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hippy
09-16-2007, 09:16 AM
I am having no end of trouble trying to get my head round solving this, so any help would be gratefully received ....

timeout := ( clkfreq / 1000 ) * mS + CNT
repeat
if timeout > CNT
NotTimedOut
else
TimedOut

It will work fine if 'timeout' and CNT are positive, but it does not handle all cases of CNT wrapround and overflows. The solution maybe obvious, but I cannot see it.

Lawson
09-16-2007, 11:41 AM

(timeout - CNT) > 0

the subtraction should be more tolerant of overflows than the greater than test.

My 2 cents
Marty

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deSilva
09-16-2007, 01:03 PM
This is how it works:

+ 2^31 <------- C -------- 0 -------- TOUT -- > -2^31
..> ..........> .....

Starting with: timeout > CNT.
subtracting "CNT" from both side gives "Lawson's formula"

You see in the diagram, that this "moves" CNT to the center = 0, now "timeout" can easily be found
either at the + (still time left) or the - (time out!) side.
This works as long as ||(timeout-CNT) < 2^31 ticks.

Post Edited (deSilva) : 9/16/2007 6:09:44 AM GMT

hippy
09-16-2007, 04:02 PM
Thanks both, it took a bit of paper working to convince myself that would work, but I believe that solves the problem.

deSilva
09-16-2007, 06:13 PM
hippy said...
Thanks both, it took a bit of paper working to convince myself that would work, but I believe that solves the problem.

I intended to save you the paperwork through my diagram http://forums.parallax.com/images/smilies/smile.gif
An even better diagram would show a circle where +/- 2^31 meet,opposite to zero - so a subtraction (or addition) is a "turn of that wheel"

hippy
09-17-2007, 12:28 AM
deSilva said...
I intended to save you the paperwork through my diagram http://forums.parallax.com/images/smilies/smile.gif

And it was appreciated. It wasn't that your answer caused any problems, just that I needed to to do the working out to make myself understand and assure myself that the theory held up to practice. Thanks, it does.