受到这个问题的启发:c++ generate (xyz) points in range
我开始怀疑是否有一种形式的模板代码可以,从这个声明:
using X = axis_limits<-10, +10>;
using Y = axis_limits<-10, +10>;
using Z = axis_limits<-10, +10>;
auto space = std::vector<point>{ generate_point_space<X, Y, Z> };
在编译时构造一个名为 space 的 vector ,其中每个 x、y、z 包含一个点,其中 begin(X) <= x < end(X)... y 和 z 等。
顺序不重要。
generate_point_space<>
的返回类型应该是 std::initializer_list<int>
或类似的编译时构造的序列。我不想生成对 push_back()
的调用序列.那太容易了:)
struct point
将具有以下形式的构造函数:
point::point(int x, int y, int z)
整数的单一维度很简单(下面的代码)。这个问题的多维方面今天超出了我的范围
#include <utility>
#include <iostream>
#include <vector>
template<int Begin, int End>
struct axis_limits
{
static constexpr int first = Begin;
static constexpr int last = End;
};
namespace details
{
template<typename Int, typename, Int Begin, bool Increasing>
struct integer_range_impl;
template<typename Int, Int... N, Int Begin>
struct integer_range_impl<Int, std::integer_sequence<Int, N...>, Begin, true> {
using type = std::integer_sequence<Int, N+Begin...>;
};
template<typename Int, Int... N, Int Begin>
struct integer_range_impl<Int, std::integer_sequence<Int, N...>, Begin, false> {
using type = std::integer_sequence<Int, Begin-N...>;
};
}
template<typename Int, Int Begin, Int End>
using integer_range = typename details::integer_range_impl<
Int,
std::make_integer_sequence<Int, (Begin<End) ? End-Begin : Begin-End>,
Begin,
(Begin<End) >::type;
template<int...Is>
std::vector<int> make_vector(std::integer_sequence<int, Is...>)
{
return std::vector<int> { Is... };
}
template<int Begin, int End>
struct axis_range
{
using sequence_type = integer_range<int, Begin, End>;
static constexpr int size = sequence_type::size();
static std::vector<int> as_vector()
{
return make_vector(sequence_type {});
}
};
template< int Begin, int End >
std::vector<int> make_axis(const axis_limits<Begin, End> &)
{
return axis_range<Begin, End>::as_vector();
}
template<class T>
void dump_vector(std::ostream& os, const std::vector<T>& v) {
const char* sep = "{ ";
for(const auto& i : v) {
os << sep << i;
sep = ", ";
}
os << " }";
}
template<class T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vec)
{
dump_vector(os, vec);
return os;
}
using namespace std;
int main()
{
using X = axis_limits<-5, +5>;
auto space = std::vector<int>(make_axis(X{}));
cout << space << endl;
return 0;
}
当前输出:
{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 }
我在找什么:
{ { -10, -10, -10 }, { -10, -10, -9 } .... { 9, 9, 8 }, { 9, 9, 9 } }
最佳答案
您可以执行以下操作:
template<int Begin, int End>
struct axis_limits
{
static constexpr int first = Begin;
static constexpr int last = End;
static constexpr int range = End - Begin + 1;
};
struct point
{
explicit point(int x, int y, int z) : x(x), y(y), z(z) {}
int x; int y; int z;
};
namespace detail
{
template <typename X, typename Y, typename Z, std::size_t... Is>
std::vector<point> generate_point_space_impl(std::index_sequence<Is...>)
{
return {point(
static_cast<int>(Is / (Z::range * Y::range)) % X::range + X::first,
static_cast<int>(Is / Z::range) % Y::range + Y::first,
static_cast<int>(Is) % Z::range + Z::first)...
};
}
}
template <typename X, typename Y, typename Z>
std::vector<point> generate_point_space()
{
return detail::generate_point_space_impl<X, Y, Z>(std::make_index_sequence<X::range * Y::range * Z::range>());
}
关于c++ - 在编译时推导整个 vector 空间,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30171531/