[Tutor] conversion confusion

[Lloyd Kvam]

int(value2add,16) requires that value2add be a string with characters
from the set: 0123456789abcdefABCDEF
It won't work with any old string, such as 'm'.

The CRC equation doesn't work with strings, it works with integers.  To turn
a single character into its integer representation, python provides the ord
function.
CRCval = ord('m') ^ int(seed,16)

You shouldn't need any hex values within your processing.  Having said that,
to turn the hex string '6d' into an integer, you'd use the int function:
int('6d',16) which equals 109
ord('m') also equals 109

If you want to post the problem code, we can probably help.  I'd expect your
code should look something like:
	input_char = serial.get_a_character()	# I made up the function name
	CRCval = ord(input_char) ^ int(seed,16)
	print "input_char(hex): %x   CRC(hex): %x" % (input_char, CRCval)


One other point.  Your device documentation may show stuff in terms of
C programming code.  In C, 'm' is actually an integer not a string.  So
a C programmer can write:
	'n' - 'm'
and get a result of 1.  In python, we'd write:
	ord('n') - ord('m')
and also get 1.

Hope this helps.  Try giving us a code snippet if necessary.

Stanfield, Vicki {D167~Indianapolis} wrote:

> Well, I still can't get this to work right. It works fine when the value2add is a small number, but when it is an 'm' or some other letter, it doesn't work. If I fake it out and send the 'm' as a string representation of its hex value ('6d'), it will work. So the problem must be that after I hexlify the 'm', it is numeric rather than a string which is what the equation which follows wants.
> 
> CRCval=int(value2add,16) ^ int(seed,16)
> 
> How does one turn a hex number 6d into a string so that the equation will happily take it? I may be going about things strangely, but the value comes in as an 'm' and needs to be a string '6d'. How do I get from one to the other?
> 
> --vicki
> 
> P.S. I apologize if I am being dense here; some parts of this come easy and some with great difficulty.
> 
> 

-- 
Lloyd Kvam
Venix Corp.
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