我不确定你的帖子最后你是否感兴趣
在：-
- 生成一个在编译时填充连续的矩阵
某些函数的值
f(i)以行优先顺序环绕矩阵，例如
Cols = 3; Rows = 3; f(i) = 2i; Vals = (1,2,3,4,5,6,7,8,9) ->
|02|04|06|
----------
|08|10|12|
----------
|14|16|18|
or:-
- 生成一个矩阵，其中连续行在编译时填充
具有某个函数的连续值
f(i)对于一些指定的初始i每行，例如
Cols = 3; f(i) = 3i; First_Vals = (4,7,10) ->
|12|15|18|
----------
|21|24|27|
----------
|30|33|36|
无论如何，有多种方法可以做到这两点，这是一种可以与
符合 C++14 的编译器。 （正如 @AndyG 所评论的，适当的
编译时矩阵的实现 - 利用标准库
- 是一个std::array of std::array.)
#include <array>
#include <utility>
namespace detail {
template<typename IntType, IntType(*Step)(IntType), IntType Start, IntType ...Is>
constexpr auto make_integer_array(std::integer_sequence<IntType,Is...>)
{
return std::array<IntType,sizeof...(Is)>{{Step(Start + Is)...}};
}
template<typename IntType, IntType(*Step)(IntType), IntType Start, std::size_t Length>
constexpr auto make_integer_array()
{
return make_integer_array<IntType,Step,Start>(
std::make_integer_sequence<IntType,Length>());
}
template<
typename IntType, std::size_t Cols,
IntType(*Step)(IntType),IntType Start, std::size_t ...Rs
>
constexpr auto make_integer_matrix(std::index_sequence<Rs...>)
{
return std::array<std::array<IntType,Cols>,sizeof...(Rs)>
{{make_integer_array<IntType,Step,Start + (Rs * Cols),Cols>()...}};
}
} // namespace detail
/*
Return a compiletime initialized matrix (`std::array` of std::array`)
of `Cols` columns by `Rows` rows. Ascending elements from [0,0]
in row-first order are populated with successive values of the
constexpr function `IntType Step(IntType i)` for `i` in
`[Start + 0,Start + (Rows * Cols))`
*/
template<
typename IntType, std::size_t Cols, std::size_t Rows,
IntType(*Step)(IntType), IntType Start
>
constexpr auto make_integer_matrix()
{
return detail::make_integer_matrix<IntType,Cols,Step,Start>(
std::make_index_sequence<Rows>());
}
/*
Return a compiletime initialized matrix (`std::array` of std::array`)
of `Cols` columns by `sizeof...(Starts)` rows. Successive rows are populated
with successive values of the constexpr function `IntType Step(IntType i)`
for `i` in `[start + 0,start + Cols)`, for `start` successively in `...Starts`.
*/
template<typename IntType, std::size_t Cols, IntType(*Step)(IntType), IntType ...Starts>
constexpr auto make_integer_matrix()
{
return std::array<std::array<IntType,Cols>,sizeof...(Starts)>
{{detail::make_integer_array<IntType,Step,Starts,Cols>()...}};
}
您可以通过附加以下内容来制作演示程序：
#include <iostream>
using namespace std;
template<typename IntType>
constexpr auto times_3(IntType i)
{
return i * 3;
}
static constexpr auto m4x6 = make_integer_matrix<int,4,6,×_3<int>,4>();
static constexpr auto m5x1 = make_integer_matrix<int,5,×_3<int>,7>();
static constexpr auto m6x5 = make_integer_matrix<int,6,×_3<int>,11,13,17,19,23>();
static_assert(m4x6[0][0] == 12,"");
int main()
{
cout << "A 4 x 6 matrix that wraps around in steps of `3i` from `i` = 4" << endl;
for (auto const & ar : m4x6) {
for (auto const & i : ar) {
cout << i << ' ';
}
cout << endl;
}
cout << endl;
cout << "A 6 x 5 matrix with rows of `3i` for initial `i` in <11,13,17,19,23>"
<< endl;
for (auto const & ar : m6x5) {
for (auto const & i : ar) {
cout << i << ' ';
}
cout << endl;
}
cout << endl;
cout << "A 5 x 1 matrix with rows of of ` 3i` for initial `i` in <7>"
<< endl;
for (auto const & ar : m5x1) {
for (auto const & i : ar) {
cout << i << ' ';
}
cout << endl;
}
return 0;
}
应该输出：
A 4 x 6 matrix that wraps around in steps of `3i` from `i` = 4
12 15 18 21
24 27 30 33
36 39 42 45
48 51 54 57
60 63 66 69
72 75 78 81
A 6 x 5 matrix with rows of `3i` for initial `i` in <11,13,17,19,23>
33 36 39 42 45 48
39 42 45 48 51 54
51 54 57 60 63 66
57 60 63 66 69 72
69 72 75 78 81 84
A 5 x 1 matrix with rows of of ` 3i` for initial `i` in <7>
21 24 27 30 33
See it 在ideone https://ideone.com/cZTplL
你也可能对此有兴趣std::实验::make_array http://en.cppreference.com/w/cpp/experimental/make_array
