I think I might have been asleep in my CS class when they talked about Bit Positions, so I am hoping someone can lend a hand.
I have a unsigned 32-bit integer (Lets use the value: 28)
According to some documentation I am going over, the value of the integer contains flags specifying various things.
Bit positions within the flag are numbered from 1 (low-order) to 32 (high-order).
All undefined flag bits are reserved and must be set to 0.
I have a Table that shows the meanings of the flags, with meaning for the numbers 1-10.
I am hoping that someone can try and explain to me what this all means and how to find the "flag" value(s) from a number like, 28, based off of bit position.
Thanks
28 converts to 11100 in binary. That means bits 1 and 2 are not set and bits 3, 4 and 5 are set.
A few points: first, anybody who's really accustomed to C will usually start the numbering at 0, not 1. Second, you can test of individual flags with the bitwise and operator (&), as in:
#define flag1 1 // 1 = 00 0001
#define flag2 2 // 2 = 00 0010
#define flag3 4 // 4 = 00 0100
#define flag4 8 // 8 = 00 1000
#define flag5 16 // 16 = 01 0000
#define flag6 32 // 32 = 10 0000
if (myvalue & flag1)
// flag1 was set
if (myvalue & flag4)
// flag4 was set
and so on. You can also check which bits are set in a loop:
#include <stdio.h>
int main() {
int myvalue = 28;
int i, iter;
for (i=1, iter=1; i<256; i<<=1, iter++)
if (myvalue & i)
printf("Flag: %d set\n", iter);
return 0;
}
should print:
Flag: 3 set
Flag: 4 set
Flag: 5 set
Instead of looping through every single bit, you can instead loop through only the set bits, which can be faster if you expect bits to be sparsely set:
Assume the bit field is in (scalar integer) variable field.
while (field){
temp = field & -field; //extract least significant bit on a 2s complement machine
field ^= temp; // toggle the bit off
//now you could have a switch statement or bunch of conditionals to test temp
//or get the index of the bit and index into a jump table, etc.
}
Works pretty well when the bit field is not limited to the size of a single data type, but could be of some arbitrary size. In that case, you can extract 32 (or whatever your register size is) bits at a time, test it against 0, and then move on to the next word.
To get an int with the value 0 or 1 representing just the nth bit from that integer, use:
int bitN = (value >> n) & 1;
But that's not usually what you want to do. A more common idiom is this:
int bitN = value & (1 << n);
In this case bitN will be 0 if the nth bit is not set, and non-zero in the case that the nth bit is set. (Specifically, it'll be whatever value comes out with just the nth bit set.)
Assuming flags is unsigned...
int flag_num = 1;
while (flags != 0)
{
if ((flags&1) != 0)
{
printf("Flag %d set\n", flags);
}
flags >>= 1;
flag_num += 1;
}
If flags is signed you should replace
flags >>= 1;
with
flags = (flags >> 1) & 0x7fffffff;
Use a log function, with base 2. In python, that would look like:
import math
position = math.log(value, 2)
If position is not an integer, then more than 1 bit was set to 1.
A slight variation of #invaliddata's answer-
unsigned int tmp_bitmap = x;
while (tmp_bitmap > 0) {
int next_psn = __builtin_ffs(tmp_bitmap) - 1;
tmp_bitmap &= (tmp_bitmap-1);
printf("Flag: %d set\n", next_psn);
}
// You can check the bit set positions of 32 bit integer.
// That's why the check is added "i != 0 && i <= val" to iterate till
// the end bit position.
void find_bit_pos(unsigned int val) {
unsigned int i;
int bit_pos;
printf("%u::\n", val);
for(i = 1, bit_pos = 1; i != 0 && i <= val; i <<= 1, bit_pos++) {
if(val & i)
printf("set bit pos: %d\n", bit_pos);
}
}
An MSVC variation of #boolAeon's answer
#include <vector>
#include <intrin.h>
std::vector<unsigned long> poppos(const unsigned long input)
{
std::vector<unsigned long> result;
result.reserve(sizeof(input) * CHAR_BIT);
unsigned long num = input;
unsigned long index = -1;
while (_BitScanForward(&index, num))
{
result.push_back(index);
num &= num - 1;
}
return result;
}
Let's say that you have an array of integers, and you want to find all the positions (32-bit positions) where the bits are set collectively i.e. for a particular bit position how many set bits you will have in total by considering all the integers. In this case what you can do is that check for every Integer and mark its set bit position :
// let arr[n] is an array of integers of size n.
int fq[33] = {0} // frequency array that will contain frequency of set bits at a particular position as 1 based indexing.
for(int i=0; i<n; i++) {
int x = arr[i];
int pos = 1; // bit position
for(int i=1; i<=pow(2,32); i= i<<1) { // i is the bit mask for checking every position and will go till 2^32 because x is an integer.
if(x & i) fq[pos]++;
pos++;
}
}
I want to create a very large array on which I write '0's and '1's. I'm trying to simulate a physical process called random sequential adsorption, where units of length 2, dimers, are deposited onto an n-dimensional lattice at a random location, without overlapping each other. The process stops when there is no more room left on the lattice for depositing more dimers (lattice is jammed).
Initially I start with a lattice of zeroes, and the dimers are represented by a pair of '1's. As each dimer is deposited, the site on the left of the dimer is blocked, due to the fact that the dimers cannot overlap. So I simulate this process by depositing a triple of '1's on the lattice. I need to repeat the entire simulation a large number of times and then work out the average coverage %.
I've already done this using an array of chars for 1D and 2D lattices. At the moment I'm trying to make the code as efficient as possible, before working on the 3D problem and more complicated generalisations.
This is basically what the code looks like in 1D, simplified:
int main()
{
/* Define lattice */
array = (char*)malloc(N * sizeof(char));
total_c = 0;
/* Carry out RSA multiple times */
for (i = 0; i < 1000; i++)
rand_seq_ads();
/* Calculate average coverage efficiency at jamming */
printf("coverage efficiency = %lf", total_c/1000);
return 0;
}
void rand_seq_ads()
{
/* Initialise array, initial conditions */
memset(a, 0, N * sizeof(char));
available_sites = N;
count = 0;
/* While the lattice still has enough room... */
while(available_sites != 0)
{
/* Generate random site location */
x = rand();
/* Deposit dimer (if site is available) */
if(array[x] == 0)
{
array[x] = 1;
array[x+1] = 1;
count += 1;
available_sites += -2;
}
/* Mark site left of dimer as unavailable (if its empty) */
if(array[x-1] == 0)
{
array[x-1] = 1;
available_sites += -1;
}
}
/* Calculate coverage %, and add to total */
c = count/N
total_c += c;
}
For the actual project I'm doing, it involves not just dimers but trimers, quadrimers, and all sorts of shapes and sizes (for 2D and 3D).
I was hoping that I would be able to work with individual bits instead of bytes, but I've been reading around and as far as I can tell you can only change 1 byte at a time, so either I need to do some complicated indexing or there is a simpler way to do it?
Thanks for your answers
If I am not too late, this page gives awesome explanation with examples.
An array of int can be used to deal with array of bits. Assuming size of int to be 4 bytes, when we talk about an int, we are dealing with 32 bits. Say we have int A[10], means we are working on 10*4*8 = 320 bits and following figure shows it: (each element of array has 4 big blocks, each of which represent a byte and each of the smaller blocks represent a bit)
So, to set the kth bit in array A:
// NOTE: if using "uint8_t A[]" instead of "int A[]" then divide by 8, not 32
void SetBit( int A[], int k )
{
int i = k/32; //gives the corresponding index in the array A
int pos = k%32; //gives the corresponding bit position in A[i]
unsigned int flag = 1; // flag = 0000.....00001
flag = flag << pos; // flag = 0000...010...000 (shifted k positions)
A[i] = A[i] | flag; // Set the bit at the k-th position in A[i]
}
or in the shortened version
void SetBit( int A[], int k )
{
A[k/32] |= 1 << (k%32); // Set the bit at the k-th position in A[i]
}
similarly to clear kth bit:
void ClearBit( int A[], int k )
{
A[k/32] &= ~(1 << (k%32));
}
and to test if the kth bit:
int TestBit( int A[], int k )
{
return ( (A[k/32] & (1 << (k%32) )) != 0 ) ;
}
As said above, these manipulations can be written as macros too:
// Due order of operation wrap 'k' in parentheses in case it
// is passed as an equation, e.g. i + 1, otherwise the first
// part evaluates to "A[i + (1/32)]" not "A[(i + 1)/32]"
#define SetBit(A,k) ( A[(k)/32] |= (1 << ((k)%32)) )
#define ClearBit(A,k) ( A[(k)/32] &= ~(1 << ((k)%32)) )
#define TestBit(A,k) ( A[(k)/32] & (1 << ((k)%32)) )
typedef unsigned long bfield_t[ size_needed/sizeof(long) ];
// long because that's probably what your cpu is best at
// The size_needed should be evenly divisable by sizeof(long) or
// you could (sizeof(long)-1+size_needed)/sizeof(long) to force it to round up
Now, each long in a bfield_t can hold sizeof(long)*8 bits.
You can calculate the index of a needed big by:
bindex = index / (8 * sizeof(long) );
and your bit number by
b = index % (8 * sizeof(long) );
You can then look up the long you need and then mask out the bit you need from it.
result = my_field[bindex] & (1<<b);
or
result = 1 & (my_field[bindex]>>b); // if you prefer them to be in bit0
The first one may be faster on some cpus or may save you shifting back up of you need
to perform operations between the same bit in multiple bit arrays. It also mirrors
the setting and clearing of a bit in the field more closely than the second implemention.
set:
my_field[bindex] |= 1<<b;
clear:
my_field[bindex] &= ~(1<<b);
You should remember that you can use bitwise operations on the longs that hold the fields
and that's the same as the operations on the individual bits.
You'll probably also want to look into the ffs, fls, ffc, and flc functions if available. ffs should always be avaiable in strings.h. It's there just for this purpose -- a string of bits.
Anyway, it is find first set and essentially:
int ffs(int x) {
int c = 0;
while (!(x&1) ) {
c++;
x>>=1;
}
return c; // except that it handles x = 0 differently
}
This is a common operation for processors to have an instruction for and your compiler will probably generate that instruction rather than calling a function like the one I wrote. x86 has an instruction for this, by the way. Oh, and ffsl and ffsll are the same function except take long and long long, respectively.
You can use & (bitwise and) and << (left shift).
For example, (1 << 3) results in "00001000" in binary. So your code could look like:
char eightBits = 0;
//Set the 5th and 6th bits from the right to 1
eightBits &= (1 << 4);
eightBits &= (1 << 5);
//eightBits now looks like "00110000".
Then just scale it up with an array of chars and figure out the appropriate byte to modify first.
For more efficiency, you could define a list of bitfields in advance and put them in an array:
#define BIT8 0x01
#define BIT7 0x02
#define BIT6 0x04
#define BIT5 0x08
#define BIT4 0x10
#define BIT3 0x20
#define BIT2 0x40
#define BIT1 0x80
char bits[8] = {BIT1, BIT2, BIT3, BIT4, BIT5, BIT6, BIT7, BIT8};
Then you avoid the overhead of the bit shifting and you can index your bits, turning the previous code into:
eightBits &= (bits[3] & bits[4]);
Alternatively, if you can use C++, you could just use an std::vector<bool> which is internally defined as a vector of bits, complete with direct indexing.
bitarray.h:
#include <inttypes.h> // defines uint32_t
//typedef unsigned int bitarray_t; // if you know that int is 32 bits
typedef uint32_t bitarray_t;
#define RESERVE_BITS(n) (((n)+0x1f)>>5)
#define DW_INDEX(x) ((x)>>5)
#define BIT_INDEX(x) ((x)&0x1f)
#define getbit(array,index) (((array)[DW_INDEX(index)]>>BIT_INDEX(index))&1)
#define putbit(array, index, bit) \
((bit)&1 ? ((array)[DW_INDEX(index)] |= 1<<BIT_INDEX(index)) \
: ((array)[DW_INDEX(index)] &= ~(1<<BIT_INDEX(index))) \
, 0 \
)
Use:
bitarray_t arr[RESERVE_BITS(130)] = {0, 0x12345678,0xabcdef0,0xffff0000,0};
int i = getbit(arr,5);
putbit(arr,6,1);
int x=2; // the least significant bit is 0
putbit(arr,6,x); // sets bit 6 to 0 because 2&1 is 0
putbit(arr,6,!!x); // sets bit 6 to 1 because !!2 is 1
EDIT the docs:
"dword" = "double word" = 32-bit value (unsigned, but that's not really important)
RESERVE_BITS: number_of_bits --> number_of_dwords
RESERVE_BITS(n) is the number of 32-bit integers enough to store n bits
DW_INDEX: bit_index_in_array --> dword_index_in_array
DW_INDEX(i) is the index of dword where the i-th bit is stored.
Both bit and dword indexes start from 0.
BIT_INDEX: bit_index_in_array --> bit_index_in_dword
If i is the number of some bit in the array, BIT_INDEX(i) is the number
of that bit in the dword where the bit is stored.
And the dword is known via DW_INDEX().
getbit: bit_array, bit_index_in_array --> bit_value
putbit: bit_array, bit_index_in_array, bit_value --> 0
getbit(array,i) fetches the dword containing the bit i and shifts the dword right, so that the bit i becomes the least significant bit. Then, a bitwise and with 1 clears all other bits.
putbit(array, i, v) first of all checks the least significant bit of v; if it is 0, we have to clear the bit, and if it is 1, we have to set it.
To set the bit, we do a bitwise or of the dword that contains the bit and the value of 1 shifted left by bit_index_in_dword: that bit is set, and other bits do not change.
To clear the bit, we do a bitwise and of the dword that contains the bit and the bitwise complement of 1 shifted left by bit_index_in_dword: that value has all bits set to one except the only zero bit in the position that we want to clear.
The macro ends with , 0 because otherwise it would return the value of dword where the bit i is stored, and that value is not meaningful. One could also use ((void)0).
It's a trade-off:
(1) use 1 byte for each 2 bit value - simple, fast, but uses 4x memory
(2) pack bits into bytes - more complex, some performance overhead, uses minimum memory
If you have enough memory available then go for (1), otherwise consider (2).
Given an integer typedef:
typedef unsigned int TYPE;
or
typedef unsigned long TYPE;
I have the following code to reverse the bits of an integer:
TYPE max_bit= (TYPE)-1;
void reverse_int_setup()
{
TYPE bits= (TYPE)max_bit;
while (bits <<= 1)
max_bit= bits;
}
TYPE reverse_int(TYPE arg)
{
TYPE bit_setter= 1, bit_tester= max_bit, result= 0;
for (result= 0; bit_tester; bit_tester>>= 1, bit_setter<<= 1)
if (arg & bit_tester)
result|= bit_setter;
return result;
}
One just needs first to run reverse_int_setup(), which stores an integer with the highest bit turned on, then any call to reverse_int(arg) returns arg with its bits reversed (to be used as a key to a binary tree, taken from an increasing counter, but that's more or less irrelevant).
Is there a platform-agnostic way to have in compile-time the correct value for max_int after the call to reverse_int_setup(); Otherwise, is there an algorithm you consider better/leaner than the one I have for reverse_int()?
Thanks.
#include<stdio.h>
#include<limits.h>
#define TYPE_BITS sizeof(TYPE)*CHAR_BIT
typedef unsigned long TYPE;
TYPE reverser(TYPE n)
{
TYPE nrev = 0, i, bit1, bit2;
int count;
for(i = 0; i < TYPE_BITS; i += 2)
{
/*In each iteration, we swap one bit on the 'right half'
of the number with another on the left half*/
count = TYPE_BITS - i - 1; /*this is used to find how many positions
to the left (and right) we gotta move
the bits in this iteration*/
bit1 = n & (1<<(i/2)); /*Extract 'right half' bit*/
bit1 <<= count; /*Shift it to where it belongs*/
bit2 = n & 1<<((i/2) + count); /*Find the 'left half' bit*/
bit2 >>= count; /*Place that bit in bit1's original position*/
nrev |= bit1; /*Now add the bits to the reversal result*/
nrev |= bit2;
}
return nrev;
}
int main()
{
TYPE n = 6;
printf("%lu", reverser(n));
return 0;
}
This time I've used the 'number of bits' idea from TK, but made it somewhat more portable by not assuming a byte contains 8 bits and instead using the CHAR_BIT macro. The code is more efficient now (with the inner for loop removed). I hope the code is also slightly less cryptic this time. :)
The need for using count is that the number of positions by which we have to shift a bit varies in each iteration - we have to move the rightmost bit by 31 positions (assuming 32 bit number), the second rightmost bit by 29 positions and so on. Hence count must decrease with each iteration as i increases.
Hope that bit of info proves helpful in understanding the code...
The following program serves to demonstrate a leaner algorithm for reversing bits, which can be easily extended to handle 64bit numbers.
#include <stdio.h>
#include <stdint.h>
int main(int argc, char**argv)
{
int32_t x;
if ( argc != 2 )
{
printf("Usage: %s hexadecimal\n", argv[0]);
return 1;
}
sscanf(argv[1],"%x", &x);
/* swap every neigbouring bit */
x = (x&0xAAAAAAAA)>>1 | (x&0x55555555)<<1;
/* swap every 2 neighbouring bits */
x = (x&0xCCCCCCCC)>>2 | (x&0x33333333)<<2;
/* swap every 4 neighbouring bits */
x = (x&0xF0F0F0F0)>>4 | (x&0x0F0F0F0F)<<4;
/* swap every 8 neighbouring bits */
x = (x&0xFF00FF00)>>8 | (x&0x00FF00FF)<<8;
/* and so forth, for say, 32 bit int */
x = (x&0xFFFF0000)>>16 | (x&0x0000FFFF)<<16;
printf("0x%x\n",x);
return 0;
}
This code should not contain errors, and was tested using 0x12345678 to produce 0x1e6a2c48 which is the correct answer.
typedef unsigned long TYPE;
TYPE reverser(TYPE n)
{
TYPE k = 1, nrev = 0, i, nrevbit1, nrevbit2;
int count;
for(i = 0; !i || (1 << i && (1 << i) != 1); i+=2)
{
/*In each iteration, we swap one bit
on the 'right half' of the number with another
on the left half*/
k = 1<<i; /*this is used to find how many positions
to the left (or right, for the other bit)
we gotta move the bits in this iteration*/
count = 0;
while(k << 1 && k << 1 != 1)
{
k <<= 1;
count++;
}
nrevbit1 = n & (1<<(i/2));
nrevbit1 <<= count;
nrevbit2 = n & 1<<((i/2) + count);
nrevbit2 >>= count;
nrev |= nrevbit1;
nrev |= nrevbit2;
}
return nrev;
}
This works fine in gcc under Windows, but I'm not sure if it's completely platform independent. A few places of concern are:
the condition in the for loop - it assumes that when you left shift 1 beyond the leftmost bit, you get either a 0 with the 1 'falling out' (what I'd expect and what good old Turbo C gives iirc), or the 1 circles around and you get a 1 (what seems to be gcc's behaviour).
the condition in the inner while loop: see above. But there's a strange thing happening here: in this case, gcc seems to let the 1 fall out and not circle around!
The code might prove cryptic: if you're interested and need an explanation please don't hesitate to ask - I'll put it up someplace.
#ΤΖΩΤΖΙΟΥ
In reply to ΤΖΩΤΖΙΟΥ 's comments, I present modified version of above which depends on a upper limit for bit width.
#include <stdio.h>
#include <stdint.h>
typedef int32_t TYPE;
TYPE reverse(TYPE x, int bits)
{
TYPE m=~0;
switch(bits)
{
case 64:
x = (x&0xFFFFFFFF00000000&m)>>16 | (x&0x00000000FFFFFFFF&m)<<16;
case 32:
x = (x&0xFFFF0000FFFF0000&m)>>16 | (x&0x0000FFFF0000FFFF&m)<<16;
case 16:
x = (x&0xFF00FF00FF00FF00&m)>>8 | (x&0x00FF00FF00FF00FF&m)<<8;
case 8:
x = (x&0xF0F0F0F0F0F0F0F0&m)>>4 | (x&0x0F0F0F0F0F0F0F0F&m)<<4;
x = (x&0xCCCCCCCCCCCCCCCC&m)>>2 | (x&0x3333333333333333&m)<<2;
x = (x&0xAAAAAAAAAAAAAAAA&m)>>1 | (x&0x5555555555555555&m)<<1;
}
return x;
}
int main(int argc, char**argv)
{
TYPE x;
TYPE b = (TYPE)-1;
int bits;
if ( argc != 2 )
{
printf("Usage: %s hexadecimal\n", argv[0]);
return 1;
}
for(bits=1;b;b<<=1,bits++);
--bits;
printf("TYPE has %d bits\n", bits);
sscanf(argv[1],"%x", &x);
printf("0x%x\n",reverse(x, bits));
return 0;
}
Notes:
gcc will warn on the 64bit constants
the printfs will generate warnings too
If you need more than 64bit, the code should be simple enough to extend
I apologise in advance for the coding crimes I committed above - mercy good sir!
There's a nice collection of "Bit Twiddling Hacks", including a variety of simple and not-so simple bit reversing algorithms coded in C at http://graphics.stanford.edu/~seander/bithacks.html.
I personally like the "Obvious" algorigthm (http://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious) because, well, it's obvious. Some of the others may require less instructions to execute. If I really need to optimize the heck out of something I may choose the not-so-obvious but faster versions. Otherwise, for readability, maintainability, and portability I would choose the Obvious one.
Here is a more generally useful variation. Its advantage is its ability to work in situations where the bit length of the value to be reversed -- the codeword -- is unknown but is guaranteed not to exceed a value we'll call maxLength. A good example of this case is Huffman code decompression.
The code below works on codewords from 1 to 24 bits in length. It has been optimized for fast execution on a Pentium D. Note that it accesses the lookup table as many as 3 times per use. I experimented with many variations that reduced that number to 2 at the expense of a larger table (4096 and 65,536 entries). This version, with the 256-byte table, was the clear winner, partly because it is so advantageous for table data to be in the caches, and perhaps also because the processor has an 8-bit table lookup/translation instruction.
const unsigned char table[] = {
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};
const unsigned short masks[17] =
{0,0,0,0,0,0,0,0,0,0X0100,0X0300,0X0700,0X0F00,0X1F00,0X3F00,0X7F00,0XFF00};
unsigned long codeword; // value to be reversed, occupying the low 1-24 bits
unsigned char maxLength; // bit length of longest possible codeword (<= 24)
unsigned char sc; // shift count in bits and index into masks array
if (maxLength <= 8)
{
codeword = table[codeword << (8 - maxLength)];
}
else
{
sc = maxLength - 8;
if (maxLength <= 16)
{
codeword = (table[codeword & 0X00FF] << sc)
| table[codeword >> sc];
}
else if (maxLength & 1) // if maxLength is 17, 19, 21, or 23
{
codeword = (table[codeword & 0X00FF] << sc)
| table[codeword >> sc] |
(table[(codeword & masks[sc]) >> (sc - 8)] << 8);
}
else // if maxlength is 18, 20, 22, or 24
{
codeword = (table[codeword & 0X00FF] << sc)
| table[codeword >> sc]
| (table[(codeword & masks[sc]) >> (sc >> 1)] << (sc >> 1));
}
}
How about:
long temp = 0;
int counter = 0;
int number_of_bits = sizeof(value) * 8; // get the number of bits that represent value (assuming that it is aligned to a byte boundary)
while(value > 0) // loop until value is empty
{
temp <<= 1; // shift whatever was in temp left to create room for the next bit
temp |= (value & 0x01); // get the lsb from value and set as lsb in temp
value >>= 1; // shift value right by one to look at next lsb
counter++;
}
value = temp;
if (counter < number_of_bits)
{
value <<= counter-number_of_bits;
}
(I'm assuming that you know how many bits value holds and it is stored in number_of_bits)
Obviously temp needs to be the longest imaginable data type and when you copy temp back into value, all the extraneous bits in temp should magically vanish (I think!).
Or, the 'c' way would be to say :
while(value)
your choice
We can store the results of reversing all possible 1 byte sequences in an array (256 distinct entries), then use a combination of lookups into this table and some oring logic to get the reverse of integer.
Here is a variation and correction to TK's solution which might be clearer than the solutions by sundar. It takes single bits from t and pushes them into return_val:
typedef unsigned long TYPE;
#define TYPE_BITS sizeof(TYPE)*8
TYPE reverser(TYPE t)
{
unsigned int i;
TYPE return_val = 0
for(i = 0; i < TYPE_BITS; i++)
{/*foreach bit in TYPE*/
/* shift the value of return_val to the left and add the rightmost bit from t */
return_val = (return_val << 1) + (t & 1);
/* shift off the rightmost bit of t */
t = t >> 1;
}
return(return_val);
}
The generic approach hat would work for objects of any type of any size would be to reverse the of bytes of the object, and the reverse the order of bits in each byte. In this case the bit-level algorithm is tied to a concrete number of bits (a byte), while the "variable" logic (with regard to size) is lifted to the level of whole bytes.
Here's my generalization of freespace's solution (in case we one day get 128-bit machines). It results in jump-free code when compiled with gcc -O3, and is obviously insensitive to the definition of foo_t on sane machines. Unfortunately it does depend on shift being a power of 2!
#include <limits.h>
#include <stdio.h>
typedef unsigned long foo_t;
foo_t reverse(foo_t x)
{
int shift = sizeof (x) * CHAR_BIT / 2;
foo_t mask = (1 << shift) - 1;
int i;
for (i = 0; shift; i++) {
x = ((x & mask) << shift) | ((x & ~mask) >> shift);
shift >>= 1;
mask ^= (mask << shift);
}
return x;
}
int main() {
printf("reverse = 0x%08lx\n", reverse(0x12345678L));
}
In case bit-reversal is time critical, and mainly in conjunction with FFT, the best is to store the whole bit reversed array. In any case, this array will be smaller in size than the roots of unity that have to be precomputed in FFT Cooley-Tukey algorithm. An easy way to compute the array is:
int BitReverse[Size]; // Size is power of 2
void Init()
{
BitReverse[0] = 0;
for(int i = 0; i < Size/2; i++)
{
BitReverse[2*i] = BitReverse[i]/2;
BitReverse[2*i+1] = (BitReverse[i] + Size)/2;
}
} // end it's all