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

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

sergiusz_mAuthor Commented:

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.

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One of a set of tools we're offering as a way to say thank you for being a part of the community.

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.

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:

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

sergiusz_mAuthor Commented:

Thanks for All
Dimitry you are genious, yes it works llike that

> 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