按给定条件拆分给定 std::variant 类型

Question

如何按给定的变体类型

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>>;

其中 V1 包括 V 中的所有算术类型，V2 包括 V?

中的所有非算术类型

V可以作为模板的参数class，例如：

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

一般来说，条件可以是一个 constexpr 变量，如下所示：

template <class T>
constexpr bool filter;

Answer 1

对于 Boost.Mp11，这是一个简短的一行（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>;

让两者更加对称

或者，

using P = mp_partition<V, std::is_arithmetic>;
using V1 = mp_first<P>;
using V2 = mp_second<P>;

Answer 2

EDIT 鉴于空变体 (std::variant<>) 格式不正确（根据 cppreference），应该使用 std::variant<std::monostate> 代替，我修改了答案（为空元组添加了 tuple2variant() 专业化）以支持 V1 或 V2 的类型列表为空的情况。

有点decltype() 精神错乱但是...如果你声明一个辅助过滤器函数如下

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 class 只是 (?) 变成

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（如果你想将 std::arithmetic 作为模板参数传递），你可以修改 filterArithm() 函数传递一个模板模板过滤器参数 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 ();

把TheAnswer变成class

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>()... ))>()));
 };

和 TA 声明也采用 std::is_arithmetic

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

下面是一个完整的编译例子，参数是std::is_arithmetic，空case是V2

#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> );
 }

Answer 3

如果出于某种原因你不想使用 Barry 的简短而合理的答案，这里有一个两者都不是的答案（感谢@xskxzr for removing the awkward "bootstrap" specialization, and to @max66 警告我不要使用空变体极端情况）：

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>;

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按给定条件拆分给定 std::variant 类型

Split a given std::variant type by a given criteria

c++

c++17

std-variant