What is the most efficient algorithm to achieve the following:
0010 0000 => 0000 0100
The conversion is from MSB->LSB to LSB->MSB. All bits must be reversed; that is, this is not endianness-swapping.
Efficient Algorithm for Bit Reversal (from MSB->LSB to LSB->MSB) in C
Viewed 253842
26 Answers
NOTE: All algorithms below are in C, but should be portable to your language of choice (just don't look at me when they're not as fast :)
Options
Low Memory (32-bit int, 32-bit machine)(from here):
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));
}
From the famous Bit Twiddling Hacks page:
Fastest (lookup table):
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]];
You can extend this idea to 64-bit ints, or trade off memory for speed (assuming your L1 Data Cache is large enough), and reverse 16 bits at a time with a 64K-entry lookup table.
Others
Simple
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
Faster (32-bit processor)
unsigned char b = x;
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;
Faster (64-bit processor)
unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
If you want to do this on a 32-bit int, just reverse the bits in each byte, and reverse the order of the bytes. That is:
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);
Results
I benchmarked the two most promising solutions, the lookup table, and bitwise-AND (the first one). The test machine is a laptop w/ 4GB of DDR2-800 and a Core 2 Duo T7500 @ 2.4GHz, 4MB L2 Cache; YMMV. I used gcc 4.3.2 on 64-bit Linux. OpenMP (and the GCC bindings) were used for high-resolution timers.
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;
}
I tried both approaches at several different optimizations, ran 3 trials at each level, and each trial reversed 100 million random unsigned ints. For the lookup table option, I tried both schemes (options 1 and 2) given on the bitwise hacks page. Results are shown below.
Bitwise AND
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
Lookup Table (option 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
Lookup Table (option 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
Conclusion
Use the lookup table, with option 1 (byte addressing is unsurprisingly slow) if you're concerned about performance. If you need to squeeze every last byte of memory out of your system (and you might, if you care about the performance of bit reversal), the optimized versions of the bitwise-AND approach aren't too shabby either.
Caveat
Yes, I know the benchmark code is a complete hack. Suggestions on how to improve it are more than welcome. Things I know about:
- 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 (
ldblew 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-bit
.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
EDIT: I also tried using uint64_t types on my machine to see if there was any performance boost. Performance was about 10% faster than 32-bit, and was nearly identical whether you were just using 64-bit types to reverse bits on two 32-bit int types at a time, or whether you were actually reversing bits in half as many 64-bit values. The assembly code is shown below (for the former case, reversing bits for two 32-bit int types at a time):
.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
Presuming that you have an array of bits, how about this: 1. Starting from MSB, push bits into a stack one by one. 2. Pop bits from this stack into another array (or the same array if you want to save space), placing the first popped bit into MSB and going on to less significant bits from there.
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();
}
Of course the obvious source of bit-twiddling hacks is here: http://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious
Generic
C code. Using 1 byte input data num as example.
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: %d\n", x);
// extract bit on the right, from LSB
y = num & (1 << i);
y = y >> i;
printf("y: %d\n", y);
var = var | (x << i); // apply x
var = var | (y << p); // apply y
}
printf("new: 0x%x\n", new);
Efficient can mean throughput or latency.
For throughout, see the answer by Anders Cedronius, it’s a good one.
For lower latency, I would recommend this code:
uint32_t reverseBits( uint32_t x )
{
#if defined(__arm__) || defined(__aarch64__)
__asm__( "rbit %0, %1" : "=r" ( x ) : "r" ( x ) );
return x;
#endif
// Flip pairwise
x = ( ( x & 0x55555555 ) << 1 ) | ( ( x & 0xAAAAAAAA ) >> 1 );
// Flip pairs
x = ( ( x & 0x33333333 ) << 2 ) | ( ( x & 0xCCCCCCCC ) >> 2 );
// Flip nibbles
x = ( ( x & 0x0F0F0F0F ) << 4 ) | ( ( x & 0xF0F0F0F0 ) >> 4 );
// Flip bytes. CPUs have an instruction for that, pretty fast one.
#ifdef _MSC_VER
return _byteswap_ulong( x );
#elif defined(__INTEL_COMPILER)
return (uint32_t)_bswap( (int)x );
#else
// Assuming gcc or clang
return __builtin_bswap32( x );
#endif
}
Compilers output: https://godbolt.org/z/5ehd89
For other web-searchers who might encounter this question, here is a summary (for C and JavaScript).
For a complete solution in JavaScript, we can first generate the table:
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;
}
This gives us BIT_REVERSAL_TABLE, which is what @MattJ posted:
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
])
Then the algorithms for 8-bit, 16-bit, and 32-bit unsigned integers can be found here:
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];
}
Note, the 32-bit version doesn't work in JavaScript (must convert to using BigInts which is straightforward), but should work in a 64-bit language:
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')}`
}
Note: This solution works in JavaScript for 32-bits:
function reverseBits32(n) {
let res = 0;
for (let i = 0; i < 32; i++) {
res = (res << 1) + (n & 1);
n = n >>> 1;
}
return res >>> 0;
}
All 3 table based solutions will work fine in C. Here is a rough C version:
#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;
}
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