CRC-15 giving wrong values - c
I am trying to create a CRC-15 check in c and the output is never correct for each line of the file. I am trying to output the CRC for each line cumulatively next to each line. I use: #define POLYNOMIAL 0xA053 for the divisor and text for the dividend. I need to represent numbers as 32-bit unsigned integers. I have tried printing out the hex values to keep track and flipping different shifts around. However, I just can't seem to figure it out! I have a feeling it has something to do with the way I am padding things. Is there a flaw to my logic?
The CRC is to be represented in four hexadecimal numbers, that sequence will have four leading 0's. For example, it will look like 0000xxxx where the x's are the hexadecimal digits. The polynomial I use is 0xA053.
I thought about using a temp variable and do 4 16 bit chunks of code per line every XOR, however, I'm not quite sure how I could use shifts to accomplish this so I settled for a checksum of the letters on the line and then XORing that to try to calculate the CRC code.
I am testing my code using the following input and padding with . until the string is of length 504 because that is what the pad character needs to be via the requirements given:
"This is the lesson: never give in, never give in, never, never, never, never - in nothing, great or small, large or petty - never give in except to convictions of honor and good sense. Never yield to force; never yield to the apparently overwhelming might of the enemy."
The CRC of the first 64 char line ("This is the lesson: never give in, never give in, never, never,) is supposed to be 000015fa and I am getting bfe6ec00.
My logic:
In CRCCalculation I add each character to a 32-bit unsigned integer and after 64 (the length of one line) I send it into the XOR function.
If it the top bit is not 1, I shift the number to the left one
causing 0s to pad the right and loop around again.
If the top bit is 1, I XOR the dividend with the divisor and then shift the dividend to the left one.
After all calculations are done, I return the dividend shifted to the left four ( to add four zeros to the front) to the calculation function
Add result to the running total of the result
Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <ctype.h>
#define POLYNOMIAL 0xA053
void crcCalculation(char *text, int length)
{
int i;
uint32_t dividend = atoi(text);
uint32_t result;
uint32_t sumText = 0;
// Calculate CRC
printf("\nCRC 15 calculation progress:\n");
i = length;
// padding
if(i < 504)
{
for(; i!=504; i++)
{
// printf("i is %d\n", i);
text[i] = '.';
}
}
// Try calculating by first line of crc by summing the values then calcuating, then add in the next line
for (i = 0; i < 504; i++)
{
if(i%64 == 0 && i != 0)
{
result = XOR(POLYNOMIAL, sumText);
printf(" - %x\n",result);
}
sumText +=(uint32_t)text[i];
printf("%c", text[i]);
}
printf("\n\nCRC15 result : %x\n", result);
}
uint32_t XOR(uint32_t divisor, uint32_t dividend)
{
uint32_t divRemainder = dividend;
uint32_t currentBit;
// Note: 4 16 bit chunks
for(currentBit = 32; currentBit > 0; --currentBit)
{
// if topbit is 1
if(divRemainder & 0x80)
{
//divRemainder = (divRemainder << 1) ^ divisor;
divRemainder ^= divisor;
printf("%x %x\n", divRemainder, divisor);
}
// else
// divisor = divisor >> 1;
divRemainder = (divRemainder << 1);
}
//return divRemainder; , have tried shifting to right and left, want to add 4 zeros to front so >>
//return divRemainder >> 4;
return divRemainder >> 4;
}
The first issue I see is the top bit check, it should be:
if(divRemainder & 0x8000)
The question doesn't state if the CRC is bit reflected (xor data into low order bits of CRC, right shift for cycle) or not (xor data into high order bits of CRC, left shift for cycle), so I can't offer help for the rest of the code.
The question doesn't state the initial value of CRC (0x0000 or 0x7fff), or if the CRC is post complemented.
The logic for a conventional CRC is:
xor a byte of data into the CRC (upper or lower bits)
cycle the CRC 8 times (or do a table lookup)
After generating the CRC for an entire message, the CRC can be appended to the message. If a CRC is generated for a message with the appended CRC and there are no errors, the CRC will be zero (or a constant value if the CRC is post complemented).
here is a typical CRC16, extracted from: <www8.cs.umu.se/~isak/snippets/crc-16.c>
#define POLY 0x8408
/*
// 16 12 5
// this is the CCITT CRC 16 polynomial X + X + X + 1.
// This works out to be 0x1021, but the way the algorithm works
// lets us use 0x8408 (the reverse of the bit pattern). The high
// bit is always assumed to be set, thus we only use 16 bits to
// represent the 17 bit value.
*/
unsigned short crc16(char *data_p, unsigned short length)
{
unsigned char i;
unsigned int data;
unsigned int crc = 0xffff;
if (length == 0)
return (~crc);
do
{
for (i=0, data=(unsigned int)0xff & *data_p++;
i < 8;
i++, data >>= 1)
{
if ((crc & 0x0001) ^ (data & 0x0001))
crc = (crc >> 1) ^ POLY;
else crc >>= 1;
}
} while (--length);
crc = ~crc;
data = crc;
crc = (crc << 8) | (data >> 8 & 0xff);
return (crc);
}
Since you want to calculate a CRC15 rather than a CRC16, the logic will be more complex as cannot work with whole bytes, so there will be a lot of bit shifting and ANDing to extract the desire 15 bits.
Note: the OP did not mention if the initial value of the CRC is 0x0000 or 0x7FFF, nor if the result is to be complemented, nor certain other criteria, so this posted code can only be a guide.
Related
How to calculate crc for byte array?
I understand how CRC for a byte can be calculated by modulo 2 division with a polynomial but I don't understand how do you calculate CRC for data consisting of byte arrays. CRC for a single byte can be calculated by this following code #define WIDTH 8 #define TOPBIT 1 << 7 #define POLYNOMIAL 0xD8 uint8_t(uint8_t const message) { uint8_t remainder = 0; remainder ^= message; for (uint8_t bit = 8; bit > 0; --bit) { if (remainder & TOPBIT) { remainder = (remainder << 1) ^ POLYNOMIAL; } else { remainder = (remainder << 1); } } return (remainder); } but what about byte array ? I found above code on this site, Author also gave the code for byte array where he just XOR'ed current remainder with next byte remainder ^= (message[byte] << (WIDTH - 8)); I don't quite understand why? why he XOR'ed to get in the next byte into remainder?
See A painless guide to CRC error detection algorithms. It has everything on CRCs, including your question. An array is treated as a single massive number so the remainder is carried over to the next byte. CRC is the remainder that is left over at the end.
Look at page on wikipedia Mathematics of cyclic redundancy checks It turns out that CRC is a linear operation meaning that crc(x^y^z) = crc(x)crc(y)crc(x) and hence the author XOR'd the remainder of previous byte with the next byte
CRC code and implementation compatibility
the mechanism or the steps for the CRC checksum is easy , but the software is somehow much complicated and there are some steps in software that are not compatible with the steps of CRC the following picture is the steps for getting the checksum of the CRC ( which is simply a modulo 2 division): the checksum is the remainder = 001 the software for calculating the CRC checksum is for a string of bits is: /* * The width of the CRC calculation and result. * Modify the typedef for a 16 or 32-bit CRC standard. */ typedef uint8_t crc; #define WIDTH (8 * sizeof(crc)) #define TOPBIT (1 << (WIDTH - 1)) crc crcSlow(uint8_t const message[], int nBytes) { crc remainder = 0; /* * Perform modulo-2 division, a byte at a time. */ for (int byte = 0; byte < nBytes; ++byte) { /* * Bring the next byte into the remainder. */ remainder ^= (message[byte] << (WIDTH - 8)); /* * Perform modulo-2 division, a bit at a time. */ for (uint8_t bit = 8; bit > 0; --bit) { /* * Try to divide the current data bit. */ if (remainder & TOPBIT) { remainder = (remainder << 1) ^ POLYNOMIAL; } else { remainder = (remainder << 1); } } } /* * The final remainder is the CRC result. */ return (remainder); } I see that there is incompatibility in the software in the part( remainder<<1 ) because the shifting will always add 0 at the right even if the following bit is not 0. and also in the part: remainder ^= (message[byte] << (WIDTH - 8)); when putting the first byte I don't see problem because the initial value is because the initial value is 0, but when inserting the next bytes why we xor every byte of them with the previous remainder
The code example appears to use a variable sized CRC, where the size of the CRC is WIDTH. POLYNOMIAL is the bottom WIDTH bits of a WIDTH+1 bit polynomial, which will have the least significant bit set to 1. Since the operations are XOR, the order in which the data bits are XOR'ed into the remainder doesn't matter, so 8 data bits can be XOR'ed into the upper bits of remainder at the same time. Then the bit at a time feedback cycle occurs for 8 bits. Since the bottom bit of POLYNOMIAL is a 1, that will keep the cycle going, as long as there are any 1 bits in the data.
How to return multiple bits from a number in C
I have a function to extract a single bit from a number: int getBit (int value, int position) { return value & (1 << position)); } But how do I do it for a range (both for signed and unsigned numbers)? For instance: get bits 10:14 from 0x12345678 (signed 0) = 0x15 int getField (int value, int hi, int lo, bool isSigned)
I suspect you might want to approach the entire problem in a different way. Rather than extracting bits, why not just use bit masks. For example, to check if the most significant bit in a byte is enabled: if(byte & 0xf0) {} To check for the least significant bit it would be: if(byte & 0x01) {} To check for multiple (or a "range") of bits, say the low order nibble: if(byte & 0x0f) {} From what you've said, I suspect this is much closer to what you want and much simpler than shifting to extract bits.
You just have to create a mask: int createMask(int a, int b){ int c = a; int mask = 0; /*First we set the lenght of the mask*/ while(c <= b){ /*Including b*/ mask <<= 1; mask = mask|1; c++; } /*Then we set the position to the mask, the first bit is in the position 0*/ c=0; while(c<a){ c++; mask <<= 1 ; } return mask; } I haven't tested the function yet, but its just for explaining a way to make a mask. And the final function may be something like this: int getBits(int value, int a, int b){ int mask = createMask(a, b); mask &= value; //Now we have to move the bits to the right while(a>0){ mask >>= 1; a--; } return mask; } An example, if you want the first 6 bits, you have to code: getBits(myValue, 0, 5). Im not sure what did you mean about the signeds and unsigneds numbers, but i hope it can help you. Srry for my english.
That was a bit of fun :) In three easy steps: shift your value right by the amount lo and decrease hi with lo. This simplifies the problem to 'get the lowest hi bits'. clip off the highest bits -- a custom mask is created on the fly. if necessary, use the highest bit to sign-extend the result (bitfiddling based on Sign extending from a constant bit width in C#). I don't know the reason for the suggested function prototype, but I would suggest using the order lo, hi rather than hi, lo. Somehow 10,14 feels more natural than the other way around, even though bits count down from high to low, when counted left to right -- the computer is supposed to make things easier for us! #include <stdio.h> #include <stdbool.h> int getField (int value, int hi, int lo, bool isSigned) { /* step 1: clip off lower bits */ value >>= lo; hi -= lo-1; /* step 2: clip off higher bits */ value &= ~(-1<<hi); /* step 3: extend sign */ if (isSigned && (value & (1<<(hi-1)))) value |= -(1<<hi); return value; } int main (void) { int i; i = getField (0x123456c8, 14,10, true); printf ("i = %d / %Xh\n", i,i); return 0; } Result: i = -11 / FFFFFFF5h which is the correct bit set: 16 12 8 4 0 <- bit position ...4 5 6 7 8 <- value 0100 0101 0110 0111 1000 <- bitwise --- -- <- mask 101 01 <- result ..111101 01 < sign extended result
Easy way to convert a string of 0's and 1's into a character? Plain C
I'm doing a steganography project where I read in bytes from a ppm file and add the least significant bit to an array. So once 8 bytes are read in, I would have 8 bits in my array, which should equal some character in a hidden message. Is there an easy way to convert an array of 0's and 1's into an ascii value? For example, the array: char bits[] = {0,1,1,1,0,1,0,0} would equal 't'. Plain C Thanks for all the answers. I'm gonna give some of these a shot.
A simple for loop would work - something like unsigned char ascii = 0; unsigned char i; for(i = 0; i < 8; i++) ascii |= (bits[7 - i] << i); There might be a faster way to do this, but this is a start at least.
I wouldn't store the bits in an array -- I'd OR them with a char. So you start off with a char value of 0: char bit = 0; When you get the first bit, OR it with what you have: bit |= bit_just_read; Keep doing that with each bit, shifting appropriately; i.e., after you get the next bit, do bit |= (next_bit << 1);. And so forth. After you read your 8 bits, bit will be the appropriate ASCII value, and you can print it out or do whatever with it you want to do.
I agree with mipadi, don't bother storing in an array first, that's kind of pointless. Since you have to loop or otherwise keep track of the array index while reading it in, you might as well do it in one go. Something like this, perhaps? bits = 0; for ( i = 0; i < 8; ++i ) { lsb = get_byte_from_ppm_somehow() & 0x01; bits <<= 1 | lsb; }
As long as the bit endian is correct, this should work and compile down pretty small. If the bit endian is backwards then you should be able to change the initial value of mask to 1, the mask shift to <<= , and you might need to have (0x0ff & mask) as the do{}while conditional if your compiler doesn't do what it's supposed to with byte sized variables. Don't forget to do something for the magic functions that I included where I didn't know what you wanted or how you did something #include <stdint.h> // needed for uint8_t ... uint8_t acc, lsb, mask; uint8_t buf[SOME_SIZE]; size_t len = 0; while (is_there_more_ppm_data()) { acc = 0; mask = 0x80; // This is the high bit do { if (!is_there_more() ) { // I don't know what you think should happen if you run out on a non-byte boundary EARLY_END_OF_DATA(); break; } lsb = 1 & get_next_ppm_byte(); acc |= lsb ? mask : 0; // You could use an if statement mask >>= 1; } while (mask); buf[len] = acc; // NOTE: I didn't worry about the running off the end of the buff, but you should. len++; }
Bit reversal of an integer, ignoring integer size and endianness
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