Mauro Salvini wrote:
>> Well, looking at the code, this should not happen. The question is:
>> how do you ensure, in the code that calls rt_task_join, that the
>> thread being created had the time to call rt_task_shadow? In other
>> words, are you sure that you are not passing rt_task_join an
>> uninitialized RT_TASK?
>
>> So, as usual, we are missing a simple standalone test showing the
>> issue you have.
>
> There was a subtle bug that realizes exactly what you explained,
> thank you!
This is not exactly a subtle bug.
> Question is: after a successful rt_task_shadow() call done
> by a pthread, we can forget that this rt-task was created as pthread
> and then we must handle it with rt_task_* procedures, is it right? In
> other words, does pthread become a RT_TASK at all, as if it was
> created with rt_task_create()?
No, an RT_TASK is also a pthread, whether created with rt_task_shadow or
rt_task_create (if you look at src/skins/native/task.c, you will see
that rt_task_create creates a thread with pthread_create which does the
same thing as rt_task_shadow). You saw that rt_task_join was using
pthread_join under the hood. And by the way, rt_task_shadow by itself is
mainly useful for turning the main thread into an RT_TASK, it is simpler
to create other RT_TASKs with rt_task_create.
--
Gilles.
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