Re: [PATCH 3/3] locking/percpu-rwsem: Avoid unnecessary writer wakeups

From: Oleg Nesterov
Date: Mon Nov 21 2016 - 07:47:33 EST


On 11/21, Peter Zijlstra wrote:
>
> On Mon, Nov 21, 2016 at 01:23:44PM +0100, Oleg Nesterov wrote:
> > > int __percpu_init_rwsem(struct percpu_rw_semaphore *sem,
> > > const char *name, struct lock_class_key *rwsem_key)
> > > {
> > > @@ -103,41 +141,11 @@ void __percpu_up_read(struct percpu_rw_semaphore *sem)
> > > __this_cpu_dec(*sem->read_count);
> > >
> > > /* Prod writer to recheck readers_active */
> > > - swake_up(&sem->writer);
> > > + if (__readers_active_check(sem))
> > > + swake_up(&sem->writer);
> >
> > Suppose we have 2 active readers which call __percpu_up_read() at the same
> > time and the pending writer sleeps.
> >
> > What guarantees that one of these readers will observe per_cpu_sum() == 0 ?
> > They both can read the old value of the remote per-cpu counter, no?
>
> In particular, you're thinking of what provides the guarantee that the
> woken CPU observes the same state the wakee saw?

No, no, I meant that afaics both readers can see per_cpu_sum() != 0 and
thus the writer won't be woken up. Till the next down_read/up_read.

Suppose that we have 2 CPU's, both counters == 1, both readers decrement.
its counter at the same time.

READER_ON_CPU_0 READER_ON_CPU_1

--ctr_0; --ctr_1;

if (ctr_0 + ctr_1) if (ctr_0 + ctr_1)
wakeup(); wakeup();

Why we can't miss a wakeup?

This patch doesn't even add a barrier, but I think wmb() won't be enough
anyway.

Oleg.