[pygtk] gobject.timeout_add timeout recalculation
In my application, I am using gobject.timeout_add to trigger periodic updates of the status of a number of machines. However, I am having a bit of trouble understanding exactly how the timeout is calculated. According to the documentation: After each call to the timeout function, the time of the next timeout is recalculated based on the current time and the given interval (it does not try to 'catch up' time lost in delays). The After each call to the timeout function,... leads me to think that the timer is reset when the timeout function returns. However, the phrase (it does not try to 'catch up' time lost in delays). makes me think that the timer is reset as soon as the timeout function thread is started. (This also seems to be confirmed by tracing my code) What I would like to happen is that the timeout gets recalculated when my timeout function is done/returns. I guess, I could add a new timeout from within my timeout function and have it [the timeout function] return False (so the old timeout is destroyed) but I am thinking that there has to be a better way to do this. I appreciate any advice on the matter. -- Mitko Haralanov == A Fortran compiler is the hobgoblin of little minis. ___ pygtk mailing list pygtk@daa.com.au http://www.daa.com.au/mailman/listinfo/pygtk Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
Re: [pygtk] gobject.timeout_add timeout recalculation
Mitko However, I am having a bit of trouble understanding exactly how Mitko the timeout is calculated. According to the documentation: Mitko After each call to the timeout function, the time of the Mitko next timeout is recalculated based on the current time and Mitko the given interval (it does not try to 'catch up' time lost Mitko in delays). Mitko The After each call to the timeout function,... leads me to Mitko think that the timer is reset when the timeout function returns. Mitko However, the phrase (it does not try to 'catch up' time lost in Mitko delays). makes me think that the timer is reset as soon as the Mitko timeout function thread is started. (This also seems to be Mitko confirmed by tracing my code) I believe it works like this: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now ... If X is 2500ms and your callback takes 2ms to execute, the interval between successive calls will actually be 2502ms. Mitko What I would like to happen is that the timeout gets recalculated Mitko when my timeout function is done/returns. I guess, I could add a Mitko new timeout from within my timeout function and have it [the Mitko timeout function] return False (so the old timeout is destroyed) Mitko but I am thinking that there has to be a better way to do this. In our applications we added an abs_timeout_add function which takes hour, minute, second and microsecond. The callback is called at the same time each day by taking into account the runtime of the callback. You could do something similar with a shorter period. Skip ___ pygtk mailing list pygtk@daa.com.au http://www.daa.com.au/mailman/listinfo/pygtk Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
Re: [pygtk] gobject.timeout_add timeout recalculation
On Thu, 15 May 2008 18:48:22 -0500 [EMAIL PROTECTED] wrote: I believe it works like this: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now call your callback if it returned true: set wakeup for X milliseconds from now ... If X is 2500ms and your callback takes 2ms to execute, the interval between successive calls will actually be 2502ms. Unfortunately, it is how it work. In fact, that how I would like it to work. Here is what my testing shows: t0: set wakeup for X milliseconds from now t0+X: set wakeup for X milliseconds from now (t0+X+X) call callback if it returned true: noop else: delete the timer Here is an example: #!/usr/bin/python import gtk import gobject import time def callback (timeout): print callback called at: %s%time.time () time.sleep (timeout) print callback returning at: %s%time.time () return True if __name__ == __main__: gobject.timeout_add (20*1000, callback, 10) gtk.main () If you run this, you get: callback called at: 1210897497.69 callback returning at: 1210897507.69 callback called at: 1210897517.69 callback returning at: 1210897527.69 callback called at: 1210897537.69 callback returning at: 1210897547.69 callback called at: 1210897557.69 Note that the time difference between the callback returning at and callback called at is not 20 seconds but 10 seconds (20 second timeout - 10 second callback runtime). -- Mitko Haralanov == A programming language is low level when its programs require attention to the irrelevant. ___ pygtk mailing list pygtk@daa.com.au http://www.daa.com.au/mailman/listinfo/pygtk Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
Re: [pygtk] gobject.timeout_add timeout recalculation
If X is 2500ms and your callback takes 2ms to execute, the interval between successive calls will actually be 2502ms. Mitko Unfortunately, it is how it work. In fact, that how I would like ^ not? Mitko it to work. Here is what my testing shows: Hmmm... It would appear the documentation is wrong. I thought in the past I remember seeing it work the other way. I can't reproduce it with the versions I have installed (2.6 and 2.10 at work and 2.12 at home). I'm installing gtk 1.2.10 on my Mac now. I'll see how it works there. (If I can get it installed...) Skip ___ pygtk mailing list pygtk@daa.com.au http://www.daa.com.au/mailman/listinfo/pygtk Read the PyGTK FAQ: http://www.async.com.br/faq/pygtk/
