::variant<std::string, std::vector<int>>;

whereV1包括来自 的所有算术类型V并V2包括来自 的所有非算术类型V？

V可以是模板类的参数，例如：

template <class V>
struct TheAnswer
{
    using V1 = ?;
    using V2 = ?;
};

一般来说，标准可以是这样的constexpr变量：

template <class T>
constexpr bool filter;
4

3 回答 3

14

使用Boost.Mp11，这是一个简短的单行（一如既往）：

using V1 = mp_filter<std::is_arithmetic, V>;
using V2 = mp_remove_if<V, std::is_arithmetic>;

您还可以使用：

using V1 = mp_copy_if<V, std::is_arithmetic>;

使两者更加对称。

或者，

using P = mp_partition<V, std::is_arithmetic>;
using V1 = mp_first<P>;
using V2 = mp_second<P>;
于 2020-03-04T14:48:22.587 回答
6

如果出于某种原因您不想使用 Barry 的简短而合理的答案，那么这两者都不是（感谢@xskxzr删除了尴尬的“引导”专业化，并感谢 @ max66警告我不要使用空变体角落案例） ：

namespace detail {
    template <class V>
    struct convert_empty_variant {
        using type = V;
    };

    template <>
    struct convert_empty_variant<std::variant<>> {
        using type = std::variant<std::monostate>;
    };

    template <class V>
    using convert_empty_variant_t = typename convert_empty_variant<V>::type;

    template <class V1, class V2, template <class> class Predicate, class V>
    struct split_variant;

    template <class V1, class V2, template <class> class Predicate>
    struct split_variant<V1, V2, Predicate, std::variant<>> {
        using matching = convert_empty_variant_t<V1>;
        using non_matching = convert_empty_variant_t<V2>;
    };

    template <class... V1s, class... V2s, template <class> class Predicate, class Head, class... Tail>
    struct split_variant<std::variant<V1s...>, std::variant<V2s...>, Predicate, std::variant<Head, Tail...>>
    : std::conditional_t<
        Predicate<Head>::value,
        split_variant<std::variant<V1s..., Head>, std::variant<V2s...>, Predicate, std::variant<Tail...>>,
        split_variant<std::variant<V1s...>, std::variant<V2s..., Head>, Predicate, std::variant<Tail...>>
    > { };
}

template <class V, template <class> class Predicate>
using split_variant = detail::split_variant<std::variant<>, std::variant<>, Predicate, V>;

在 Wandbox 上现场观看

于 2020-03-04T14:59:08.023 回答
2

编辑鉴于空变体 ( std::variant<>) 格式不正确（根据cppreference）并且应该使用它std::variant<std::monostate>，我修改了答案（tuple2variant()为空元组添加了专门化）以支持V1or类型列表V2为空时的情况。

这有点decltype()谵妄，但是......如果你声明一个辅助过滤器几个函数，如下所示

template <bool B, typename T>
constexpr std::enable_if_t<B == std::is_arithmetic_v<T>, std::tuple<T>>
   filterArithm ();

template <bool B, typename T>
constexpr std::enable_if_t<B != std::is_arithmetic_v<T>, std::tuple<>>
   filterArithm ();

和一个元组到变体函数（专门针对空元组，以避免空std::variant）

std::variant<std::monostate> tuple2variant (std::tuple<> const &);

template <typename ... Ts>
std::variant<Ts...> tuple2variant (std::tuple<Ts...> const &);

你的班级只是（？）成为

template <typename ... Ts>
struct TheAnswer<std::variant<Ts...>>
 {
   using V1 = decltype(tuple2variant(std::declval<
                 decltype(std::tuple_cat( filterArithm<true, Ts>()... ))>()));
   using V2 = decltype(tuple2variant(std::declval<
                 decltype(std::tuple_cat( filterArithm<false, Ts>()... ))>()));
 };

如果你想要更通用的东西（如果你想std::arithmetic作为模板参数传递），你可以修改filterArithm()传递模板模板过滤器参数的函数F（重命名filterType()）

template <template <typename> class F, bool B, typename T>
constexpr std::enable_if_t<B == F<T>::value, std::tuple<T>>
   filterType ();

template <template <typename> class F, bool B, typename T>
constexpr std::enable_if_t<B != F<T>::value, std::tuple<>>
   filterType ();

班级TheAnswer成为

template <typename, template <typename> class>
struct TheAnswer;

template <typename ... Ts, template <typename> class F>
struct TheAnswer<std::variant<Ts...>, F>
 {
   using V1 = decltype(tuple2variant(std::declval<decltype(
                 std::tuple_cat( filterType<F, true, Ts>()... ))>()));
   using V2 = decltype(tuple2variant(std::declval<decltype(
                 std::tuple_cat( filterType<F, false, Ts>()... ))>()));
 };

并且TA声明也采取std::is_arithmetic

using TA = TheAnswer<std::variant<bool, char, std::string, int, float,
                                  double, std::vector<int>>,
                     std::is_arithmetic>;

以下是一个完整的编译示例，带有std::is_arithmeticas 参数和一个V2空案例

#include <tuple>
#include <string>
#include <vector>
#include <variant>
#include <type_traits>

std::variant<std::monostate> tuple2variant (std::tuple<> const &);

template <typename ... Ts>
std::variant<Ts...> tuple2variant (std::tuple<Ts...> const &);

template <template <typename> class F, bool B, typename T>
constexpr std::enable_if_t<B == F<T>::value, std::tuple<T>>
   filterType ();

template <template <typename> class F, bool B, typename T>
constexpr std::enable_if_t<B != F<T>::value, std::tuple<>>
   filterType ();

template <typename, template <typename> class>
struct TheAnswer;

template <typename ... Ts, template <typename> class F>
struct TheAnswer<std::variant<Ts...>, F>
 {
   using V1 = decltype(tuple2variant(std::declval<decltype(
                 std::tuple_cat( filterType<F, true, Ts>()... ))>()));
   using V2 = decltype(tuple2variant(std::declval<decltype(
                 std::tuple_cat( filterType<F, false, Ts>()... ))>()));
 };

int main ()
 {
   using TA = TheAnswer<std::variant<bool, char, std::string, int, float,
                                     double, std::vector<int>>,
                        std::is_arithmetic>;
   using TB = TheAnswer<std::variant<bool, char, int, float, double>,
                        std::is_arithmetic>;

   using VA1 = std::variant<bool, char, int, float, double>;
   using VA2 = std::variant<std::string, std::vector<int>>;
   using VB1 = VA1;
   using VB2 = std::variant<std::monostate>;

   static_assert( std::is_same_v<VA1, TA::V1> );
   static_assert( std::is_same_v<VA2, TA::V2> );
   static_assert( std::is_same_v<VB1, TB::V1> );
   static_assert( std::is_same_v<VB2, TB::V2> );
 }
于 2020-03-04T14:59:04.177 回答