M. Koehrer wrote:
> Hi!
>
> I am just playing around with Xenomai to understand it.
> And I have one question concerning the entry function that will be called from
> rt_task_start (Native API).
> What happens if this entry function reaches its end or a return within the
> function is
> called?
>
> I have made a simple example that lead to strange effects.
> In my example (see complete C Code in attachement) I create to tasks
> that do the same.
> The entry functions look like:
>
> void taska(void *cookie)
> {
> RTIME delay;
> int i;
> printf("Hi, I am task A %s\n", (char*)cookie);
> delay = 100000;
>
> for (i=0; i<200; i++)
> {
> rt_task_sleep(delay);
> }
>
> printf("This is the end of A\n");
> // rt_task_delete(0);
> }
> When I do not place the rt_task_delete(0) at the end of the function
> my second task will never reach the end with the second printf.
> Whenever I use rt_task_delete(0) at the end of the function it works
> perfectly.
Quite inconsistent, sounds like a bug. Does the second rt_task_join just
rush through and main terminates?
>
> My question is now: What happens if the task's entry function returns?
pthread_exit() is invoked implicitly, which implies rt_task_delete(NULL).
>
> Thanks for any feedback on that question.
Is this issue SMP-related? What happens if you force all your tasks to
the same CPU? Or if you run on a !CONFIG_SMP box?
Jan
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