sched: Clarify ordering between task_rq_lock() and move_queued_task()
There was a wee bit of confusion around the exact ordering here; clarify things. Reported-by: Kirill Tkhai <ktkhai@parallels.com> Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Oleg Nesterov <oleg@redhat.com> Cc: Paul E. McKenney <paulmck@linux.vnet.ibm.com> Link: http://lkml.kernel.org/r/20150217121258.GM5029@twins.programming.kicks-ass.net Signed-off-by: Ingo Molnar <mingo@kernel.org>
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@ -341,6 +341,22 @@ static struct rq *task_rq_lock(struct task_struct *p, unsigned long *flags)
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raw_spin_lock_irqsave(&p->pi_lock, *flags);
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rq = task_rq(p);
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raw_spin_lock(&rq->lock);
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/*
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* move_queued_task() task_rq_lock()
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*
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* ACQUIRE (rq->lock)
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* [S] ->on_rq = MIGRATING [L] rq = task_rq()
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* WMB (__set_task_cpu()) ACQUIRE (rq->lock);
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* [S] ->cpu = new_cpu [L] task_rq()
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* [L] ->on_rq
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* RELEASE (rq->lock)
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*
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* If we observe the old cpu in task_rq_lock, the acquire of
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* the old rq->lock will fully serialize against the stores.
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*
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* If we observe the new cpu in task_rq_lock, the acquire will
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* pair with the WMB to ensure we must then also see migrating.
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*/
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if (likely(rq == task_rq(p) && !task_on_rq_migrating(p)))
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return rq;
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raw_spin_unlock(&rq->lock);
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