Re: simple transport protocol

Roman wrote:
) Hello, Willem!
) You wrote on Mon, 26 May 2008 06:18:53 +0000 (UTC):
)
) W> Suppose the checksum is a simple sum modulo 256.
) W> We want to send the values: 2 5 8 10 15 20
) W> So, the total of what we send is this:
)
) W> length | data | crc
)
)
) W> 0 7 2 5 8 10 15 20 196
)
) W> What I do _is_ sum % 256. Read it again, more carefully.
) Alright, from the example above, considering the values are decimal:
) 2+5+8+10+15+20 = 60, 60 mod 256 = 60, not 196. How then you get 196?

Here's the bit that I wrote that you did not quote:

) >Now, if we calculate the checksum on the seven received bytes:
) >2+5+8+10+15+20+196 (modulo 256) = 0
) >
) >This 0 means the check succeeded, no compare necessary.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
.

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