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.
My question is now: What happens if the task's entry function returns?
Thanks for any feedback on that question.
Regards
Mathias
--
Mathias Koehrer
[EMAIL PROTECTED]
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multitasks.c.gz
Description: PostScript document
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