如何用强盗转换这个类型列表?
How to transform this list of types with brigand?
我有以下类型列表:
using ComponentList = brigand::list<TransformComponent, ObjectComponent, BodyComponent>
如何将之前的列表转换成如下所示的新列表:
using ComponentHandleList = brigand::list<entityx::ComponentHandle<TransformComponent>, entityx::ComponentHandle<ObjectComponent>, entityx::ComponentHandle<BodyComponent>>;
我基本上想生成一个新列表,其中包含前一个列表中每个元素的包装器。
我尝试使用:
using ComponentHandleList = brigand::transform<ComponentList, AddComponentHandle<brigand::_1>>;
但我对元编程知之甚少,未能实现 struct AddComponentHandle<>
,它将类型 T
转换为 entityx::ComponentHandle<T>
。 Brigand 通过显示使用 std::add_pointer<>
将每种类型 T
转换为 T*
的示例,包括有关转换类型列表的相关 documentation。这与我想做的相同,只是从 T
到 entityx::ComponentHandle<T>
。如何转换第一个列表,使其看起来像第二个列表?
您可以使用
进行转换
template<typename, template<typename...> class>
struct apply {};
template<template<typename...> class T, template<typename...> class List, typename ... Ts>
struct apply<List<Ts...>, T> {
using type = List<T<Ts>...>;
};
可以这样用
template<typename...>
struct A {};
template<typename...>
struct B {};
int main()
{
static_assert(std::is_same<apply<A<int, bool>, B>::type, A<B<int>, B<bool>>>::value);
}
#include <brigand/sequences/list.hpp>
#include <brigand/algorithms/transform.hpp>
#include <type_traits>
struct TransformComponent{};
struct ObjectComponent{};
struct BodyComponent{};
namespace entityx
{
template <typename Component>
struct ComponentHandle{};
}
using ComponentList = brigand::list<TransformComponent, ObjectComponent, BodyComponent>;
template <typename Comp>
using AddComponentHandle = entityx::ComponentHandle<Comp>;
template <typename Comp>
struct AddComponentHandle2
{
using type = entityx::ComponentHandle<Comp>;
};
using ComponentHandleList = brigand::transform<ComponentList, brigand::bind<AddComponentHandle,brigand::_1>>;
using ComponentHandleList2 = brigand::transform<ComponentList, AddComponentHandle2<brigand::_1>>;
int main()
{
static_assert(std::is_same<ComponentHandleList,
brigand::list<
entityx::ComponentHandle<TransformComponent>,
entityx::ComponentHandle<ObjectComponent>,
entityx::ComponentHandle<BodyComponent>
>
>::value
);
static_assert(std::is_same<ComponentHandleList2,
brigand::list<
entityx::ComponentHandle<TransformComponent>,
entityx::ComponentHandle<ObjectComponent>,
entityx::ComponentHandle<BodyComponent>
>
>::value
);
}
我有以下类型列表:
using ComponentList = brigand::list<TransformComponent, ObjectComponent, BodyComponent>
如何将之前的列表转换成如下所示的新列表:
using ComponentHandleList = brigand::list<entityx::ComponentHandle<TransformComponent>, entityx::ComponentHandle<ObjectComponent>, entityx::ComponentHandle<BodyComponent>>;
我基本上想生成一个新列表,其中包含前一个列表中每个元素的包装器。 我尝试使用:
using ComponentHandleList = brigand::transform<ComponentList, AddComponentHandle<brigand::_1>>;
但我对元编程知之甚少,未能实现 struct AddComponentHandle<>
,它将类型 T
转换为 entityx::ComponentHandle<T>
。 Brigand 通过显示使用 std::add_pointer<>
将每种类型 T
转换为 T*
的示例,包括有关转换类型列表的相关 documentation。这与我想做的相同,只是从 T
到 entityx::ComponentHandle<T>
。如何转换第一个列表,使其看起来像第二个列表?
您可以使用
进行转换template<typename, template<typename...> class>
struct apply {};
template<template<typename...> class T, template<typename...> class List, typename ... Ts>
struct apply<List<Ts...>, T> {
using type = List<T<Ts>...>;
};
可以这样用
template<typename...>
struct A {};
template<typename...>
struct B {};
int main()
{
static_assert(std::is_same<apply<A<int, bool>, B>::type, A<B<int>, B<bool>>>::value);
}
#include <brigand/sequences/list.hpp>
#include <brigand/algorithms/transform.hpp>
#include <type_traits>
struct TransformComponent{};
struct ObjectComponent{};
struct BodyComponent{};
namespace entityx
{
template <typename Component>
struct ComponentHandle{};
}
using ComponentList = brigand::list<TransformComponent, ObjectComponent, BodyComponent>;
template <typename Comp>
using AddComponentHandle = entityx::ComponentHandle<Comp>;
template <typename Comp>
struct AddComponentHandle2
{
using type = entityx::ComponentHandle<Comp>;
};
using ComponentHandleList = brigand::transform<ComponentList, brigand::bind<AddComponentHandle,brigand::_1>>;
using ComponentHandleList2 = brigand::transform<ComponentList, AddComponentHandle2<brigand::_1>>;
int main()
{
static_assert(std::is_same<ComponentHandleList,
brigand::list<
entityx::ComponentHandle<TransformComponent>,
entityx::ComponentHandle<ObjectComponent>,
entityx::ComponentHandle<BodyComponent>
>
>::value
);
static_assert(std::is_same<ComponentHandleList2,
brigand::list<
entityx::ComponentHandle<TransformComponent>,
entityx::ComponentHandle<ObjectComponent>,
entityx::ComponentHandle<BodyComponent>
>
>::value
);
}