Lawrence D'Oliveiro wrote: >Hendrik van Rooyen wrote: > >> import time >> while True: >> end_time = time.time() + 5 >> while time.time() < end_time: >> do_the_in_between_stuff() >> do_the_every_five_second_stuff() > >Maybe I'm dense, but ... where do you stop the busy-waiting?
The above is an endless loop, doing something every five seconds, and something else in the in between times. I don't think you are dense - its a good point, leading to the following analysis: The do_the_in_between_stuff call either takes less than, or more than, or exactly five seconds to do. In the unlikely exact case, there is no problem, we have a synchronous machine. (assuming the every five seconds code takes zero time - else five has to be diminished to 5 less the time of the periodic routine, but the argument stays the same) If it takes more than five seconds, we fail in our endeavour to do something every five seconds, and we need threads or interrupts or something else more advanced to get some apparent simultaneity. If it takes less than five seconds, we will enter it more than once. I wrote the above under the twin assumptions that multiple entry does not matter, and that the routine is short enough in time that the overrun error at the end does not significantly change the timing of the five second interval call. If it perversely takes say 4 seconds to do the in between stuff, the above code will do the five second stuff at intervals of eight seconds or so. - Conversely, if the in between code takes a millisec, then we will be almost spot-on. Now if multiple entry does matter, and it has to be done only once in between calls to the five second stuff, then it properly belongs with the five second stuff, and we are back to sleeping for time.sleep(5 - time_to_do_the_work) to get a call every five seconds. Obviously, if time_to_do_the_work is bigger than 5, then you can't get there from here. And these, as far as I can see, are the only alternatives. - Hendrik -- http://mail.python.org/mailman/listinfo/python-list
