Hi Peter,
Thanks for looking. I'll try to reply on Monday, just one note...
On 07/30, Peter Zijlstra wrote:
>
> On Tue, Jul 21, 2015 at 09:22:47PM +0200, Oleg Nesterov wrote:
>
> > +static int cpu_stop_queue_two_works(int cpu1, struct cpu_stop_work *work1,
> > + int cpu2, struct cpu_stop_work *work2)
> > +{
> > + struct cpu_stopper *stopper1 = per_cpu_ptr(&cpu_stopper, cpu1);
> > + struct cpu_stopper *stopper2 = per_cpu_ptr(&cpu_stopper, cpu2);
> > + int err;
> > +retry:
> > + spin_lock_irq(&stopper1->lock);
> > + spin_lock_nested(&stopper2->lock, SINGLE_DEPTH_NESTING);
> > + /*
> > + * If we observe both CPUs active we know _cpu_down() cannot yet have
> > + * queued its stop_machine works and therefore ours will get executed
> > + * first. Or its not either one of our CPUs that's getting unplugged,
> > + * in which case we don't care.
> > + */
> > + err = -ENOENT;
> > + if (!cpu_active(cpu1) || !cpu_active(cpu2))
> > + goto unlock;
> > +
> > + WARN_ON(!stopper1->enabled || !stopper2->enabled);
> > + /*
> > + * Ensure that if we race with stop_cpus() the stoppers won't
> > + * get queued up in reverse order, leading to system deadlock.
> > + */
> > + err = -EDEADLK;
> > + if (stop_work_pending(stopper1) != stop_work_pending(stopper2))
> > + goto unlock;
>
> You could DoS/false positive this by running stop_one_cpu() in a loop,
> and thereby 'always' having work pending on one but not the other.
IIRC no. I am pretty sure stop_one_cpu() doesn't use stopper->stop_work,
only stop_machine() does.
Oleg.
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