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SubjectRe: [PATCH 3/7] barrier: a scalable synchonisation barrier
On 02/01, Paul E. McKenney wrote:
> > >
> > > void synchronize_qrcu(struct qrcu_struct *qp)
> > > {
> > > int idx;
> > >
> > > smp_mb();
> > >
> > > if (atomic_read(qp->ctr[0]) + atomic_read(qp->ctr[1]) <= 1) {
> > > smp_rmb();
> > > if (atomic_read(qp->ctr[0]) +
> > > atomic_read(qp->ctr[1]) <= 1)
> > > goto out;
> > > }
> > >
> > > mutex_lock(&qp->mutex);
> > > idx = qp->completed & 0x1;
> > > atomic_inc(qp->ctr + (idx ^ 0x1));
> > > /* Reduce the likelihood that qrcu_read_lock() will loop */
> > > smp_mb__after_atomic_inc();
> > > qp->completed++;
> > >
> > > atomic_dec(qp->ctr + idx);
> > > __wait_event(qp->wq, !atomic_read(qp->ctr + idx));
> > > mutex_unlock(&qp->mutex);
> > > out:
> > > smp_mb();
> > > }
> > >
> > > For the first "if" to give a false positive, a concurrent switch had
> > > to have happened. For example, qp->ctr[0] was zero and qp->ctr[1]
> > > was two at the time of the first atomic_read(), but then qp->completed
> > > switched so that both qp->ctr[0] and qp->ctr[1] were one at the time
> > > of the second atomic_read. The only way the second "if" can give us a
> > > false positive is if there was another change to qp->completed in the
> > > meantime -- but that means that all of the pre-existing qrcu_read_lock()
> > > holders must have gotten done, otherwise the second switch could not
> > > have happened. Yes, you do incur three memory barriers on the fast
> > > path, but the best you could hope for with your approach was two of them
> > > (unless I am confused about how you were using barrier_sync()).

Yes. Without synchronize_qrcu() in between, one of the counters should be == 0,
another >= 1. == 1 means we have no active readers. So the false positive really
means a concurrent switch. And we can check twice - excellent idea!

> > While doing qrcu, somehow I convinced myself we can't optimize out taking
> > qp->mutex. Now I think I was wrong. Good!
>
> Me, I didn't want to worry about it unless someone needed it. Which
> it now appears they do. ;-)

No. I do remember I tried hard to optimize out taking qp->mutex, but failed.
So I decided it is not possible. And now you show that I just don't have enough
brains! (of course, I hate you :)

> > Q: you deleted "if (atomic_read(qp->ctr + idx) == 1)" fastpath under ->mutex,
> > was this needed for this optimization to work? I am asking because I can't
> > understand how it can make any difference.
>
> Before, we held the lock, so we could just check the single current
> element. Now we don't hold the lock, so we need to check both elements.
> So I replaced the "if (atomic_read(qp->ctr + idx) == 1)" with the
> nested "if" statements that test both elements.

Ah, my question was different. The current version of qrcu does

mutex_lock(&qp->mutex);

idx = qp->completed & 0x1;
if (atomic_read(qp->ctr + idx) == 1) // fast path
return;

...

and it seems to me that we can retain this fastpath even with your optimization,
no? Surely, it is not so important, but it is nearly free.

Paul, could you make a patch? (I'll do rcutorture test tomorrow, but I only have
P-4 ht).

Peter, do you think you can use qrcu?

Oleg.

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