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On Sun, 5 Nov 2006, Oleg Nesterov wrote: > > When task->array != NULL, try_to_wake_up() just goes to "out_running" and sets > task->state = TASK_RUNNING. > > In that case hrtimer_wakeup() does: > > timeout->task = NULL; <----- [1] > > spin_lock(runqueues->lock); > > task->state = TASK_RUNNING; <----- [2] > > from Documentation/memory-barriers.txt > > Memory operations that occur before a LOCK operation may appear to > happen after it completes. > > This means that [2] may be completed before [1], and Yes. On x86 (and x86-64) you'll never see this, because writes are always seen in order regardless, and in addition, the spin_lock is actually totally serializing anyway. On most other architectures, the spin_lock will serialize all the writes too, but it's not guaranteed, so in theory you're right. I suspect no actual architecture will do this, but hey, when talking memory ordering, safe is a lot better than sorry. That said, since "task->state" in only tested _inside_ the runqueue lock, there is no race that I can see. Since we've gotten the runqueue lock in order to even check task-state, the processor that _sets_ task state must not only have done the "spin_lock()", it must also have done the "spin_unlock()", and _that_ will not allow either the timeout or the task state to haev leaked out from under it (because that would imply that the critical region leaked out too). So I don't think the race exists anyway - the schedule() will return immediately (because it will see TASK_RUNNING), and we'll just retry. Linus - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/ | ||||||||||||
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