Re: [PATCH 1/2] rt: Don't call schedule_work_on in preemption disabled context

[Sebastian Andrzej Siewior <bigeasy@linutronix.de>]

* Yang Shi | 2013-10-04 09:36:41 [-0700]:

>>>--- a/mm/memcontrol.c
>>>+++ b/mm/memcontrol.c
>>>@@ -2453,8 +2453,11 @@ static void drain_all_stock(struct mem_cgroup *root_memcg, bool sync)
>>>		if (!test_and_set_bit(FLUSHING_CACHED_CHARGE, &stock->flags)) {
>>>			if (cpu == curcpu)
>>>				drain_local_stock(&stock->work);
>>>-			else
>>>+			else {
>>>+				preempt_enable();
>>>				schedule_work_on(cpu, &stock->work);
>>>+				preempt_disable();
>>>+			}
>>>		}
>>What ensures that you don't switch CPUs between preempt_enable() &
>>preempt_disable() and is curcpu != smp_processor_id() ?
>
>drain_all_stock is called by drain_all_stock_async or
>drain_all_stock_sync, and the call in both is protected by mutex:
>
>if (!mutex_trylock(&percpu_charge_mutex))
>                return;
>        drain_all_stock(root_memcg, false);
>        mutex_unlock(&percpu_charge_mutex);
>
>
>So, I suppose this should be able to protect from migration?

preempt_disable() ensures that the task executing drain_all_stock() is
not moved from cpu1 to cpu5. Lets say we run cpu1, on first invocation
we get we get moved from cpu1 to cpu5 after preempt_enable(). On the
second run we have (1 == 1) and invoke drain_local_stock() the argument
is ignored so we execute drain_local_stock() with data of cpu5. Later we
schedule work for cpu5 again but we never did it for cpu1.

The code here is robust enough that nothing bad happens if
drain_local_stock() is invoked twice on one CPU and the system probably
survives it if one CPU is skipped. However I would prefer not to have
such an example in the queue where it seems that it is okay to just
enable preemption and invoke schedule_work_on() because it breaks the
assumptions which are made by get_cpu().

>Thanks,
>Yang

Sebastian