bitXor 8 bytes of data

Neil Allan neil at yippo.co.uk
Wed Dec 9 14:48:51 EST 2009


After some deep thought today, I achieved my desired goal by xoring each 
individual byte. I could swear I tried this before, but anyhoo it's working 
now!

put byte 5 to 12 of tPointData into tEncLat -- read out lats, lons and keys 
from 129 bytes of tPointData

put byte 13 to 20 of tPointData into tEncLon


put Byte 127 of tPointData into tKey1 -- get Encryption bytes from 
tPointData

put byte 128 of tPointData into tKey2

put byte 129 of tPointData into tKey3

etcetera...


-- decrypt lat

replace byte 4 of tEncLat with numToByte((byteToNum(byte 4 of tEncLat) 
bitXor (byteToNum(tKey1)))) in tEncLat

replace byte 5 of tEncLat with numToByte((byteToNum(byte 5 of tEncLat) 
bitXor (byteToNum(tKey2)))) in tEncLat

replace byte 6 of tEncLat with numToByte((byteToNum(byte 6 of tEncLat) 
bitXor (byteToNum(tKey3)))) in tEncLat


Thanks anyway, I was starting to panic after promising an alpha before 
Christmas!!

Regards

Neil

From: "Alex Tweedly" <alex at tweedly.net>
To: "How to use Revolution" <use-revolution at lists.runrev.com>
Sent: Wednesday, December 09, 2009 12:04 AM
Subject: Re: bitXor 8 bytes of data


> Neil Allan wrote:
>> Does anyone have a way of doing a "simple" bitwise xor  with two 8 byte 
>> signed floating point numbers?
>>
>> I believe the revTalk bitXor command can only accept non signed 
>> "numbers".
>>
>> I have tried performing the bitXor byte by byte on the two strings using 
>> the byteToNum() function on each byte individually but the result is 
>> quite different to what one would get banging the hex equivalent of the 
>> string into "calculator" on windows.
>>
>> Calculator gives me the correct answer. I set it to hex mode then just 
>> type in the 8 bytes of hex, hit xor, type in the other hex string and "="
>>
>> Any Ideas?
>>
> bitXor() will do what you want. Can you show a more complete code
> fragment (including the data you used) and we'll see if it's something
> about how you're setting the data up ?
>
> Thanks
> -- Alex.
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