这是我的问题。 我在 C++ 中有两个短整数:

short a;
short b;

它们的位表示形式可以写成

a = a0 a1 a2 a3 a4 ... a15
b = b0 b1 b2 b3 b4 ... b15

其中 a0、b0、a1、b1 等表示两个短整数的单个位。 现在，我想知道是否有一种有效的方法可以生成以下形式的 int:

a0 b0 a1 b1 a2 b2 ... a15 b15

我知道我可以迂腐地使用循环并手动对每一位进行位掩码，但我想知道是否有更有效的方法来做到这一点。

非常感谢

最佳答案

这是一种使用查找表的方法:

static const unsigned short MortonTable256[256] = 
{
  0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015, 
  0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055, 
  0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115, 
  0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155, 
  0x0400, 0x0401, 0x0404, 0x0405, 0x0410, 0x0411, 0x0414, 0x0415, 
  0x0440, 0x0441, 0x0444, 0x0445, 0x0450, 0x0451, 0x0454, 0x0455, 
  0x0500, 0x0501, 0x0504, 0x0505, 0x0510, 0x0511, 0x0514, 0x0515, 
  0x0540, 0x0541, 0x0544, 0x0545, 0x0550, 0x0551, 0x0554, 0x0555, 
  0x1000, 0x1001, 0x1004, 0x1005, 0x1010, 0x1011, 0x1014, 0x1015, 
  0x1040, 0x1041, 0x1044, 0x1045, 0x1050, 0x1051, 0x1054, 0x1055, 
  0x1100, 0x1101, 0x1104, 0x1105, 0x1110, 0x1111, 0x1114, 0x1115, 
  0x1140, 0x1141, 0x1144, 0x1145, 0x1150, 0x1151, 0x1154, 0x1155, 
  0x1400, 0x1401, 0x1404, 0x1405, 0x1410, 0x1411, 0x1414, 0x1415, 
  0x1440, 0x1441, 0x1444, 0x1445, 0x1450, 0x1451, 0x1454, 0x1455, 
  0x1500, 0x1501, 0x1504, 0x1505, 0x1510, 0x1511, 0x1514, 0x1515, 
  0x1540, 0x1541, 0x1544, 0x1545, 0x1550, 0x1551, 0x1554, 0x1555, 
  0x4000, 0x4001, 0x4004, 0x4005, 0x4010, 0x4011, 0x4014, 0x4015, 
  0x4040, 0x4041, 0x4044, 0x4045, 0x4050, 0x4051, 0x4054, 0x4055, 
  0x4100, 0x4101, 0x4104, 0x4105, 0x4110, 0x4111, 0x4114, 0x4115, 
  0x4140, 0x4141, 0x4144, 0x4145, 0x4150, 0x4151, 0x4154, 0x4155, 
  0x4400, 0x4401, 0x4404, 0x4405, 0x4410, 0x4411, 0x4414, 0x4415, 
  0x4440, 0x4441, 0x4444, 0x4445, 0x4450, 0x4451, 0x4454, 0x4455, 
  0x4500, 0x4501, 0x4504, 0x4505, 0x4510, 0x4511, 0x4514, 0x4515, 
  0x4540, 0x4541, 0x4544, 0x4545, 0x4550, 0x4551, 0x4554, 0x4555, 
  0x5000, 0x5001, 0x5004, 0x5005, 0x5010, 0x5011, 0x5014, 0x5015, 
  0x5040, 0x5041, 0x5044, 0x5045, 0x5050, 0x5051, 0x5054, 0x5055, 
  0x5100, 0x5101, 0x5104, 0x5105, 0x5110, 0x5111, 0x5114, 0x5115, 
  0x5140, 0x5141, 0x5144, 0x5145, 0x5150, 0x5151, 0x5154, 0x5155, 
  0x5400, 0x5401, 0x5404, 0x5405, 0x5410, 0x5411, 0x5414, 0x5415, 
  0x5440, 0x5441, 0x5444, 0x5445, 0x5450, 0x5451, 0x5454, 0x5455, 
  0x5500, 0x5501, 0x5504, 0x5505, 0x5510, 0x5511, 0x5514, 0x5515, 
  0x5540, 0x5541, 0x5544, 0x5545, 0x5550, 0x5551, 0x5554, 0x5555
};

unsigned short x; // Interleave bits of x and y, so that all of the
unsigned short y; // bits of x are in the even positions and y in the odd;
unsigned int z;   // z gets the resulting 32-bit Morton Number.

z = MortonTable256[y >> 8]   << 17 | 
    MortonTable256[x >> 8]   << 16 |
    MortonTable256[y & 0xFF] <<  1 | 
    MortonTable256[x & 0xFF];

查找表将 8 位二进制数 abcdefgh 转换为 0a0b0c0d0e0f0g0h。该代码适用于 16 位输入(和 32 位输出)，但可以很容易地推广到更广泛的输入。

代码取自Bit Twiddling Hacks .请参阅其他两种方法的链接。

作为背景，这种比特交织称为 Morton code并且是一种在保留点的局部性的同时将多个维度组合成一个维度的方法。

关于c++ - C++中的按位扩展，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27216984/