用于多个组件的索引技巧
Indices trick used for several components
考虑这个完全有效的代码:
#include <type_traits>
template <typename T, typename IndexPack> struct Make;
template <typename T, template <T...> class P, T... Indices>
struct Make<T, P<Indices...>> {
using type = P<(Indices+1)..., (-3*Indices)..., (Indices-1)...>;
};
template <int...> class Pack;
int main() {
static_assert (std::is_same<Make<int, Pack<1,2,3,4>>::type,
Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>>::value, "false");
}
我实际想要的输出是
Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>
而不是 Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>
。我第一次尝试
using type = P<(Indices+1, -3*Indices, Indices-1)...>;
但是编译器简单地将其理解为无用的逗号运算符。获得我想要的东西所需的语法是什么?如果没有这样的语法,最简洁的方法是什么,请记住使用 Indices
3 次只是一个示例(我们可能希望使用它超过 3 次)。请不要告诉我我必须编写一个帮助程序来提取单个包,然后 "interlace" 所有元素。这种噩梦般的方法不可能是最好的解决方案(而且这种解决方案也只有在我们确切知道要提取多少包的情况下才有效)。
会定义
template <typename T, template <T...> class P, T I>
struct Component {
using type = P<I+1, -3*I, I-1>;
};
有什么帮助吗?对此进行包扩展?
是的,你可以递归连接:
template <typename, typename, typename> struct Concat;
template <typename T, template <T...> class P, T... A, T... B>
struct Concat<T, P<A...>, P<B...>> {
using type = P<A..., B...>;
};
template <typename T, typename IndexPack> struct Make;
template <typename T, template <T...> class P, T... I, T F >
struct Make<T, P<F, I...>> {
using type = typename Concat<T,
typename Make<T, P<F>>::type,
typename Make<T, P<I...>>::type>::type;
};
template <typename T, template <T...> class P, T I>
struct Make<T, P<I>> {
using type = P<I+1, -3*I, I-1>;
};
这是受 Columbo 解决方案的启发。它使用了我最初寻求的包扩展语法,即
using type = typename Merge<T, typename Component<T, P, Indices>::type...>::type;
因此,现在Make
是可以复用的,先用Triple
,再用Quadruple
,所以可以同时扩展Indices
的任意数量的用法.这里 Component
是传给 Make
:
的模板-模板-模板参数
#include <type_traits>
template <typename T, typename... Packs> struct Merge;
template <typename T, template <T...> class P1, template <T...> class P2, T... Is, T... Js>
struct Merge<T, P1<Is...>, P2<Js...>> {
using type = P1<Is..., Js...>;
};
template <typename T, typename Pack1, typename Pack2, typename... Packs>
struct Merge<T, Pack1, Pack2, Packs...> {
using type = typename Merge<T, Pack1, typename Merge<T, Pack2, Packs...>::type>::type;
};
template <typename T, template <T...> class P, T I>
struct Triple {
using type = P<I+1, -3*I, I-1>;
};
template <typename T, template <T...> class P, T I>
struct Quadruple {
using type = P<I+1, -3*I, I-1, I>;
};
template <typename T, typename IndexPack,
template <typename U, template <U...> class P, U I> class Component> struct Make;
template <typename T, template <T...> class Z, T... Indices,
template <typename U, template <U...> class P, U I> class Component>
struct Make<T, Z<Indices...>, Component> {
using type = typename Merge<T, typename Component<T, Z, Indices>::type...>::type;
};
template <int...> class Pack;
int main() {
static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Triple>::type,
Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>>::value, "false");
static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Quadruple>::type,
Pack<2,-3,0,1, 3,-6,1,2, 4,-9,2,3, 5,-12,3,4>>::value, "false");
}
考虑这个完全有效的代码:
#include <type_traits>
template <typename T, typename IndexPack> struct Make;
template <typename T, template <T...> class P, T... Indices>
struct Make<T, P<Indices...>> {
using type = P<(Indices+1)..., (-3*Indices)..., (Indices-1)...>;
};
template <int...> class Pack;
int main() {
static_assert (std::is_same<Make<int, Pack<1,2,3,4>>::type,
Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>>::value, "false");
}
我实际想要的输出是
Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>
而不是 Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>
。我第一次尝试
using type = P<(Indices+1, -3*Indices, Indices-1)...>;
但是编译器简单地将其理解为无用的逗号运算符。获得我想要的东西所需的语法是什么?如果没有这样的语法,最简洁的方法是什么,请记住使用 Indices
3 次只是一个示例(我们可能希望使用它超过 3 次)。请不要告诉我我必须编写一个帮助程序来提取单个包,然后 "interlace" 所有元素。这种噩梦般的方法不可能是最好的解决方案(而且这种解决方案也只有在我们确切知道要提取多少包的情况下才有效)。
会定义
template <typename T, template <T...> class P, T I>
struct Component {
using type = P<I+1, -3*I, I-1>;
};
有什么帮助吗?对此进行包扩展?
是的,你可以递归连接:
template <typename, typename, typename> struct Concat;
template <typename T, template <T...> class P, T... A, T... B>
struct Concat<T, P<A...>, P<B...>> {
using type = P<A..., B...>;
};
template <typename T, typename IndexPack> struct Make;
template <typename T, template <T...> class P, T... I, T F >
struct Make<T, P<F, I...>> {
using type = typename Concat<T,
typename Make<T, P<F>>::type,
typename Make<T, P<I...>>::type>::type;
};
template <typename T, template <T...> class P, T I>
struct Make<T, P<I>> {
using type = P<I+1, -3*I, I-1>;
};
这是受 Columbo 解决方案的启发。它使用了我最初寻求的包扩展语法,即
using type = typename Merge<T, typename Component<T, P, Indices>::type...>::type;
因此,现在Make
是可以复用的,先用Triple
,再用Quadruple
,所以可以同时扩展Indices
的任意数量的用法.这里 Component
是传给 Make
:
#include <type_traits>
template <typename T, typename... Packs> struct Merge;
template <typename T, template <T...> class P1, template <T...> class P2, T... Is, T... Js>
struct Merge<T, P1<Is...>, P2<Js...>> {
using type = P1<Is..., Js...>;
};
template <typename T, typename Pack1, typename Pack2, typename... Packs>
struct Merge<T, Pack1, Pack2, Packs...> {
using type = typename Merge<T, Pack1, typename Merge<T, Pack2, Packs...>::type>::type;
};
template <typename T, template <T...> class P, T I>
struct Triple {
using type = P<I+1, -3*I, I-1>;
};
template <typename T, template <T...> class P, T I>
struct Quadruple {
using type = P<I+1, -3*I, I-1, I>;
};
template <typename T, typename IndexPack,
template <typename U, template <U...> class P, U I> class Component> struct Make;
template <typename T, template <T...> class Z, T... Indices,
template <typename U, template <U...> class P, U I> class Component>
struct Make<T, Z<Indices...>, Component> {
using type = typename Merge<T, typename Component<T, Z, Indices>::type...>::type;
};
template <int...> class Pack;
int main() {
static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Triple>::type,
Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>>::value, "false");
static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Quadruple>::type,
Pack<2,-3,0,1, 3,-6,1,2, 4,-9,2,3, 5,-12,3,4>>::value, "false");
}