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
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.
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
您可能希望使用标准模板库。它可能比上面提到的代码慢。然而，在我看来，这似乎更清楚，更容易理解。
#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;
}
我很好奇原始旋转有多快。 在我的机器(i7@2600)上，1,500,150,000次迭代的平均值为27.28 ns(在131,071个64位整数的随机集上)。
优点:占用内存少，代码简单。我想说它也没有那么大。对于任何输入(128个算术SHIFT运算+ 64个逻辑and运算+ 64个逻辑OR运算)，所需的时间都是可预测的常量。
我比较了@Matt J获得的最佳时间，他有公认的答案。如果我没有看错他的答案，他得到的最好结果是0.631739秒，100万次迭代，这导致平均每次旋转631 ns。
我使用的代码片段如下:
unsigned long long reverse_long(unsigned long long x)
{
return (((x >> 0) & 1) << 63) |
(((x >> 1) & 1) << 62) |
(((x >> 2) & 1) << 61) |
(((x >> 3) & 1) << 60) |
(((x >> 4) & 1) << 59) |
(((x >> 5) & 1) << 58) |
(((x >> 6) & 1) << 57) |
(((x >> 7) & 1) << 56) |
(((x >> 8) & 1) << 55) |
(((x >> 9) & 1) << 54) |
(((x >> 10) & 1) << 53) |
(((x >> 11) & 1) << 52) |
(((x >> 12) & 1) << 51) |
(((x >> 13) & 1) << 50) |
(((x >> 14) & 1) << 49) |
(((x >> 15) & 1) << 48) |
(((x >> 16) & 1) << 47) |
(((x >> 17) & 1) << 46) |
(((x >> 18) & 1) << 45) |
(((x >> 19) & 1) << 44) |
(((x >> 20) & 1) << 43) |
(((x >> 21) & 1) << 42) |
(((x >> 22) & 1) << 41) |
(((x >> 23) & 1) << 40) |
(((x >> 24) & 1) << 39) |
(((x >> 25) & 1) << 38) |
(((x >> 26) & 1) << 37) |
(((x >> 27) & 1) << 36) |
(((x >> 28) & 1) << 35) |
(((x >> 29) & 1) << 34) |
(((x >> 30) & 1) << 33) |
(((x >> 31) & 1) << 32) |
(((x >> 32) & 1) << 31) |
(((x >> 33) & 1) << 30) |
(((x >> 34) & 1) << 29) |
(((x >> 35) & 1) << 28) |
(((x >> 36) & 1) << 27) |
(((x >> 37) & 1) << 26) |
(((x >> 38) & 1) << 25) |
(((x >> 39) & 1) << 24) |
(((x >> 40) & 1) << 23) |
(((x >> 41) & 1) << 22) |
(((x >> 42) & 1) << 21) |
(((x >> 43) & 1) << 20) |
(((x >> 44) & 1) << 19) |
(((x >> 45) & 1) << 18) |
(((x >> 46) & 1) << 17) |
(((x >> 47) & 1) << 16) |
(((x >> 48) & 1) << 15) |
(((x >> 49) & 1) << 14) |
(((x >> 50) & 1) << 13) |
(((x >> 51) & 1) << 12) |
(((x >> 52) & 1) << 11) |
(((x >> 53) & 1) << 10) |
(((x >> 54) & 1) << 9) |
(((x >> 55) & 1) << 8) |
(((x >> 56) & 1) << 7) |
(((x >> 57) & 1) << 6) |
(((x >> 58) & 1) << 5) |
(((x >> 59) & 1) << 4) |
(((x >> 60) & 1) << 3) |
(((x >> 61) & 1) << 2) |
(((x >> 62) & 1) << 1) |
(((x >> 63) & 1) << 0);
}
对于其他可能遇到这个问题的网络搜索者，这里有一个总结(针对C和JavaScript)。
对于JavaScript的完整解决方案，我们可以首先生成表:
const BIT_REVERSAL_TABLE = new Array(256)
for (var i = 0; i < 256; ++i) {
var v = i, r = i, s = 7;
for (v >>>= 1; v; v >>>= 1) {
r <<= 1;
r |= v & 1;
--s;
}
BIT_REVERSAL_TABLE[i] = (r << s) & 0xff;
}
这给了我们BIT_REVERSAL_TABLE，这是@MattJ发布的:
const BIT_REVERSAL_TABLE = new Uint8Array([
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
])
8位、16位和32位无符号整数的算法可以在这里找到:
function reverseBits8(n) {
return BIT_REVERSAL_TABLE[n]
}
function reverseBits16(n) {
return (BIT_REVERSAL_TABLE[(n >> 8) & 0xff] |
BIT_REVERSAL_TABLE[n & 0xff] << 8)
}
function reverseBits32(n) {
return (BIT_REVERSAL_TABLE[n & 0xff] << 24) |
(BIT_REVERSAL_TABLE[(n >>> 8) & 0xff] << 16) |
(BIT_REVERSAL_TABLE[(n >>> 16) & 0xff] << 8) |
BIT_REVERSAL_TABLE[(n >>> 24) & 0xff];
}
注意，32位版本不能在JavaScript中工作(必须转换为使用bigint，这很简单)，但应该可以在64位语言中工作:
log8(0b11000100)
log16(0b1110001001001100)
log32(0b11110010111110111100110010101011)
// 0b11000100 => 0b00100011
// 0b1110001001001100 => 0b0011001001000111
// doesn't work in JS it seems:
// 0b11110010111110111100110010101011 => 0b0-101010110011000010000010110001
function log8(n) {
console.log(`${bits(n, 8)} => ${bits(reverseBits8(n), 8)}`)
}
function log16(n) {
console.log(`${bits(n, 16)} => ${bits(reverseBits16(n), 16)}`)
}
function log32(n) {
console.log(`${bits(n, 32)} => ${bits(reverseBits32(n), 32)}`)
}
function bits(n, size) {
return `0b${n.toString(2).padStart(size, '0')}`
}
注意:这个解决方案适用于JavaScript的32位:
function reverseBits32(n) {
let res = 0;
for (let i = 0; i < 32; i++) {
res = (res << 1) + (n & 1);
n = n >>> 1;
}
return res >>> 0;
}
所有3个基于表格的解决方案都可以在C中正常工作。下面是一个粗略的C版本:
#include <stdlib.h>
static uint8_t* BIT_REVERSAL_TABLE;
uint8_t*
make_bit_reversal_table() {
uint8_t *table = malloc(256 * sizeof(uint8_t));
uint8_t i;
for (i = 0; i < 256 ; ++i) {
uint8_t v = i;
uint8_t r = i;
uint8_t s = 7;
for (v = v >> 1; v; v = v >> 1) {
r <<= 1;
r |= v & 1;
--s;
}
table[i] = (r << s) & 0xff;
}
return table;
}
uint8_t
reverse_bits_8(uint8_t n) {
return BIT_REVERSAL_TABLE[n];
}
uint16_t
reverse_bits_16(uint16_t n)
{
return (BIT_REVERSAL_TABLE[(n >> 8) & 0xff]
| BIT_REVERSAL_TABLE[n & 0xff] << 8);
}
uint32_t
reverse_bits_32(uint32_t n) {
return (BIT_REVERSAL_TABLE[n & 0xff] << 24)
| (BIT_REVERSAL_TABLE[(n >> 8) & 0xff] << 16)
| (BIT_REVERSAL_TABLE[(n >> 16) & 0xff] << 8)
| BIT_REVERSAL_TABLE[(n >> 24) & 0xff];
}
int
main(void) {
BIT_REVERSAL_TABLE = make_bit_reversal_table();
return 0;
}
另一个基于循环的解决方案，在数量较低时快速退出(在c++中用于多种类型)
template<class T>
T reverse_bits(T in) {
T bit = static_cast<T>(1) << (sizeof(T) * 8 - 1);
T out;
for (out = 0; bit && in; bit >>= 1, in >>= 1) {
if (in & 1) {
out |= bit;
}
}
return out;
}
或者C语言中unsigned int
unsigned int reverse_bits(unsigned int in) {
unsigned int bit = 1u << (sizeof(T) * 8 - 1);
unsigned int out;
for (out = 0; bit && in; bit >>= 1, in >>= 1) {
if (in & 1)
out |= bit;
}
return out;
}
