Messages in this thread | | | Date | Thu, 3 Dec 2009 08:57:32 +0100 | Subject | Re: Should we use preempt_disable() in sleep_on_common()? | From | Dmitry Adamushko <> |
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2009/12/3 liu pf <kernelfans@gmail.com>: > Hi: > > I am puzzled with the following scenario. Could anyone enlighten me? > > Thanks > pfliu > > > static long __sched > sleep_on_common(wait_queue_head_t *q, int state, long timeout) > { > unsigned long flags; > wait_queue_t wait; > > init_waitqueue_entry(&wait, current); > > __set_current_state(state); > > ==========>suppose that after task A set state=TASK_INTERRUPTIBLE > , it is preempted by task B. > > spin_lock_irqsave(&q->lock, flags); > ............................................................... > } > > asmlinkage void __sched schedule(void) > { > ....................................................................................................... > if (prev->state && !(preempt_count() & PREEMPT_ACTIVE)) { > if (unlikely(signal_pending_state(prev->state, prev))) > prev->state = TASK_RUNNING; > else > deactivate_task(rq, prev, 1); > =============>This will remove task A from rq, but there are no > wait queue referring to A, so we lose A. > switch_count = &prev->nvcsw; > }
In this case, (preempt_count() & PREEMPT_ACTIVE) == 1(see preempt_schedule_irq() and other use-cases of PREEMPT_ACTIVE) so we don't enter this block.
i.e. a preempted task stays on its queue (with state != TASK_RUNNING but that doesn't really matter).
-- Dmitry
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