[英]Create cartesian product expansion of two variadic, non-type template parameter packs
可以說,我有
foo
,該模板foo
將每個列表中的一個值作為參數 如何創建變量foo
的可變參數包,並使用兩個列表元素的笛卡爾乘積進行參數化?
這是我的意思:
template<int ...>
struct u_list {};
template<char ...>
struct c_list {};
template<int, char >
struct foo {};
template<class ...>
struct bar {};
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using result_t = /* magic happens*/
using ref_t = bar<
foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3>
>;
static_assert(std::is_same<result_t, ref_t >::value, "");
我正在尋找一種可以在c ++ 11中工作並且不使用除c ++ 11標准庫之外的任何庫的解決方案。 我還擁有c ++ 14的人工版本的index_sequence
/ make_index_sequence
並且可以簡化數組,並可以將非類型參數列表作為數組提供。
到目前為止,我發現的最接近的是: 如何創建類型列表的笛卡爾積? 。 因此,原則上(我尚未測試過),應該可以將非類型參數包轉換為類型參數包,然后在鏈接的帖子中應用解決方案,但是我希望能夠有一個更簡單/更短的解決方案這行:
template<int... Ints, char ... Chars>
auto magic(u_list<Ints...>, c_list<Chars...>)
{
//Doesn't work, as it tries to expand the parameter packs in lock step
return bar<foo<Ints,Chars>...>{};
}
using result_t = decltype(magic(int_vals{}, char_vals{}));
您可以執行以下操作:
template <int... Is>
using u_list = std::integer_sequence<int, Is...>;
template <char... Cs>
using c_list = std::integer_sequence<char, Cs...>;
template<int, char> struct foo {};
template<class ...> struct bar {};
template <std::size_t I, typename T, template <typename, T...> class C, T ... Is>
constexpr T get(C<T, Is...> c)
{
constexpr T values[] = {Is...};
return values[I];
}
template <std::size_t I, typename T>
constexpr auto get_v = get<I>(T{});
template<int... Ints, char ... Chars, std::size_t ... Is>
auto cartesian_product(u_list<Ints...>, c_list<Chars...>, std::index_sequence<Is...>)
-> bar<foo<
get_v<Is / sizeof...(Chars), u_list<Ints...> >,
get_v<Is % sizeof...(Chars), c_list<Chars...> >
>...
>;
template<int... Ints, char ... Chars>
auto cartesian_product(u_list<Ints...> u, c_list<Chars...> c)
-> decltype(cartesian_product(u, c, std::make_index_sequence<sizeof...(Ints) * sizeof...(Chars)>()));
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using result_t = decltype(cartesian_product(int_vals{}, char_vals{}));
標准部分的可能實現:
template <typename T, T ... Is> struct integer_sequence{};
template <std::size_t ... Is>
using index_sequence = integer_sequence<std::size_t, Is...>;
template <std::size_t N, std::size_t... Is>
struct make_index_sequence : make_index_sequence<N - 1, N - 1, Is...> {};
template <std::size_t... Is>
struct make_index_sequence<0u, Is...> : index_sequence<Is...> {};
並改變答案:
template <std::size_t I, typename T, template <typename, T...> class C, T ... Is>
constexpr T get(C<T, Is...> c)
{
using array = T[];
return array{Is...}[I];
}
template<int... Ints, char ... Chars, std::size_t ... Is>
auto cartesian_product(u_list<Ints...>, c_list<Chars...>, index_sequence<Is...>)
-> bar<foo<
get<Is / sizeof...(Chars)>(u_list<Ints...>{}),
get<Is % sizeof...(Chars)>(c_list<Chars...>{})
>...
>;
在我看來,在純類型領域中進行模板元編程要容易得多。
從非類型模板參數范圍轉移到類型范圍,然后再返回,這需要一些工作,但這意味着您正在使用通用元編程實用程序,而不是針對您的問題的通用元編程實用程序。
因此,我將您的問題簡化為一系列類型上的笛卡爾積。
這是我的打字包:
template<class...Ts>struct types {
using type=types; // makes inheriting from it useful
static constexpr std::size_t size = sizeof...(Ts);
};
首先我們編寫fmap
。 Fmap接受一個函數和一個列表,並返回應用了該函數的列表中每個元素的列表。
template<template<class...>class Z, class List>
struct fmap {};
template<template<class...>class Z, class List>
using fmap_t = typename fmap<Z,List>::type;
template<template<class...>class Z, class...Ts>
struct fmap<Z, types<Ts...>>:
types<Z<Ts>...>
{};
和fapply
。 fapply也接受一個函數和一個列表,但是將該函數應用於整個列表元素集。
template<template<class...>class Z, class List>
struct fapply {};
template<template<class...>class Z, class List>
using fapply_t=typename fapply<Z,List>::type;
template<template<class...>class Z, class...Ts>
struct fapply<Z, types<Ts...>> {
using type=Z<Ts...>;
};
碰巧的是,部分應用fapply
非常有用:
template<template<class...>class Z>
struct applier {
template<class List>
using apply = fapply_t<Z,List>;
};
我們希望能夠合並列表:
template<class...>
struct cat:types<> {};
template<class...As, class...Bs, class...Cs>
struct cat<types<As...>, types<Bs...>, Cs...>:
cat<types<As..., Bs...>, Cs...>
{};
template<class...As>
struct cat<types<As...>>:types<As...>{};
template<class...Ts>using cat_t=typename cat<Ts...>::type;
然后,這里是cart_product_t:
template<class A, class B>
struct cart_product {};
template<class A, class B>
using cart_product_t = typename cart_product<A,B>::type;
template<class A, class... Bs>
struct cart_product<types<A>, types<Bs...>>:
types< types<A, Bs>... >
{};
// reduce cart_product to cart_product on a one element list on the lhs:
template<class...As, class... Bs>
struct cart_product<types<As...>, types<Bs...>>:
fapply_t<
cat_t,
fmap_t<
applier<cart_product_t>::template apply,
types<
types< types<As>, types<Bs...> >...
>
>
>
{};
特定於您的問題的類型:
template<int...>struct u_list {};
template<char...>struct c_list {};
template<int, char>struct foo {};
template<class...>struct bar{};
一種將值列表提升為類型的工具:
template<class> struct lift {};
template<int...is> struct lift<u_list<is...>>:
types< std::integral_constant<int, is>... >
{};
template<char...is> struct lift<c_list<is...>>:
types< std::integral_constant<char, is>... >
{};
template<class T>using lift_t=typename lift<T>::type;
lower_to_foo采用一對類型,並將它們轉換為foo:
template<class I, class C>
using lower_to_foo = foo<I::value, C::value>;
現在我們將它們放在一起:
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using product = cart_product_t< lift_t<int_vals>, lift_t<char_vals> >;
static_assert( product::size == 6, "should be 6" );
using result_t = fapply_t< bar, fmap_t< applier<lower_to_foo>::template apply, product > >;
using ref_t = bar<
foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3>
>;
ref_t test = result_t{}; // gives better error messages than static_assert
static_assert(std::is_same<result_t, ref_t >::value, "");
鮑勃是你叔叔
cat
, fmap
和fapply
在函數式編程中都是相對標准的函數。 applier
僅允許您編寫使用元素而不是列表的模板映射函數(這是部分應用的fapply
)。
現場例子 。
現在,還記得我怎么說模板元編程對類型更容易嗎?
注意所有那些模板模板參數嗎? 如果它們是類型,它將變得更加容易。
template<template<class...>class Z>
struct ztemplate {
template<class...Ts>using apply=Z<Ts...>;
};
您可以使用ztemplate
上的類型標簽和operator()
進行ztemplate
風格的constexpr元編程,這很有趣。
以類型列表交叉產品為基礎
#include <iostream>
#include <typeinfo>
#include <cxxabi.h>
template<int ...> struct u_list {};
template<char ...> struct c_list {};
template<int, char > struct foo {};
template<typename...> struct type_list {};
我們用row
擴展char...
包
template<int I, char... Cs>
struct row
{
typedef type_list<foo<I,Cs>...> type;
};
template <typename... T> struct concat;
template <typename... S, typename... T>
struct concat<type_list<S...>, type_list<T...>>
{
using type = type_list<S..., T...>;
};
我們想要concat
的額外專業化突破基本案例
template <typename... T>
struct concat<type_list<T...>, void>
{
using type = type_list<T...>;
};
template<typename I, typename C>
struct cross_product;
基本情況:沒有更多的整數
template<char... Cs>
struct cross_product<u_list<>, c_list<Cs...>>
{
using type = void;
};
遞歸的情況:一個int,然后是一包int
template<int I, int... Is, char... Cs>
struct cross_product<u_list<I, Is...>, c_list<Cs...>>
{
using type = typename concat<typename row<I,Cs...>::type, typename cross_product<u_list<Is...>, c_list<Cs...>>::type>::type;
};
int main()
{
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using result_t = cross_product<int_vals, char_vals>::type;
using ref_t = type_list<
foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3>
>;
static_assert(std::is_same<result_t, ref_t >::value, "");
return 0;
}
以下是我的2美分...
如果您想要一個通用的解決方案,我看到的更大的問題是,從int_vals
和char_vals
類型中char_vals
並不容易(在C ++ 11中;在C ++ 17中更簡單)提取包含值的類型( int
和char
)。
因此,我想您必須將它們與foo
和bar
一起傳遞給magic<>
(如果您不希望對foo
和bar
硬編碼)。
所以對magic<>
的調用變成了(以我的方式)
using result_t
= typename magic<int, char, foo, bar, int_vals, char_vals>::type;
以下是我的解決方案的完整示例。
#include <type_traits>
template <int...> struct u_list {};
template <char...> struct c_list {};
template <int, char> struct foo {};
template <typename ...> struct bar {};
template <typename T1, typename T2, T1 t1, T2 ... T2s>
struct midProd
{ };
template <typename T1, typename T2, template <T1, T2> class, typename...>
struct magicHelper;
template <typename T1, typename T2,
template <T1, T2> class ResIn,
template <typename...> class ResOut,
typename ... R>
struct magicHelper<T1, T2, ResIn, ResOut<R...>>
{ using type = ResOut<R...>; };
template <typename T1, typename T2,
template <T1, T2> class ResIn,
template <typename...> class ResOut,
typename ... R, T1 ts1, T2 ... ts2, typename ... MpS>
struct magicHelper<T1, T2, ResIn, ResOut<R...>,
midProd<T1, T2, ts1, ts2...>, MpS...>
{ using type = typename magicHelper<T1, T2, ResIn,
ResOut<R..., ResIn<ts1, ts2>...>, MpS...>::type; };
template <typename T1, typename T2,
template <T1, T2> class,
template <typename...> class,
typename, typename>
struct magic;
template <typename T1, typename T2,
template <T1, T2> class ResIn,
template <typename...> class ResOut,
template <T1...> class C1, template <T2...> class C2,
T1 ... ts1, T2 ... ts2>
struct magic<T1, T2, ResIn, ResOut, C1<ts1...>, C2<ts2...>>
{ using type = typename magicHelper<T1, T2, ResIn, ResOut<>,
midProd<T1, T2, ts1, ts2...>...>::type ; };
int main ()
{
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using result_t
= typename magic<int, char, foo, bar, int_vals, char_vals>::type;
using ref_t = bar< foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3> >;
static_assert(std::is_same<result_t, ref_t >::value, "");
}
顯然,如果您希望對某些類型( u_list
, c_list
, foo
和bar
)進行硬編碼,則解決方案將變得更加簡單
#include <type_traits>
template <int...> struct u_list {};
template <char...> struct c_list {};
template <int, char> struct foo {};
template <typename ...> struct bar {};
template <int, char...> struct midProd {};
template <typename...>
struct magicH;
template <typename ... R>
struct magicH<bar<R...>>
{ using type = bar<R...>; };
template <typename ... R, int i, char ... cs, typename ... MpS>
struct magicH<bar<R...>, midProd<i, cs...>, MpS...>
{ using type = typename magicH<bar<R..., foo<i, cs>...>, MpS...>::type; };
template <typename, typename>
struct magic;
template <int ... is, char ... cs>
struct magic<u_list<is...>, c_list<cs...>>
{ using type = typename magicH<bar<>, midProd<is, cs...>...>::type; };
int main ()
{
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
using result_t = typename magic<int_vals, char_vals>::type;
using ref_t = bar< foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3> >;
static_assert(std::is_same<result_t, ref_t >::value, "");
}
與C ++ 17中的其他代碼相同:
// Type your code here, or load an example.
#include <type_traits>
template<int ...>
struct u_list {};
template<char ...>
struct c_list {};
template<int, char >
struct foo {};
template<class ...>
struct bar {};
using int_vals = u_list<1, 5, 7>;
using char_vals = c_list<-3, 3>;
template<class... Args> struct type_list{
template<class> struct make_concat;
template<class ...Xs>
struct make_concat<type_list<Xs...>>{
using type = type_list<Args...,Xs...>;
};
template<class T>
using concat = typename make_concat<T>::type;
template<template<class...>class TT>
using applied_to = TT<Args...>;
};
template<
template<auto,auto> class C
,class X,class Y,class Yit=Y>
struct cart_prod;
template<template<auto,auto> class C,
template<auto...> class Xt,
template<auto...> class Yt,
class Yit,
auto Xi,auto...Xis,auto Yi,auto...Yis>
struct cart_prod<C,Xt<Xi,Xis...>,Yt<Yi,Yis...>,Yit>{
using type = typename type_list<class C<Xi,Yi>>
::template concat<typename cart_prod<C,Xt<Xi,Xis...>,Yt<Yis...>,Yit>::type>;
};
template<template<auto,auto> class C,
template<auto...> class Xt,
template<auto...> class Yt,
class Yit,
auto Xi,auto...Xis,auto Yi>
struct cart_prod<C,Xt<Xi,Xis...>,Yt<Yi>,Yit>{
using type = typename type_list<class C<Xi,Yi>>
::template concat<typename cart_prod<C,Xt<Xis...>,Yit,Yit>::type>;
};
template<template<auto,auto> class C,
template<auto...> class Xt,
template<auto...> class Yt,
class Yit,
auto Xi,auto Yi>
struct cart_prod<C,Xt<Xi>,Yt<Yi>,Yit>{
using type = type_list<class C<Xi,Yi>>;
};
using result_t = cart_prod<foo,int_vals,char_vals>::type::applied_to<bar>;
using ref_t = bar<
foo<1, -3>, foo<1, 3>,
foo<5, -3>, foo<5, 3>,
foo<7, -3>, foo<7, 3>
>;
static_assert(std::is_same<result_t, ref_t >::value, "");
另一個(但更短)的解決方案可能是
template<typename Ret,typename R>
auto magic( bar<u_list<>, R>, Ret result, R ) { return result; }
template<int I, int... Ints, typename... Foos, typename R>
auto magic( bar<u_list<I,Ints...>, c_list<>>, bar<Foos...>, R rollback ) { return magic(
bar<u_list<Ints...>,R>{}, bar<Foos...>{}, rollback );}
template<int I, int... Ints, char J, char ... Chars, typename... Foos, typename R >
auto magic( bar<u_list<I,Ints...>, c_list<J,Chars...>>, bar<Foos...>, R rollback ) { return magic(
bar<u_list<I,Ints...>, c_list<Chars...>>{},
bar<Foos...,foo<I,J>>{},
rollback );}
using result_t = decltype(magic( bar<int_vals,char_vals>{}, bar<>{}, char_vals{} ));
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.