Messages in this thread | | | Date | Thu, 10 Jan 2002 01:29:57 -0500 | From | kevin@koconnor ... | Subject | Re: lock order in O(1) scheduler |
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On Thu, Jan 10, 2002 at 12:26:08AM -0500, Robert Love wrote: > On Thu, 2002-01-10 at 00:10, kevin@koconnor.net wrote: > > > I was unable to figure out what the logic of the '(smp_processor_id() < > > p->cpu)' test is.. (Why should the CPU number of the process being awoken > > matter?) My best guess is that this is to enforce a locking invariant - > > but if so, isn't this test backwards? If p->cpu > current->cpu then > > p->cpu's runqueue is locked first followed by this_rq - locking greatest to > > least, where the rest of the code does least to greatest.. > > Not so sure of the validity, but it is to respect lock order. [...]
Right. It is a confusing though - depending on the value of p->cpu relative to current->cpu, the synchronous flag may be ignored. Since the relationship between p->cpu and current->cpu is essentially random, this makes the behavior of the synchronous flag random - why bother?
I can see that re-grabbing the locks in the proper order would be a pain. Also, after a quick grep, it appears only fs/pipe.c is interested in the wake_up_sync() variant. Maybe this feature should just be culled?
-Kevin
The try_to_wake_up function in pre11 (just for reference):
static int try_to_wake_up(task_t * p, int synchronous) { unsigned long flags; int success = 0; runqueue_t *rq;
lock_task_rq(rq, p, flags); p->state = TASK_RUNNING; if (!p->array) { if (!rt_task(p) && synchronous && (smp_processor_id() < p->cpu)) { spin_lock(&this_rq()->lock); p->cpu = smp_processor_id(); activate_task(p, this_rq()); spin_unlock(&this_rq()->lock); } else { activate_task(p, rq); if ((rq->curr == rq->idle) || (p->prio < rq->curr->prio)) resched_task(rq->curr); } success = 1; } unlock_task_rq(rq, p, flags); return success; }
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