Checking for CNT timeout...

Kye

Mmm, I forgot how to do this...

  result := cnt
  
  repeat until(...)
 
    if((cnt - result) > (clkfreq / 200))
      .. Do stuff

So, I backup a copy of CNT in result. Then I wait until their difference is over 5ms apart... or (clkfreq / 200).

However, in spin the ">" operator is signed so this comparison does not always work... So, How do I fix this?

Thanks,

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Nyamekye,

W9GFO

Use "||" to make (cnt-result) an absolute value.

Rich H

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Dave Hein

cnt - result will produce a 32-bit signed result independent of the starting value of cnt.· The difference will increase from 0 to $7fffffff, and then become negative at $80000000.· The statement if ((cnt - result) > (clkfreq / 200)) will work just fine.

JonnyMac

If you're waiting until the difference between now and when you first grabbed count, why not just wait five milliseconds with standard delay? -- or are you doing something else in between? Otherwise, as others have pointed out, make the result of now minus the cnt capture positive:

t0 := cnt
...
if (||(cnt - t0) > (clkfreq / 200)) ...

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Jon McPhalen
Hollywood, CA

Kye

||(cnt - to) works fine. Thanks,

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Nyamekye,

Dave Hein

The || is unnecessary.· It will work with or without it.

Bobb Fwed

I use a few different methods that can be helpful:

  wait := cnt       
  REPEAT
    IF ((cnt - wait) => length)
      wait += length
      '' do stuff once every [i]length[/i]
    '' continually do stuff

This one allows you to do a continuously repeating process, then every length cycles something specific is executed.

Then there is the repetitive timing loop:

  wait := cnt
  REPEAT
    '' do stuff once every [i]length[/i]
    waitcnt(wait += length)

and here is the one I think you are thinking of:

  wait := cnt       
  REPEAT UNTIL ((cnt - wait) => length)
    '' continually do stuff for [i]length[/i] cycles

EDIT: looks like I misunderstood the questions...oh well, that is still good information above

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Post Edited (Bobb Fwed) : 6/16/2010 2:30:41 PM GMT

ErNa

Maybe, "checking for CNT timeout" points more precisely to the problem than: "Will the universe oscillate", but I think, I was talking about the same problem: http://forums.parallax.com/showthread.php?p=912500.

The essence is: Never use : if a > CNT, always: if (a - CNT) > 0

I hope I'm right. I didn't systematically verify that, but since I changed the If statement, my program didn't hang!

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Post Edited (ErNa) : 6/16/2010 7:43:01 AM GMT

kuroneko

Dave Hein said...
cnt - result will produce a 32-bit signed result independent of the starting value of cnt. The difference will increase from 0 to $7fffffff, and then become negative at $80000000. The statement if ((cnt - result) > (clkfreq / 200)) will work just fine.

Since we're talking SPIN here you could simply shift right the result by 1 (e.g. (cnt-result)>>1), it's not really important whether you have 12.5ns or 25ns resolution (@80MHz). But at least you're safe for about 53sec unless 26sec are enough [noparse]:)[/noparse] Using || is actually dangerous as e.g. $FFFFFFFF is downsized to 1.

Kye

The difference will never be over 400,000 so there really is no problem.

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Nyamekye,