D中的缓存友好的多维数组迭代

3 回答

• 4

  外部循环应为 DEPTH ，如以下程序所示：

  import std.stdio;
  
  void main() {
    enum X = 6, Y = 4, Z = 2;
    ubyte[X][Y][Z] root;
  
    foreach (i, ref level3; root) {
      foreach (j, ref level2; level3) {
        foreach (k, ref level1; level2) {
          writefln("%s %s %s: 0x%x", i, j, k, &level1);
        }
      }
    }   
  }
  

  这打印：

  0 0 0: 0x7fbfc8b9f0
  0 0 1: 0x7fbfc8b9f1
  0 0 2: 0x7fbfc8b9f2
  0 0 3: 0x7fbfc8b9f3
  0 0 4: 0x7fbfc8b9f4
  0 0 5: 0x7fbfc8b9f5
  0 1 0: 0x7fbfc8b9f6
  0 1 1: 0x7fbfc8b9f7
  0 1 2: 0x7fbfc8b9f8
  0 1 3: 0x7fbfc8b9f9
  0 1 4: 0x7fbfc8b9fa
  0 1 5: 0x7fbfc8b9fb
  0 2 0: 0x7fbfc8b9fc
  0 2 1: 0x7fbfc8b9fd
  0 2 2: 0x7fbfc8b9fe
  0 2 3: 0x7fbfc8b9ff
  0 2 4: 0x7fbfc8ba00
  0 2 5: 0x7fbfc8ba01
  0 3 0: 0x7fbfc8ba02
  0 3 1: 0x7fbfc8ba03
  0 3 2: 0x7fbfc8ba04
  0 3 3: 0x7fbfc8ba05
  0 3 4: 0x7fbfc8ba06
  0 3 5: 0x7fbfc8ba07
  1 0 0: 0x7fbfc8ba08
  1 0 1: 0x7fbfc8ba09
  1 0 2: 0x7fbfc8ba0a
  1 0 3: 0x7fbfc8ba0b
  1 0 4: 0x7fbfc8ba0c
  1 0 5: 0x7fbfc8ba0d
  1 1 0: 0x7fbfc8ba0e
  1 1 1: 0x7fbfc8ba0f
  1 1 2: 0x7fbfc8ba10
  1 1 3: 0x7fbfc8ba11
  1 1 4: 0x7fbfc8ba12
  1 1 5: 0x7fbfc8ba13
  1 2 0: 0x7fbfc8ba14
  1 2 1: 0x7fbfc8ba15
  1 2 2: 0x7fbfc8ba16
  1 2 3: 0x7fbfc8ba17
  1 2 4: 0x7fbfc8ba18
  1 2 5: 0x7fbfc8ba19
  1 3 0: 0x7fbfc8ba1a
  1 3 1: 0x7fbfc8ba1b
  1 3 2: 0x7fbfc8ba1c
  1 3 3: 0x7fbfc8ba1d
  1 3 4: 0x7fbfc8ba1e
  1 3 5: 0x7fbfc8ba1f
  

  请注意，地址以线性方式增加，并且最快变化的索引是 X . 所以内部循环代表 X ，外部循环代表 Z .

  回复于 2024-04-24T13:03:51+08:00
• 3

  yaz ' answer is perfectly fine. But to memoize the semantics, it might help to think of array declarations being ' left associative' - 即 ubyte value[X][Y] 声明了一个

  （元素类型的长度为Y的数组（元素类型的长度为X的数组（ubyte）））

  或 (((ubyte value)[X])[Y]) .

  回复于 2024-04-24T13:03:51+08:00
• 0

  我相信所有D编译器都会在优化过程中自动展开 . 至少那是有人在IRC上传递给我的未经证实的信息 .

  从逻辑上讲，第一个循环应该是您知道不会有很多值的维度 . 内部循环应该是你提到的那些尺寸将成千上万 .

  回复于 2024-04-24T13:03:51+08:00