C ++中的按位擴展

Question

這是我的問題。 我在C ++中有兩個簡短的整數：

short a;
short b;

它們的位表示可以放在表單中

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

其中a0，b0，a1，b1等代表兩個短整數的單個位。 現在，我想知道是否有一種有效的方法來生成表單中的int：

a0 b0 a1 b1 a2 b2 ... a15 b15

我知道我可以迂腐地使用循環並手動對每一位進行位掩碼，但我想知道是否有更有效的方法來完成它。

非常感謝你

Answer 1

這是一種方法，帶有查找表：

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];

查找表將8位二進制數abcdefgh為0a0b0c0d0e0f0g0h 。 該代碼適用於16位輸入（和32位輸出），但可以很容易地推廣到更寬的輸入。

代碼取自Bit Twiddling Hacks 。 請參閱其他兩種方法的鏈接。

對於背景，這種位的交錯被稱為莫頓碼，並且是一種將多個維度組合成一個同時保留點的局部性的方法。

Answer 2

我會使用for循環並使算法工作。

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;
}

一旦您的代碼正常工作，請調高優化級別。 編譯器可以執行一些驚人的優化，例如循環展開。

您還可以在網上搜索“bit twiddling”，看看是否有任何類似的情況。

Answer 3

這是我怎么做的..

#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;
}

輸出：

a:      1010101111001101
b:      0001001000110100
merged: 10001001100011101010010110110010