实现以下目标最有效的算法是什么:
0010 0000 => 0000 0100
从MSB->LSB转换为LSB->MSB。所有位都必须反转;也就是说，这不是字节交换。
当前回答
// Purpose: to reverse bits in an unsigned short integer
// Input: an unsigned short integer whose bits are to be reversed
// Output: an unsigned short integer with the reversed bits of the input one
unsigned short ReverseBits( unsigned short a )
{
// declare and initialize number of bits in the unsigned short integer
const char num_bits = sizeof(a) * CHAR_BIT;
// declare and initialize bitset representation of integer a
bitset<num_bits> bitset_a(a);
// declare and initialize bitset representation of integer b (0000000000000000)
bitset<num_bits> bitset_b(0);
// declare and initialize bitset representation of mask (0000000000000001)
bitset<num_bits> mask(1);
for ( char i = 0; i < num_bits; ++i )
{
bitset_b = (bitset_b << 1) | bitset_a & mask;
bitset_a >>= 1;
}
return (unsigned short) bitset_b.to_ulong();
}
void PrintBits( unsigned short a )
{
// declare and initialize bitset representation of a
bitset<sizeof(a) * CHAR_BIT> bitset(a);
// print out bits
cout << bitset << endl;
}
// Testing the functionality of the code
int main ()
{
unsigned short a = 17, b;
cout << "Original: ";
PrintBits(a);
b = ReverseBits( a );
cout << "Reversed: ";
PrintBits(b);
}
// Output:
Original: 0000000000010001
Reversed: 1000100000000000
其他回答
注意:下面所有的算法都是用C语言编写的，但是应该可以移植到你所选择的语言中(当它们没有那么快的时候不要看着我:)
选项
低内存(32位int, 32位机器)(从这里):
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));
}
来自著名的Bit Twiddling Hacks页面:
最快(查找表):
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]];
您可以将此想法扩展到64位整数，或者为了速度而牺牲内存(假设L1数据缓存足够大)，并使用一个64k条目查找表一次反向16位。
其他人
简单的
unsigned int v; // input bits to be reversed
unsigned int r = v & 1; // 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
更快(32位处理器)
unsigned char b = x;
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;
更快(64位处理器)
unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
如果您想在32位整型上执行此操作，只需反转每个字节中的位，并反转字节的顺序。那就是:
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
我对两种最有希望的解决方案进行了基准测试，查找表和按位and(第一个)。测试机器是一台带有4GB DDR2-800和酷睿2 Duo T7500 @ 2.4GHz, 4MB L2缓存的笔记本电脑;YMMV。我在64位Linux上使用gcc 4.3.2。OpenMP(和GCC绑定)用于高分辨率计时器。
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 seconds\n", 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 seconds\n", end-start);
free(ints);
free(ints2);
return 0;
}
我在几个不同的优化中尝试了这两种方法，在每个级别上进行了3次试验，每次试验逆转了1亿个随机无符号整数。对于查找表选项，我尝试了按位hacks页面上给出的两种方案(选项1和2)。结果如下所示。
位和
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
查阅表(选项1)
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
查找表(选项2)
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
结论
如果您关心性能，请使用选项1(字节寻址速度慢，这是意料中事)的查找表。如果您需要从系统中挤出最后一个字节的内存(如果您关心位反转的性能，您可能会这样做)，那么按位- and方法的优化版本也不会太糟糕。
警告
是的，我知道基准代码完全是一种hack。关于如何改进它的建议非常受欢迎。我知道的事情:
I don't have access to ICC. This may be faster (please respond in a comment if you can test this out). A 64K lookup table may do well on some modern microarchitectures with large L1D. -mtune=native didn't work for -O2/-O3 (ld blew up with some crazy symbol redefinition error), so I don't believe the generated code is tuned for my microarchitecture. There may be a way to do this slightly faster with SSE. I have no idea how, but with fast replication, packed bitwise AND, and swizzling instructions, there's got to be something there. I know only enough x86 assembly to be dangerous; here's the code GCC generated on -O3 for option 1, so somebody more knowledgable than myself can check it out:
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)
addq $4, %rsi
cmpq $400000000, %rsi
jne .L3
编辑:我还尝试在我的机器上使用uint64_t类型，看看是否有任何性能提升。性能比32位快10%左右，无论您是一次使用64位类型对两个32位整型反转位，还是实际上将64位值的一半反转位，性能都几乎相同。汇编代码如下所示(对于前一种情况，一次为两个32位整型反转位):
.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)
addq $8, %rsi
cmpq $400000000, %rsi
jne .L3
似乎许多其他帖子都关心速度(即最好=最快)。 简单性怎么样?考虑:
char ReverseBits(char character) {
char reversed_character = 0;
for (int i = 0; i < 8; i++) {
char ith_bit = (c >> i) & 1;
reversed_character |= (ith_bit << (sizeof(char) - 1 - i));
}
return reversed_character;
}
并希望聪明的编译器将为您优化。
如果你想反转一个更长的位列表(包含sizeof(char) * n位)，你可以使用这个函数得到:
void ReverseNumber(char* number, int bit_count_in_number) {
int bytes_occupied = bit_count_in_number / sizeof(char);
// first reverse bytes
for (int i = 0; i <= (bytes_occupied / 2); i++) {
swap(long_number[i], long_number[n - i]);
}
// then reverse bits of each individual byte
for (int i = 0; i < bytes_occupied; i++) {
long_number[i] = ReverseBits(long_number[i]);
}
}
这将把[10000000,10101010]反向转换为[01010101,00000001]。
Anders Cedronius的答案为那些拥有支持AVX2的x86 CPU的人提供了一个很好的解决方案。对于没有AVX支持的x86平台或非x86平台，以下任何一种实现都应该工作良好。
第一个代码是经典二进制分区方法的一个变体，编码的目的是最大限度地利用shift-plus-logic习惯用法，这种习惯用法在各种ARM处理器上都很有用。此外，它使用动态掩码生成，这对于需要多个指令来加载每个32位掩码值的RISC处理器是有益的。x86平台的编译器应该在编译时而不是运行时使用常量传播来计算所有掩码。
/* Classic binary partitioning algorithm */
inline uint32_t brev_classic (uint32_t a)
{
uint32_t m;
a = (a >> 16) | (a << 16); // swap halfwords
m = 0x00ff00ff; a = ((a >> 8) & m) | ((a << 8) & ~m); // swap bytes
m = m^(m << 4); a = ((a >> 4) & m) | ((a << 4) & ~m); // swap nibbles
m = m^(m << 2); a = ((a >> 2) & m) | ((a << 2) & ~m);
m = m^(m << 1); a = ((a >> 1) & m) | ((a << 1) & ~m);
return a;
}
在“计算机编程艺术”的第4A卷中，D. Knuth展示了反转位的聪明方法，这比经典的二进制分区算法所需的操作少得令人惊讶。一个这样的32位操作数算法，我在TAOCP中找不到，在Hacker’s Delight网站上的这个文档中显示。
/* Knuth's algorithm from http://www.hackersdelight.org/revisions.pdf. Retrieved 8/19/2015 */
inline uint32_t brev_knuth (uint32_t a)
{
uint32_t t;
a = (a << 15) | (a >> 17);
t = (a ^ (a >> 10)) & 0x003f801f;
a = (t + (t << 10)) ^ a;
t = (a ^ (a >> 4)) & 0x0e038421;
a = (t + (t << 4)) ^ a;
t = (a ^ (a >> 2)) & 0x22488842;
a = (t + (t << 2)) ^ a;
return a;
}
使用Intel编译器C/ c++编译器13.1.3.198，上述两个函数都能很好地自动向量化XMM寄存器。它们也可以手动向量化，而不需要很多努力。
在我的IvyBridge Xeon E3 1270v2上，使用自动向量化代码，1亿uint32_t字在0.070秒内使用brev_classic()位反转，0.068秒使用brev_knuth()位反转。我注意确保我的基准测试不受系统内存带宽的限制。
当然，玩弄比特的黑客的明显来源是: http://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious
伪代码中的位反转
源-要反转的>字节b00101100 Destination ->反转，也需要为unsigned类型，这样符号位就不会向下传播
复制到临时，因此原始不受影响，还需要为unsigned类型，以便符号位不会自动移位
bytecopy = b0010110
循环8://执行8次 测试字节拷贝是否< 0(负)
set bit8 (msb) of reversed = reversed | b10000000
else do not set bit8
shift bytecopy left 1 place
bytecopy = bytecopy << 1 = b0101100 result
shift result right 1 place
reversed = reversed >> 1 = b00000000
8 times no then up^ LOOP8
8 times yes then done.
