The task queue, by default, is only 8 tasks deep. A task queue overflow would cause the task not to execute.
-Joe On 1/23/06, Martin Gercke <[EMAIL PROTECTED]> wrote: > Ok, > > so there must be a limit for tasks that are waiting in the queue, right? > If the loop is a running task then at the end of this task the waiting > tasks should be run by the scheduler. > Correct? > I am just wondering why the receive-task isn't run at all. > If the interrupt isn't directly processing the data and instead posting > a task of course as you said the running task and all waiting tasks will > be run first. > > Martin > > > Joe Polastre wrote: > > >Because interrupts post tasks, and if a task is halting execution, you > >never get the task posted by the interrupt. > > > >-Joe > > > >On 1/23/06, Martin Gercke <[EMAIL PROTECTED]> wrote: > > > > > >>Hmm, > >> > >>I don't really get that. > >>Isn't an interrupt supposed to interrupt a running task immediately? > >>Why is that not the case - instead of the loop there could be any other > >>task running. > >>What makes this example so special? > >> > >>Martin > >> > >> > >>Joe Polastre wrote: > >> > >> > >> > >>>Remove the loop and use Timers instead. You have completely > >>>eliminated any concurrency in the system with that loop. > >>> > >>>-Joe > >>> > _______________________________________________ > Tinyos-help mailing list > [email protected] > https://mail.millennium.berkeley.edu/cgi-bin/mailman/listinfo/tinyos-help > _______________________________________________ Tinyos-help mailing list [email protected] https://mail.millennium.berkeley.edu/cgi-bin/mailman/listinfo/tinyos-help
