创建两个可变非类型模板参数包的笛卡尔乘积扩展

您可以执行以下操作：

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{}));

演示

标准部分的可能实现：

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...> {};

并更改答案：

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...>{})
>...
>;

演示 C++11

在我看来，在纯类型领域进行模板元编程要容易得多。

从非类型模板参数的状态移动到类型状态并再次返回需要一些工作，但这意味着您使用的是通用元编程实用程序，而不是特定于您的问题的实用程序。

因此，我将把你的问题简化为类型列表中的笛卡尔乘积。

这是我的类型包：

template<class...Ts>struct types {
using type=types; // makes inheriting from it useful
static constexpr std::size_t size = sizeof...(Ts);
};

首先我们写fmap. Fmap 采用一个函数和一个列表，并返回应用了该函数的列表的每个元素的列表。

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 也接受一个函数和一个列表，但将该函数应用于整个列表元素集。

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，部分应用fapply非常有用：

template<template<class...>class Z>
struct applier {
template<class List>
using apply = fapply_t<Z,List>;
};

我们希望能够连接列表：

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，这里是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...> >...
>
>
>
{};

特定于您的问题的类型：

template<int...>struct u_list {};
template<char...>struct c_list {};
template<int, char>struct foo {};
template<class...>struct bar{};

将值列表提升为类型的工具：

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 采用一对类型，并将它们转换为 foo：

template<class I, class C>
using lower_to_foo = foo<I::value, C::value>;

现在我们把它们放在一起：

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, "");

鲍勃是你的叔叔。

cat、fmap和fapply都是函数式编程中相对标准的函数。applier只允许您编写模板映射函数，采用元素而不是列表(这是一个部分应用的fapply)。

活生生的例子。

现在，还记得我说过模板元编程使用类型更容易吗？

注意到所有这些模板模板参数了吗？ 如果它们是类型，它会变得更容易。

template<template<class...>class Z>
struct ztemplate {
template<class...Ts>using apply=Z<Ts...>;
};

你可以一直到HANA风格的ConstexPR元编程，带有类型标签和operator()ztemplate和其他乐趣。

以类型列表交叉乘积为基础

#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 {};

我们用row扩展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...>;
};

我们希望concat的额外专业化来打破基本情况

template <typename... T>
struct concat<type_list<T...>, void>
{
using type = type_list<T...>;
};
template<typename I, typename C>
struct cross_product;

基本情况：不再有整数

template<char... Cs>
struct cross_product<u_list<>, c_list<Cs...>>
{
using type = void;
};

递归大小写：一个 int，后跟一包 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;
}

住在科利鲁！

以下是我的 2 美分...

如果你想要一个通用的解决方案，我看到的更大的问题是，从int_vals和char_vals类型中并不容易(在C++11中;在C++17中更简单)提取所包含值的类型(int和char)。

所以我想你必须将它们与foo和bar一起传递给magic<>(如果你不想foo和bar硬编码)。

所以magic<>的呼唤成为(以我的方式)

using result_t
= typename magic<int, char, foo, bar, int_vals, char_vals>::type; 

以下是我的解决方案的完整工作示例。

#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, "");
}

显然，如果你更喜欢硬编码某些类型(u_list、c_list、foo和bar)，解决方案会变得简单得多。

#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, "");
}

与 C++17 中的其他人相同：

// 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, "");

另一个(但更短的)解决方案可能是

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{} ));