Conversion of hex pairs to chars...best way?

T. R. Ponn alptex2 at orwell.net
Thu May 22 18:03:01 EDT 2003


Richard...thanks, but I failed to mention that these files are typically 
binary (which of course, isn't a problem) or come to my app in the form 
of a Motorola S-Record or Intel Hex.  My app opens these files, does the 
necessary checking..prepares them as packets...then sends them via write 
to socket to a remote embedded system we've designed.  So, I don't 
believe the base64Encode/Decode will do the job for me.

BTW...we're really excited about this new product and my app has been in 
beta for a few months now...I just can't stand *not* tinkering.  Much 
praise to Rev and this list...without both, we wouldn't have been able 
to do this!  I will be posting our app to the Rev site a soon as it's 
released.

Best Regards,

Tim Ponn



Richard Gaskin wrote:

>T. R. Ponn wrote:
>
>  
>
>>Here's something I've done MANY times with HC over the years and I'm
>>hoping there's a less time consuming way to accomplish it with Rev:
>>
>>I have the need to convert mass quantities of lines of data in this form:
>>
>>"5265766F6C7574696F6E", as an example.  They are pairs (base16) of:
>>
>>"82_101_118_111_108_117_116_105_111_110", in base10 (underscores added
>>here for clarification).  And they become:
>>
>>"Revolution" in ASCII chars.
>>
>>Now what I've *been* doing in Rev is something akin to:
>>
>>put empty into whatEverOut
>>put "5265766F6C7574696F6E" into whatEver
>>repeat with i=1 to the number of chars of whatEver
>>put numToChar(baseConvert((char ((i*2)-1) to (i*2) of
>>whatEver),16,10)) after whatEverOut
>>end repeat
>>put whatEverOut
>>
>>Is this a good way to do this, or is there something that can be done
>>with a "repeat for each" form?  This ( similar) is used soooooo much in
>>my app that it's bound to make a significant difference.  Oh...and
>>occasionally I have to do this routine "in reverse".
>>    
>>
>
>Would using base64encode and base64decode functions provide you with a
>similarly useful result?
>
>  
>

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