# algorithm - 在C里面，最佳的位反转算法( 从MSB >LSB到LSB >MSB )

`0010 0000 => 0000 0100`

## 选项

``````
unsigned int
reverse(register unsigned int x)
{
x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
return((x >> 16) | (x << 16));

}

``````

``````
static const unsigned char BitReverseTable256[] =
{
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};

unsigned int v; // reverse 32-bit value, 8 bits at time
unsigned int c; // c will get v reversed

// Option 1:
c = (BitReverseTable256[v & 0xff] << 24) |
(BitReverseTable256[(v >> 8) & 0xff] << 16) |
(BitReverseTable256[(v >> 16) & 0xff] << 8) |
(BitReverseTable256[(v >> 24) & 0xff]);

// Option 2:
unsigned char * p = (unsigned char *) &v;
unsigned char * q = (unsigned char *) &c;
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];

``````

## 其他

``````
unsigned int v; // input bits to be reversed
unsigned int r = v; // r will be reversed bits of v; first get LSB of v
int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end

for (v >>= 1; v; v >>= 1)
{
r <<= 1;
r |= v & 1;
s--;
}
r <<= s; // shift when v's highest bits are zero

``````

``````
unsigned char b = x;
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;

``````

``````
unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;

``````

``````
unsigned int toReverse;
unsigned int reversed;
unsigned char inByte0 = (toReverse & 0xFF);
unsigned char inByte1 = (toReverse & 0xFF00) >> 8;
unsigned char inByte2 = (toReverse & 0xFF0000) >> 16;
unsigned char inByte3 = (toReverse & 0xFF000000) >> 24;
reversed = (reverseBits(inByte0) << 24) | (reverseBits(inByte1) << 16) | (reverseBits(inByte2) << 8) | (reverseBits(inByte3);

``````

## 结果

reverse.c

``````
#include <stdlib.h>
#include <stdio.h>
#include <omp.h>

unsigned int
reverse(register unsigned int x)
{
x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
return((x >> 16) | (x << 16));

}

int main()
{
unsigned int *ints = malloc(100000000*sizeof(unsigned int));
unsigned int *ints2 = malloc(100000000*sizeof(unsigned int));
for(unsigned int i = 0; i < 100000000; i++)
ints[i] = rand();

unsigned int *inptr = ints;
unsigned int *outptr = ints2;
unsigned int *endptr = ints + 100000000;
// Starting the time measurement
double start = omp_get_wtime();
// Computations to be measured
while(inptr != endptr)
{
(*outptr) = reverse(*inptr);
inptr++;
outptr++;
}
// Measuring the elapsed time
double end = omp_get_wtime();
// Time calculation (in seconds)
printf("Time: %f secondsn", end-start);

free(ints);
free(ints2);

return 0;
}

``````

reverse_lookup.c

``````
#include <stdlib.h>
#include <stdio.h>
#include <omp.h>

static const unsigned char BitReverseTable256[] =
{
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};

int main()
{
unsigned int *ints = malloc(100000000*sizeof(unsigned int));
unsigned int *ints2 = malloc(100000000*sizeof(unsigned int));
for(unsigned int i = 0; i < 100000000; i++)
ints[i] = rand();

unsigned int *inptr = ints;
unsigned int *outptr = ints2;
unsigned int *endptr = ints + 100000000;
// Starting the time measurement
double start = omp_get_wtime();
// Computations to be measured
while(inptr != endptr)
{
unsigned int in = *inptr;

// Option 1:
//*outptr = (BitReverseTable256[in & 0xff] << 24) |
// (BitReverseTable256[(in >> 8) & 0xff] << 16) |
// (BitReverseTable256[(in >> 16) & 0xff] << 8) |
// (BitReverseTable256[(in >> 24) & 0xff]);

// Option 2:
unsigned char * p = (unsigned char *) &(*inptr);
unsigned char * q = (unsigned char *) &(*outptr);
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];

inptr++;
outptr++;
}
// Measuring the elapsed time
double end = omp_get_wtime();
// Time calculation (in seconds)
printf("Time: %f secondsn", end-start);

free(ints);
free(ints2);

return 0;
}

``````

``````
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -o reverse reverse.c
mrj10@mjlap:~/code\$ ./reverse
Time: 2.000593 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 1.938893 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 1.936365 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O2 -o reverse reverse.c
mrj10@mjlap:~/code\$ ./reverse
Time: 0.942709 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 0.991104 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 0.947203 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O3 -o reverse reverse.c
mrj10@mjlap:~/code\$ ./reverse
Time: 0.922639 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 0.892372 seconds
mrj10@mjlap:~/code\$ ./reverse
Time: 0.891688 seconds

``````

``````
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.201127 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.196129 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.235972 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.633042 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.655880 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.633390 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.652322 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.631739 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 0.652431 seconds

``````

``````
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.671537 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.688173 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.664662 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.049851 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.048403 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.085086 seconds
mrj10@mjlap:~/code\$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.082223 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.053431 seconds
mrj10@mjlap:~/code\$ ./reverse_lookup
Time: 1.081224 seconds

``````

## 提醒

• 我没有访问ICC的权限，这更快。
• 对于具有大型L1D的一些现代微体系结构，64K查找表可能表现良好。
• -mtune=native没有为-o2/-o3 (ld重定义错误)工作，所以，我不相信生成的代码对我的微体系结构有所调优，
• 使用SSE可能让它更快。
• 这是在-O3上为选项1生成的代码GCC，所以可以查看它：

32位

``````
.L3:
movl (%r12,%rsi), %ecx
movzbl %cl, %eax
movzbl BitReverseTable256(%rax), %edx
movl %ecx, %eax
shrl \$24, %eax
mov %eax, %eax
movzbl BitReverseTable256(%rax), %eax
sall \$24, %edx
orl %eax, %edx
movzbl %ch, %eax
shrl \$16, %ecx
movzbl BitReverseTable256(%rax), %eax
movzbl %cl, %ecx
sall \$16, %eax
orl %eax, %edx
movzbl BitReverseTable256(%rcx), %eax
sall \$8, %eax
orl %eax, %edx
movl %edx, (%r13,%rsi)
cmpq \$400000000, %rsi
jne .L3

``````

``````
.L3:
movq (%r12,%rsi), %rdx
movq %rdx, %rax
shrq \$24, %rax
andl \$255, %eax
movzbl BitReverseTable256(%rax), %ecx
movzbq %dl,%rax
movzbl BitReverseTable256(%rax), %eax
salq \$24, %rax
orq %rax, %rcx
movq %rdx, %rax
shrq \$56, %rax
movzbl BitReverseTable256(%rax), %eax
salq \$32, %rax
orq %rax, %rcx
movzbl %dh, %eax
shrq \$16, %rdx
movzbl BitReverseTable256(%rax), %eax
salq \$16, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq \$16, %rdx
movzbl BitReverseTable256(%rax), %eax
salq \$8, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq \$8, %rdx
movzbl BitReverseTable256(%rax), %eax
salq \$56, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq \$8, %rdx
movzbl BitReverseTable256(%rax), %eax
andl \$255, %edx
salq \$48, %rax
orq %rax, %rcx
movzbl BitReverseTable256(%rdx), %eax
salq \$40, %rax
orq %rax, %rcx
movq %rcx, (%r13,%rsi)
cmpq \$400000000, %rsi
jne .L3

``````

bitflip 400000000字节的时间： 0.050 082秒 ！

``````
//Bitflip using AVX2 - The fastest Intel based bitflip in the world!!
//Made by Anders Cedronius 2014 (anders.cedronius (you know what) gmail.com)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <omp.h>

using namespace std;

#define DISPLAY_HEIGHT 4
#define DISPLAY_WIDTH 32
#define NUM_DATA_BYTES 400000000

//Constants (first we got the mask, then the high order nibble look up table and last we got the low order nibble lookup table)
__attribute__ ((aligned(32))) static unsigned char k1[32*3]={
0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,
0x00,0x08,0x04,0x0c,0x02,0x0a,0x06,0x0e,0x01,0x09,0x05,0x0d,0x03,0x0b,0x07,0x0f,0x00,0x08,0x04,0x0c,0x02,0x0a,0x06,0x0e,0x01,0x09,0x05,0x0d,0x03,0x0b,0x07,0x0f,
0x00,0x80,0x40,0xc0,0x20,0xa0,0x60,0xe0,0x10,0x90,0x50,0xd0,0x30,0xb0,0x70,0xf0,0x00,0x80,0x40,0xc0,0x20,0xa0,0x60,0xe0,0x10,0x90,0x50,0xd0,0x30,0xb0,0x70,0xf0
};

//The data to be bitflipped (+32 to avoid the quantization out of memory problem)
__attribute__ ((aligned(32))) static unsigned char data[NUM_DATA_BYTES+32]={};

extern"C" {
void bitflipbyte(unsigned char[],unsigned int,unsigned char[]);
}

int main()
{

for(unsigned int i = 0; i <NUM_DATA_BYTES; i++)
{
data[i] = rand();
}

printf ("rnData in(start):rn");
for (unsigned int j = 0; j <4; j++)
{
for (unsigned int i = 0; i <DISPLAY_WIDTH; i++)
{
printf ("0x%02x,",data[i+(j*DISPLAY_WIDTH)]);
}
printf ("rn");
}

printf ("rnNumber of 32-byte chunks to convert: %drn",(unsigned int)ceil(NUM_DATA_BYTES/32.0));

double start_time = omp_get_wtime();
bitflipbyte(data,(unsigned int)ceil(NUM_DATA_BYTES/32.0),k1);
double end_time = omp_get_wtime();

printf ("rnData out:rn");
for (unsigned int j = 0; j <4; j++)
{
for (unsigned int i = 0; i <DISPLAY_WIDTH; i++)
{
printf ("0x%02x,",data[i+(j*DISPLAY_WIDTH)]);
}
printf ("rn");
}
printf("rnrnTime to bitflip %d bytes: %f secondsrnrn",NUM_DATA_BYTES, end_time-start_time);

//return with no errors
return 0;
}

``````

pritf debuging的。

``````
bits 64
global bitflipbyte

bitflipbyte:
vmovdqa ymm2, [rdx]
vmovdqa ymm3, [rdx]
vmovdqa ymm4, [rdx]
bitflipp_loop:
vmovdqa ymm0, [rdi]
vpand ymm1, ymm2, ymm0
vpandn ymm0, ymm2, ymm0
vpsrld ymm0, ymm0, 4h
vpshufb ymm1, ymm4, ymm1
vpshufb ymm0, ymm3, ymm0
vpor ymm0, ymm0, ymm1
vmovdqa [rdi], ymm0
dec rsi
jnz bitflipp_loop
ret

``````

``````
Illustrated in the example below.

Ex : If Input is 00101010 ==> Expected output is 01010100

1. Divide the input into 2 halves
0010 --- 1010

2. Swap the 2 Halves
1010 0010

3. Repeat the same for each half.
10 -- 10 --- 00 -- 10
10 10 10 00

1-0 -- 1-0 --- 1-0 -- 0-0
0 1 0 1 0 1 0 0

Done! Output is 01010100

``````

``````
int reverse_bits_recursive(unsigned int num, unsigned int numBits)
{
unsigned int reversedNum;;

mask = (0x1 << (numBits/2)) - 1;

if (numBits == 1) return num;
reversedNum = reverse_bits_recursive(num >> numBits/2, numBits/2) |
reverse_bits_recursive((num & mask), numBits/2) << numBits/2;
return reversedNum;
}

int main()
{
unsigned int reversedNum;
unsigned int num;

num = 0x55;
reversedNum = reverse_bits_recursive(num, 8);
printf ("Bit Reversal Input = 0x%x Output = 0x%xn", num, reversedNum);

num = 0xabcd;
reversedNum = reverse_bits_recursive(num, 16);
printf ("Bit Reversal Input = 0x%x Output = 0x%xn", num, reversedNum);

num = 0x123456;
reversedNum = reverse_bits_recursive(num, 24);
printf ("Bit Reversal Input = 0x%x Output = 0x%xn", num, reversedNum);

num = 0x11223344;
reversedNum = reverse_bits_recursive(num,32);
printf ("Bit Reversal Input = 0x%x Output = 0x%xn", num, reversedNum);
}

``````

``````
Bit Reversal Input = 0x55 Output = 0xaa
Bit Reversal Input = 0xabcd Output = 0xb3d5
Bit Reversal Input = 0x123456 Output = 0x651690
Bit Reversal Input = 0x11223344 Output = 0x22cc4488

``````

``````
Stack stack = new Stack();
Bit[] bits = new Bit[] { 0, 0, 1, 0, 0, 0, 0, 0 };

for (int i = 0; i < bits.Length; i++)
{
stack.push(bits[i]);
}

for (int i = 0; i < bits.Length; i++)
{
bits[i] = stack.pop();
}

``````

``````
unsigned int reverse(unsigned int t)
{
size_t n = t;//store in 64 bit number for call to BSWAP
__asm__("BSWAP %0" :"=r"(n) :"0"(n));
n>> = ((sizeof(size_t) - sizeof(unsigned int)) * 8);
n = ((n & 0xaaaaaaaa)>> 1) | ((n & 0x55555555) <<1);
n = ((n & 0xcccccccc)>> 2) | ((n & 0x33333333) <<2);
n = ((n & 0xf0f0f0f0)>> 4) | ((n & 0x0f0f0f0f) <<4);
return n;
}

``````

``````
size_t reverse(size_t n, unsigned int bytes)
{
__asm__("BSWAP %0" :"=r"(n) :"0"(n));
n>> = ((sizeof(size_t) - bytes) * 8);
n = ((n & 0xaaaaaaaa)>> 1) | ((n & 0x55555555) <<1);
n = ((n & 0xcccccccc)>> 2) | ((n & 0x33333333) <<2);
n = ((n & 0xf0f0f0f0)>> 4) | ((n & 0x0f0f0f0f) <<4);
return n;
}

``````

``````
n = reverse(n, sizeof(char));//only reverse 8 bits
n = reverse(n, sizeof(short));//reverse 16 bits
n = reverse(n, sizeof(int));//reverse 32 bits
n = reverse(n, sizeof(size_t));//reverse 64 bits

``````

``````
//returns n rotated right by b bits
int rotRight(int n, int b)
{
if(b >= 0)
b = b%(8*sizeof(int));
else
b = 8*sizeof(int) - (-b)%(8*sizeof(int));
return (n >>> b) | (n << (8*sizeof(int) - b));
}

//returns n rotated left by b bits
int rotLeft(int n, int b)
{
if(b >= 0)
b = b%(8*sizeof(int));
else
b = 8*sizeof(int) - (-b)%(8*sizeof(int));
return (n << b) | (n >>> (8*sizeof(int) - b));
}

``````

``````
private Byte BitReverse(Byte bData)
{
Byte[] lookup = { 0, 8, 4, 12,
2, 10, 6, 14 ,
1, 9, 5, 13,
3, 11, 7, 15 };
Byte ret_val = (Byte)(((lookup[(bData & 0x0F)]) << 4) + lookup[((bData & 0xF0) >> 4)]);
return ret_val;
}

``````

C代码。以1字节输入数据num为例。

``````
unsigned char num = 0xaa; // 1010 1010 (aa) -> 0101 0101 (55)
int s = sizeof(num) * 8; // get number of bits
int i, x, y, p;
int var = 0; // make var data type to be equal or larger than num

for (i = 0; i < (s / 2); i++) {
// extract bit on the left, from MSB
p = s - i - 1;
x = num & (1 << p);
x = x >> p;
printf("x: %dn", x);

// extract bit on the right, from LSB
y = num & (1 << i);
y = y >> i;
printf("y: %dn", y);

var = var | (x << i); // apply x
var = var | (y << p); // apply y
}

printf("new: 0x%xn", new);

``````

``````
#include<bitset>
#include<iostream>

template<size_t N>
const std::bitset<N> reverse(const std::bitset<N>& ordered)
{
std::bitset<N> reversed;
for(size_t i = 0, j = N - 1; i < N; ++i, --j)
reversed[j] = ordered[i];
return reversed;
};

// test the function
int main()
{
unsigned long num;
const size_t N = sizeof(num)*8;

std::cin >> num;
std::cout << std::showbase << std::hex;
std::cout << "ordered = " << num << std::endl;
std::cout << "reversed = " << reverse<N>(num).to_ulong() << std::endl;
std::cout << "double_reversed = " << reverse<N>(reverse<N>(num)).to_ulong() << std::endl;
}

``````