c++ - 切片整数参数包

Question

我有一个获取整数参数包的类 [a,b,...,y,z]。我需要将它扩展成两个包 [a,...,y] 和 [b,...,z] - 第一次和一次删除最后。最终产品应该是一堆二维数组，大小为 [a,b]、[b,c] 等，直到 [x, y],[y,z]。我正在尝试使用这样的东西:

std::tuple< std::array< std::array< int, /*RemoveFirst*/Ints>, /*RemoveLast*/Ints>...>

我也对其他解决方案持开放态度。

示例:

template <int... Ints>
struct S {
  std::tuple< std::array< std::array< int, /*RemoveFirst*/Ints>, /*RemoveLast*/Ints>...> t;
};
int main() {
  S<2,3,4> a;
  std::get<0>(a.t)[0][0] = 42;
  // explenation:
  // a.t               is tuple<array<array<int,3>,2>,array<array<int,4>,3>>
  // get<0>(a.t)       is array<array<int,3>,2>
  // get<0>(a.t)[0]    is array<int,3>
  // get<0>(a.t)[0][0] is int
}

Answer 1

我提出以下代码

#include <array>
#include <tuple>
#include <utility>
#include <type_traits>

template <typename T, std::size_t ... Is>
auto constexpr bar (std::index_sequence<Is...> const &)
 { return
      std::tuple<
         std::array<
            std::array<int,
                       std::tuple_element_t<Is+1u, T>::value>,
            std::tuple_element_t<Is, T>::value>...
      >{}; }

template <std::size_t ... Is>
auto constexpr foo ()
 { return bar<std::tuple<std::integral_constant<std::size_t, Is>...>>
             (std::make_index_sequence<sizeof...(Is)-1u>{}); }

int main ()
 {
   constexpr auto f = foo<2u, 3u, 5u, 7u, 11u>();

   using typeTuple = std::tuple<
      std::array<std::array<int, 3u>, 2u>,
      std::array<std::array<int, 5u>, 3u>,
      std::array<std::array<int, 7u>, 5u>,
      std::array<std::array<int, 11u>, 7u>>;

   static_assert( std::is_same<typeTuple const, decltype(f)>::value, "!" );
 }