On 09/18/2013 09:53 AM, Sven Barth wrote:
I wouldn't use the timeout constant for this. If you have two timers of which the greatest common denominator is 1, but nevertheless rather large (e.g. two primes) then you'd loop unnecessarily (I know this is a constructed example, but nevertheless one should care for this!). I'd instead suggest to implement the timerloop of each timer as a thread that waits its timeout time on a event (so that it can also be stopped) and then notifies the mainthread using Queue().
Right you are. (I just wanted to keep this as simple as possible for a starter.)
You simply could use a sleep in the timer loop,. This would be arch independent out of the box. But it would impose accumulative delays depending on the CPU performance.
An more advanced timer implementation would use arch depending OS-based timers. This would allow for a much more exact timing.
-Michael _______________________________________________ fpc-pascal maillist - fpc-pascal@lists.freepascal.org http://lists.freepascal.org/mailman/listinfo/fpc-pascal
