Need help in understanding some of the existing code in RTEMS

[Richi Dubey]

Hi,

I had a small question. The scheduler struct inside percpu.h looks like:
------------------------------------------------------------------------------------------------------
    struct {
      /**
       * @brief The scheduler control of the scheduler owning this
processor.
       *
       * This pointer is NULL in case this processor is currently not used
by a
       * scheduler instance.
       */
      const struct _Scheduler_Control *control;

      /**
       * @brief The scheduler context of the scheduler owning this
processor.
       *
       * This pointer is NULL in case this processor is currently not used
by a
       * scheduler instance.
       */
      const struct Scheduler_Context *context;

      /**
       * @brief The idle thread for this processor in case it is online and
       * currently not used by a scheduler instance.
       */
      struct _Thread_Control *idle_if_online_and_unused;
    } Scheduler;
------------------------------------------------------------------------------------------------------
So, does this mean a CPU when active is always either executing an idle
thread and is not being used by a scheduler (so has a thread attribute in
the idle_if_online_and_unused), or is used by a scheduler and is executing
a task ( which can not be an idle task)? Another equivalent question is do
we have an idle scheduler node, like we have idle predefined threads that
run on a CPU?

Please let me know,
Thanks.

On Tue, Jul 14, 2020 at 8:28 PM Richi Dubey <richidubey at gmail.com> wrote:

> I understand. Thank you.
>
> On Tue, Jul 14, 2020 at 7:05 PM Sebastian Huber <
> sebastian.huber at embedded-brains.de> wrote:
>
>> On 14/07/2020 13:37, Richi Dubey wrote:
>>
>> >     Here we remove the affine ready queue if it
>> >     exists from the chain of affine queues since now an affine thread is
>> >     scheduled on a processor.
>> >
>> > Why are we removing the entire affine queue corresponding to a
>> > CPU when a single node of the queue gets scheduled?
>> Because the highest priority affine thread is now a schedule one.
>>
>
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