位反转的最佳算法（C语言实现）

http://blog.jobbole.com/70993/

green_t 提问：

实现如下转换的最佳算法是什么？

0010 0000 => 0000 0100

具体的转换是从MSB->LSB 到 LSB->MSB，所有的位都必须反转，那意味着，这并不是字节顺序的交换。

LSB(Least Significant Bit)，意为最低有效位；MSB(Most Significant Bit)，意为最高有效位。

 

最佳答案（来自）

注意：下面的算法都用C实现，但应该可以迁移到其它语言(只是不那么快的时候可别找我)。

可选方案

内存占用少(32位int，32位机器)(来源于)

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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));   }

最快(查找表)

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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]];

来自于著名的

你可以扩展这个算法到64位int的场景，或者为了更快的速度而牺牲多一些的内存(假设你的L1数据缓存足够大)，有一个64K的查找表且每次反转16位。

其它方案

简单

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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

更快(32位处理器)

1 2	unsigned char b = x; b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;

更快(64位处理器)

1 2	unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;

如果你想在32位int环境这样做，那么只需要把每一个byte反转，然后再颠倒byte的次序即可。如下:

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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

我测试了两个最有效的方案，查找表和按位与(第一个方案)。测试机器为一台笔记本电脑，配置为4G DDR2内存，2.4GHz的
双核T7500处理器，4MB的L2缓存。我使用的是gcc 4.3.2，64位Linux。OpenMP(外加GCC绑定)被用来提高计时器的调度能力。

reverse.c

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#include #include #include   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

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#include #include #include   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; }

在不同的优化级别(Optimizations)，两个方案我都尝试了，每个级别跑3个案例，每个案例反转
1亿个随机的无符号整数。对于查找表方案，bitwise hacks page上面的两种方法(Option 1 and Option 2)我都测试过。
结果如下:

按位与

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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

查找表(Option 1)

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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(Option 2)

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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

如果你比较在意性能，那么使用查找表Option 1(Byte的寻址不出意外的慢)。如果你需要尽可能的利用完每一个Byte内存
(且你也在意bit反转的性能)，那么优化后的按位与方案也还不赖。

附加说明

我知道上面的代码只是一个粗略的版本，非常欢迎大家提供一些优化的建议。以下是我知道的几点:

• 我没有权限访问ICC，那可能更快些(如果你可以测试请在评论中回复)。
• 在一些L1缓存比较大的现代机器上面，64K的查找表可能工作得更好。
• -mtune=native对 -O2/-O3(发生符号重定义的错误)无效，所以我不相信产生的代码是为我的微架构而优化。
• SSE环境下应该有一种方法处理得更快。我不知道怎么做，但又更快的内存复制，批量的按位与，调整的指令集，
  总是有一些手段的。
• 我知道仅仅x86的指令集是危险的，下面是GCC在-O3环境产生的代码，所以比我更厉害的大牛可以检查一下。

32-bit

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.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

更改: 我也尝试在自己机器上使用uint64，看看是否性能有所提高。相对于32-bit性能大概提高了10%。
无论你是每次用64-bit类型去反转2个32-bit的int，或者实际上看作64-bit并分两次来反转，性能都大致相当。
代码如下(对于前者，每次反转2个32-bit的int):

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.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