lock order in O(1) scheduler

From: kevin@koconnor.net
Date: Thu Jan 10 2002 - 00:10:02 EST

Hi Ingo,

I was looking through the new O(1) scheduler (found in linux-2.5.2-pre11),
when I came upon the following code in try_to_wake_up():

        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 {

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..

Also, this code in set_cpus_allowed() looks bogus:

        if (target_cpu < smp_processor_id()) {
                spin_lock_irq(&target_rq->lock);
                spin_lock(&this_rq->lock);
        } else {
                spin_lock_irq(&target_rq->lock);
                spin_lock(&this_rq->lock);
        }

The lock order is the same regardless of the if statement..

-Kevin 

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