lock order in O(1) scheduler

kevin@koconnor.net
Thu, 10 Jan 2002 00:10:02 -0500


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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