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From: Roger Stafford on 18 Nov 2005 14:34
In article
<4299367.1132302243123.JavaMail.jakarta(a)nitrogen.mathforum.org>, Aaron
<aaronshchang(a)hotmail.com> wrote:

> Hi Roger,
>
> I was actually using that "Painless" guide parallelly when writing this
crc programme. I took to referring to the value hex 29B1 as the reference
value to the string '123456789' just because a check using this string on
a few online CRC-CCITT calculator delivered hex29B1 as the resulting value
and not E5CC as delivered in the paper.. I will check again their
differences especially with the difference between the results, 29B1 and
E5CC as cited in the "
http://www.joegeluso.com/software/articles/ccitt.htm" article.
> Thanks for your help
--------------------
Hi Aaron,

I am sticking my neck out here because I don't have the 'bitshift',
'bitxor', 'bitget', 'unit16', or 'unit8' functions to try this out.
However, it seems to me the following might work. I have followed the
outline of your previous code. If it's right, it should give E5CC (if we
are to believe the website I mentioned.)

a001 = uint16(hex2dec('1021')) %CRC16-CCITT polynomial
str = uint8([hex2dec('FF') hex2dec('FF') '123456789']); % <= Put 'FFFF' here
crc = uint16(0);
for i = 1:length(str)
crc = bitxor(crc,bitshift(uint16(str(i)),8));
for j = 1:8
droppedbit=bitget(crc,16);
crc = bitshift(crc,1);
if droppedbit
crc = bitxor(crc,a001);
end
end
end
..

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford
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