Mic*_*hel 5 c++ templates template-meta-programming
我有一个类采用几个模板参数:
template<typename... ELEMENTS>
class MyContainer;
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根据定义,MyContainer<A, B, C>是与 不同的类型MyContainer<B, A, C>。
但在这种情况下,我想避免这种情况:MyContainer<B, A, C>应该被认为与MyContainer<A, B, C>.
所以我认为“忽略”顺序的一种方法是标准化参数的顺序。有一些模板元编程的魔法,将变换<B, A, C>或C, A, B>成<A, B, C>。
但我找不到任何方法来实现这一目标。你能想到一个聪明的技巧来实现这一目标吗?
ELEMENTS作为模板参数传递从同一个基类继承所有,所以我可以加我想有任何静态成员。正如评论中提到的,要对类型进行排序,您需要一个constexpr谓词。这可以手动为每种类型分配一个static constexpr值,然后您可以比较该值。如果这还不够好,并且您正在使用 GCC、Clang 或 MSVC,则可以使用它来获取constexpr类型的可比值。
使用它作为比较器和constexpr合并排序,我能够让它工作(编译器资源管理器):
#include <type_traits>
#include <string_view>
template <typename T>
constexpr auto type_name() noexcept {
std::string_view name = "Error: unsupported compiler", prefix, suffix;
#ifdef __clang__
name = __PRETTY_FUNCTION__;
prefix = "auto type_name() [T = ";
suffix = "]";
#elif defined(__GNUC__)
name = __PRETTY_FUNCTION__;
prefix = "constexpr auto type_name() [with T = ";
suffix = "]";
#elif defined(_MSC_VER)
name = __FUNCSIG__;
prefix = "auto __cdecl type_name<";
suffix = ">(void) noexcept";
#else
static_assert(false, "Unsupported compiler!");
#endif
name.remove_prefix(prefix.size());
name.remove_suffix(suffix.size());
return name;
}
template <class... Ts>
struct list;
template <template <class...> class Ins, class...> struct instantiate;
template <template <class...> class Ins, class... Ts>
struct instantiate<Ins, list<Ts...>> {
using type = Ins<Ts...>;
};
template <template <class...> class Ins, class... Ts>
using instantiate_t = typename instantiate<Ins, Ts...>::type;
template <class...> struct concat;
template <class... Ts, class... Us>
struct concat<list<Ts...>, list<Us...>>
{
using type = list<Ts..., Us...>;
};
template <class... Ts>
using concat_t = typename concat<Ts...>::type;
template <int Count, class... Ts>
struct take;
template <int Count, class... Ts>
using take_t = typename take<Count, Ts...>::type;
template <class... Ts>
struct take<0, list<Ts...>> {
using type = list<>;
using rest = list<Ts...>;
};
template <class A, class... Ts>
struct take<1, list<A, Ts...>> {
using type = list<A>;
using rest = list<Ts...>;
};
template <int Count, class A, class... Ts>
struct take<Count, list<A, Ts...>> {
using type = concat_t<list<A>, take_t<Count - 1, list<Ts...>>>;
using rest = typename take<Count - 1, list<Ts...>>::rest;
};
template <class... Types>
struct sort_list;
template <class... Ts>
using sorted_list_t = typename sort_list<Ts...>::type;
template <class A>
struct sort_list<list<A>> {
using type = list<A>;
};
template <class Left, class Right>
static constexpr bool less_than = type_name<Left>() < type_name<Right>();
template <class A, class B>
struct sort_list<list<A, B>> {
using type = std::conditional_t<less_than<A, B>, list<A, B>, list<B, A>>;
};
template <class...>
struct merge;
template <class... Ts>
using merge_t = typename merge<Ts...>::type;
template <class... Bs>
struct merge<list<>, list<Bs...>> {
using type = list<Bs...>;
};
template <class... As>
struct merge<list<As...>, list<>> {
using type = list<As...>;
};
template <class AHead, class... As, class BHead, class... Bs>
struct merge<list<AHead, As...>, list<BHead, Bs...>> {
using type = std::conditional_t<less_than<AHead, BHead>,
concat_t<list<AHead>, merge_t<list<As...>, list<BHead, Bs...>>>,
concat_t<list<BHead>, merge_t<list<AHead, As...>, list<Bs...>>>
>;
};
template <class... Types>
struct sort_list<list<Types...>> {
static constexpr auto first_size = sizeof...(Types) / 2;
using split = take<first_size, list<Types...>>;
using type = merge_t<
sorted_list_t<typename split::type>,
sorted_list_t<typename split::rest>>;
};
template <class... Ts>
struct MyActualContainer {
};
template <class... Ts>
using MyContainer = instantiate_t<MyActualContainer, sorted_list_t<list<Ts...>>>;
struct a {
};
struct b {
};
struct c {
};
static_assert(std::is_same_v<
MyContainer<a, b, c, c, c>,
MyContainer<c, b, c, a, c>>);
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请注意,这MyContainer是一个别名,用于对其参数进行排序并MyActualContainer使用排序后的参数进行实例化,而不是容器本身。
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