C++中的按位扩展

Bitwise expansion in C++

本文关键字：扩展 C++      更新时间：2023-10-16

这是我的问题。我C++有两个短整数：

short a;
short b;

它们的位表示可以放在以下形式中

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来生成以下形式的 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个位进行位屏蔽，但我想知道是否有更有效的方法可以做到这一点。

谢谢

这是一种方法，带有查找表：

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宽的输入。

该代码取自Bit Twiddling Hacks。有关其他两种方法，请参阅链接。

对于背景，这种位的交错称为莫顿码，是一种将多个维度组合成一个同时保留点局部性的方法。

我会使用for循环并使算法正常工作。

uint32_t result = 0;
uint16_t a;
uint16_t b;
const unsigned int bits_to_process = 16;
for (unsigned int i = 0; i < bits_to_process; ++i)
{
  result = a & 1;
  result << 1;
  result = b & 1;
  result << 1;
  a = a >> 1;
  b = b >> 1;
}

代码正常工作后，请调高优化级别。 编译器可能会执行一些惊人的优化，例如循环展开。

你也可以在网上搜索"有点扭动"，看看这些案例中是否有任何像你的一样。

这是我的做法。

#include <iostream>
#include <bitset>
using namespace std;
inline unsigned int
move_bit(unsigned short x, int pos, int count)
{
    return (x & (1 << pos)) << count;
}
inline unsigned int
merge_bits(unsigned short a, unsigned short b)
{
    unsigned int res{};
    for(int i=0; i<16; i++)
        res |= (move_bit(a, i, i+1) | move_bit(b, i, i));
    return res;
}
int main()
{
    unsigned short a = 0xabcd;
    unsigned short b = 0x1234;
    unsigned int   c = merge_bits(a, b);
    cout << "a:      " << bitset<16>(a) << endl
         << "b:      " << bitset<16>(b) << endl
         << "merged: " << bitset<32>(c) << endl;
}

输出：

a:      1010101111001101
b:      0001001000110100
merged: 10001001100011101010010110110010
相关文章：