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Index › Programming Questions & Help › Syntax Questions › Overcoming second() 0-59 variables to 'loop'
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Overcoming second() 0-59 variables to 'loop' (Read 614 times)
setsquare
Overcoming second() 0-59 variables to 'loop'
Nov 14th, 2009, 6:22am  
Hi again

Am looking to loop an event based on the second() method which from referring to documentation: values from 0-59.

Am implementing a 'tick' sound in time with a second hand.
This has successfully implemented except once second hand reaches '0' at top again the sound stops. I think it has something to do with the 0-59 variables but am unsure as to how to overcome.

Here is the code:

Code:
  if(second() > lastTick)
  {
  snip.trigger();
  lastTick = second();
  }


The sound stops at 0 (at top). Small buffer underrun but this *should* not matter.

Have been thinking of implementing things like:

Code:
  if(second() > lastTick)
  {
[b]  firstTick = second();[/b]
  snip.trigger();
  lastTick = second();
  }


to no avail.

thanks in advance, calum
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JR
Re: Overcoming second() 0-59 variables to 'loop'
Reply #1 - Nov 14th, 2009, 6:48am  
Try it like this:

Code:

if(second() >= (lastTick+1)%60)
  {
  snip.trigger();
  lastTick = second();
  }
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setsquare
Re: Overcoming second() 0-59 variables to 'loop'
Reply #2 - Nov 14th, 2009, 7:01am  
The code works except at around 57-59 second mark the sound overlaps and makes an awful sound. I'm thinking the fact of the second() resetting itself may be an issue.

Is there a workaround for this? Perhaps a debugger?
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setsquare
Re: Overcoming second() 0-59 variables to 'loop'
Reply #3 - Nov 14th, 2009, 7:31am  
woo! solved it. the > symbol not required.

here code:

Code:
if(second() == (lastTick+1)%60)
 {
  snip.trigger();  
  lastTick = second();
  }


thanks again
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