最優雅的方式來拆分C ++ TypeList

Question

在“現代”C ++中，我有一個類型列表：

template <typename... T> struct TypeList {};

我想根據謂詞拆分類型列表，例如std::is_floating_point 。 更確切地說，我的完整工作示例是：

#include <iostream>
#include <type_traits>

template <typename... T> struct TypeList {};

// SplitTypeList<> implementation defined at the end of this post...

template <typename T>
void printType()
{
  std::cout << "\n" << __PRETTY_FUNCTION__;
}

int main()
{
  struct A
  {
  };

  using typeList = TypeList<int, double, float, A, int>;

  using splited_typeList = SplitTypeList<std::is_floating_point, typeList>;

  using float_typeList = splited_typeList::predicate_is_true_typeList_type;
  using other_typeList = splited_typeList::predicate_is_false_typeList_type;

  printType<float_typeList>();
  printType<other_typeList>();
}

打印：

g++ -std=c++17 typeList.cpp -o typeList; ./typeList

void printType() [with T = TypeList<double, float>]
void printType() [with T = TypeList<int, main()::A, int>]

我的問題 ：您是否想過可能更短/更優雅的解決方案，只使用C ++（C ++ 17沒有問題）和STL？ （我不想使用像Boost，Hana這樣的輔助庫......）。

（ 我的動機 ：我不想錯過一兩行/超級優雅的解決方案，因為我將在其他地方廣泛使用此功能）

---

我目前的實施是：

namespace Details
{
  template <template <typename> class PREDICATE,
            typename... TYPELIST_PREDICATE_IS_TRUE,
            typename... TYPELIST_PREDICATE_IS_FALSE>
  constexpr auto splitTypeList(TypeList<TYPELIST_PREDICATE_IS_TRUE...>,
                               TypeList<TYPELIST_PREDICATE_IS_FALSE...>,
                               TypeList<>)
  {
    return std::make_pair(TypeList<TYPELIST_PREDICATE_IS_TRUE...>(),
                          TypeList<TYPELIST_PREDICATE_IS_FALSE...>());
  }

  template <template <typename> class PREDICATE,
            typename... TYPELIST_PREDICATE_IS_TRUE,
            typename... TYPELIST_PREDICATE_IS_FALSE,
            typename T,
            typename... TAIL>
  constexpr auto splitTypeList(TypeList<TYPELIST_PREDICATE_IS_TRUE...>,
                               TypeList<TYPELIST_PREDICATE_IS_FALSE...>,
                               TypeList<T, TAIL...>)
  {
    if constexpr (PREDICATE<T>::value)
    {
      return splitTypeList<PREDICATE>(
          TypeList<TYPELIST_PREDICATE_IS_TRUE..., T>(),
          TypeList<TYPELIST_PREDICATE_IS_FALSE...>(),
          TypeList<TAIL...>());
    }
    else
    {
      return splitTypeList<PREDICATE>(
          TypeList<TYPELIST_PREDICATE_IS_TRUE...>(),
          TypeList<TYPELIST_PREDICATE_IS_FALSE..., T>(),
          TypeList<TAIL...>());
    }
  }

  template <template <typename> class PREDICATE, typename... T>
  constexpr auto splitTypeList(TypeList<T...>)
  {
    return splitTypeList<PREDICATE>(
        TypeList<>(), TypeList<>(), TypeList<T...>());
  }
}

template <template <typename> class PREDICATE, typename TYPELIST>
struct SplitTypeList;

template <template <typename> class PREDICATE, typename... TAIL>
struct SplitTypeList<PREDICATE, TypeList<TAIL...>>
{
  using pair_type = decltype(
      Details::splitTypeList<PREDICATE>(std::declval<TypeList<TAIL...>>()));
  using predicate_is_true_typeList_type = typename pair_type::first_type;
  using predicate_is_false_typeList_type = typename pair_type::second_type;
};

---

僅僅是為了好奇，一個指向TypeList的歷史指針（Andrei Alexandrescu，2002年2月1日）： http ： //www.drdobbs.com/generic-programmingtypelists-and-applica/184403813

Answer 1

這樣的事情可能會更簡單/更短

template< bool, template<typename> class, class... Vs >
auto FilterImpl( TypeList<>, TypeList<Vs...> v ) { return v; }

template< bool Include, template<typename> class P, class T, class... Ts, class... Vs >
auto FilterImpl( TypeList<T,Ts...>, TypeList<Vs...> ) { return FilterImpl<Include,P>(
  TypeList<Ts...>{} ,
  std::conditional_t< Include == P<T>::value, TypeList<T,Vs...>, TypeList<Vs...> >{}
  ); }

template <template <typename> class PREDICATE, typename TYPELIST>
struct SplitTypeList
{
  using predicate_is_true_typeList_type = decltype(FilterImpl<true,PREDICATE>( TYPELIST{}, TypeList<>{} ));
  using predicate_is_false_typeList_type = decltype(FilterImpl<false,PREDICATE>( TYPELIST{}, TypeList<>{} ));
};

Answer 2

我不是說以下方式更好或更優雅。

這是不同的，這是我的方式。

僅使用可變參數模板類的特化; 沒有功能。

應該也適用於C ++ 11。

希望這個例子有幫助。

#include <tuple>
#include <type_traits>

template <template <typename> class Pred, typename>
struct PredValFirst : public std::false_type
 { };

template <template <typename> class Pred,
          template <typename...> class C,
          typename T0, typename ... Ts>
struct PredValFirst<Pred, C<T0, Ts...>> : public Pred<T0>
 { };


template <template <typename> class Pred, typename List,
          typename = std::tuple<>, typename = std::tuple<>,
          bool = PredValFirst<Pred, List>::value>
struct SplitTypeList;

template <template <typename> class Pred, template <typename...> class C,
          typename T0, typename ... Ts, typename ... Tt, typename Lf>
struct SplitTypeList<Pred, C<T0, Ts...>, std::tuple<Tt...>, Lf, true>
   : SplitTypeList<Pred, C<Ts...>, std::tuple<Tt..., T0>, Lf>
 { };

template <template <typename> class Pred, template <typename...> class C,
          typename T0, typename ... Ts, typename Lt, typename ... Tf>
struct SplitTypeList<Pred, C<T0, Ts...>, Lt, std::tuple<Tf...>, false>
   : SplitTypeList<Pred, C<Ts...>, Lt, std::tuple<Tf..., T0>>
 { };

template <template <typename> class Pred, template <typename...> class C,
          typename ... Tt, typename ... Tf>
struct SplitTypeList<Pred, C<>, std::tuple<Tt...>, std::tuple<Tf...>, false>
 {
   using types_true  = C<Tt...>;
   using types_false = C<Tf...>;
 };

template <typename...>
struct TypeList
 { };

struct A
 { };

int main ()
 {
   using typeList = TypeList<int, double, float, A, int>;

   using splited_typeList = SplitTypeList<std::is_floating_point, typeList>;

   using float_typeList = splited_typeList::types_true;
   using other_typeList = splited_typeList::types_false;

   static_assert( std::is_same<float_typeList,
                               TypeList<double, float>>{}, "!");
   static_assert( std::is_same<other_typeList,
                               TypeList<int, A, int>>{}, "!");
 }