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SubjectRe: lock order in O(1) scheduler
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;
}

--
------------------------------------------------------------------------
| Kevin O'Connor "BTW, IMHO we need a FAQ for |
| kevin@koconnor.net 'IMHO', 'FAQ', 'BTW', etc. !" |
------------------------------------------------------------------------
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