?
Solved

CRC-8 ? Polynomal calculation for RS-232 device!

Posted on 2004-04-19
10
Medium Priority
?
2,253 Views
Last Modified: 2012-05-04
Hi!
I have a big problem because I have a heat device and I wan't to know how it generate CRC checksum.
data send/retrieve by this device are as below:
The last byte is a CRC.
7 0 0 0 105 1 51 ; CRC=51
4 0 0 16             ; CRC=16
7 2 0 0 0 4 196
8 0 0 151 4 11 2 234
7 0 0 0 108 1 57
4 0 11 27
7 0 0 0 100 1 41
4 0 5 21
7 0 0 0 33 1 163
4 0 2 18
7 0 0 0 22 3 207
6 0 2 228 0 185
7 0 0 0 23 3 205
6 0 2 228 0 185
7 0 0 0 23 3 205
6 0 2 241 0 147
7 0 0 0 25 2 208
5 0 4 176 144
7 0 0 0 1 2 224
5 0 2 249 213
7 0 0 0 4 2 234
5 0 3 44 2
7 0 0 0 14 1 253

where first byte signs that sequence has x bytes lenght. Maximum bytes lenght is 8 and minimum is 4.
I have asm code of this exe driver but there's no comment and almost 10 MB so I'll put it on my desk!
The last byte is a CRC, I almost quess how this CRC is calculated because:
for eq.:
7 0 0 0 105 1 51 : so the CRC is 51

five steps:
1. 2*7^0=14
2.2*14^0=28
3. 56^0=56
4.2*56^105=25
5. 2*25^1=51  ; CRC!!!!!

so the alghoritm for eq. in Matlab would be:

function f=countcrc(tab)
%inicjuje poczatkowa sume crc:
crc=bitxor( bitshift(tab(1),1) ,tab(2) );
N=tab(1)-3;
for i=1:N,
crc=bitxor( bitshift(crc,1) ,tab(i+2) );    
end
f=crc;

but the problem is when the  a xor b >255
8 0 0 151 4 11 2 234
 6 0 2 228 0 185
6 0 2 241 0 147
7 0 0 0 158 1 196
8 0 3 118 252 137 0 73

And here I don't know how these values where xor is greather than 255 are generated

Please help me!!!!!!!

Thank's a lot for any advice!!!!!
Sergiusz
0
Comment
Question by:sergiusz_m
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 4
  • 3
  • 2
  • +1
10 Comments
 
LVL 12

Expert Comment

by:stefan73
ID: 10858267
Hi sergiusz_m,
> And here I don't know how these values where xor is greather than 255
> are generated

CRC-8 only uses the lowest 8 bits, so use in your matlab notation
crc=bitand(crc,255)

In ASM you achieve the same just by storing your result as a byte, or by using the "and" instruction.

Cheers,
Stefan
0
 

Author Comment

by:sergiusz_m
ID: 10858364
I know about it but I'm not asking about it!
look at this example:
8 0 0 151 4 11 2 234

so I use 6 steps:

1. 2*8^0=16
2. 2*16^0=32
3. 2*32^151=215 ; a.... ant there is problem because 2*215=430 >255
and now I'd like to know how to go to 234?? So what next in steep 4,5,6????
4. sth ^4= sth2 ; and now if 2*sth2^4 <255 in step 5 use 2*sth2^11 and etc..
5.
6.

there are others examples that makes xor >255:
 6 0 2 228 0 185
6 0 2 241 0 147
7 0 0 0 158 1 196

for example:
7 0 0 0 158 1 196

five steps:
1.2*7^0=14
2.2*14^0=28
3.2*28^0=56
4.2*56^158=238 ; and 2*238>255 but 196 xor 1=197 so I can quess 197 in step 5
5. 197 ^1=196

but how to go to 197 ?????? xor table or what????

So the control querende would be if ( (2*a^b) &128 !=0) ....... and what here????

The solution could be also in C/C++/Java not only in matlab.

Thanks for all
sergiusz_m
0
 
LVL 12

Expert Comment

by:stefan73
ID: 10858699
sergiusz_m,
OK. The only code I found so far was a CRC16, but it should illustrate nicely:
http://www.eagleairaust.com.au/code/crc16.htm

// Update the CRC for transmitted and received data using
// the CCITT 16bit algorithm (X^16 + X^12 + X^5 + 1).

    unsigned char ser_data;
    static unsigned int crc;

    crc  = (unsigned char)(crc >> 8) | (crc << 8);
    crc ^= ser_data;
    crc ^= (unsigned char)(crc & 0xff) >> 4;
    crc ^= (crc << 8) << 4;
    crc ^= ((crc & 0xff) << 4) << 1;

...but this should give you a good first start.

Stefan
0
Free Tool: SSL Checker

Scans your site and returns information about your SSL implementation and certificate. Helpful for debugging and validating your SSL configuration.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 

Author Comment

by:sergiusz_m
ID: 10861329
OK Stefan thanks, this CRC looks very nice but how and where can I use it in my code????

could you type mi sample alghoritm how to use it for this data: 7 0 0 0 158 1 196 ? how to obtain this 196 using this?

Thanks a lot and you'll take a points.

Sergiusz
0
 
LVL 5

Expert Comment

by:twobitadder
ID: 10862250
C code for crc:
http://ozma.ssl.berkeley.edu/~dbb/fuse/crc-code.htm

A description of the shifting/xoring (after the basics on bits :) )
http://www.4d.com/docs/CMU/CMU79909.HTM
0
 

Author Comment

by:sergiusz_m
ID: 10862448
twobitadder that's OK but I have my own case described below, so don't write how to write crc in C or how CRC works when I know nothing about used polyval in this chesksum and about behaviour when xor > 255. I've spend  hours sitting on it and I don't know what is "played" where xor >255 - the solution for it is my problem solution!!!!!! I only know (alghoritm in matlab see below) what if xor is lower than 255 so 50% of alghoritm.

Questionis still open!

Regards,
Sergiusz
0
 
LVL 11

Accepted Solution

by:
dimitry earned 500 total points
ID: 10865177
I think I understand this CRC, but check me, because I may be wrong.
If your current (CRC & 0x80) == 0x80 (it means that next shift will be > 255) then you need to XOR it with 0x19.
Take a look at the code:

#include <stdio.h>

#define updcrc(cp, crc) ((unsigned char)(crc << 1) ^ (unsigned char)cp)

unsigned char array[] = { 8, 0, 0, 151, 4, 11, 2, 234 };
// unsigned char array[] = { 7, 0, 0, 0, 105, 1, 51 };
// unsigned char array[] = { 7, 0, 0, 0, 1, 2, 224 };
// unsigned char array[] = { 7, 0, 0, 0, 4, 2, 234 };
// unsigned char array[] = { 7, 0, 0, 0, 158, 1, 196 };
// unsigned char array[] = { 8, 0, 3, 118, 252, 137, 0, 73 };
// unsigned char array[] = { 6, 0, 2, 241, 0, 147 };
// unsigned char array[] = { 7, 2, 0, 0, 0, 4, 196 };
// unsigned char array[] = { 5, 0, 3, 44, 2 };
int main( void )
{
  unsigned short mycrc = 0;
  int i;

  printf("Result = [%x]\n", array[0] );  
  for(i=1,mycrc=array[0];( i < (sizeof(array)-1) );i++) {
    if( mycrc & 0x80 )
      mycrc = (updcrc( array[i], mycrc )) ^ 0x19;
    else
      mycrc = updcrc( array[i], mycrc );
    printf("Result = [%x] -> %x\n", array[i], mycrc );
  }
  if( mycrc != array[i] )
    printf("Error CRC calculation %x != %x\n", mycrc, array[i]);
  printf("Result = %d\n", (unsigned short)mycrc );
  return( 0 );
}
0
 
LVL 12

Expert Comment

by:stefan73
ID: 10876365
sergiusz_m,
Byte overflows are ignored. If you look at Dimitry's code, you'll see that all intermediate values are casted to unsigned char in the updcrc macro. That's very much like doing "& 0xff".

In other words: You just use the lower 8 bits and discard all others.

Stefan
0
 

Author Comment

by:sergiusz_m
ID: 10877506
Thanks for All
Dimitry you are genious, yes it works llike that

THANKS A LOO.......T.

Sergiusz!
0
 
LVL 11

Expert Comment

by:dimitry
ID: 10880236
You are welcome.
0

Featured Post

Free Tool: Port Scanner

Check which ports are open to the outside world. Helps make sure that your firewall rules are working as intended.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Check out the latest tech news, community articles, and expert highlights in August's newsletter.
Hey fellow admins! This time, I have a little fairy tale for you. As many tales do, it starts boring and then gets pretty gory. I hope you like it. TL;DR: It is about an important security matter, you should read it if you run or administer Windows …
Monitoring a network: how to monitor network services and why? Michael Kulchisky, MCSE, MCSA, MCP, VTSP, VSP, CCSP outlines the philosophy behind service monitoring and why a handshake validation is critical in network monitoring. Software utilized …
If you’ve ever visited a web page and noticed a cool font that you really liked the look of, but couldn’t figure out which font it was so that you could use it for your own work, then this video is for you! In this Micro Tutorial, you'll learn yo…
Suggested Courses

771 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question