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Re: Round robin scheduler question


On Wed, Aug 20, 2003 at 07:41:57PM +0530, sandeep wrote:
> >     HAL_LSBIT_INDEX(index, queue_map);
> >
> >     Cyg_RunQueue *queue = &run_queue[index];
> >
> >     CYG_ASSERT( !queue->empty(), "Queue for index empty");
> >
> >     Cyg_Thread *thread = queue->get_head();
> >
> > +    if ( thread != current) {
> > +            cyg_priority pri    = current->priority;
> > +            Cyg_Scheduler *sched = &Cyg_Scheduler::scheduler;
> > +            Cyg_RunQueue *current_queue = &sched->run_queue[pri];
> > +            current_queue->rotate();
> > +    }
> >
> > #endif
> 

> if timeslice was over earlier then timeslice_cpu would have already
> got called before this schedule call. in that case rotate will be
> done twice, and outcomes could be many -- breaking of round-robin
> nature at that level, some thread not getting chance to execute ever
> even being in runqueue all the times.  please correct me, if I got
> it wrong.

Yes, you are right. If there are two runable threads, this will break.

Im sure Nick will be along soon with a nice neat answer.

   Andrew

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