Re: [RFC 2/2] srcu: Remove memory barrier "E" as it is not required

From: Joel Fernandes
Date: Sun Dec 18 2022 - 18:27:33 EST


On Sun, Dec 18, 2022 at 10:42:43PM +0100, Frederic Weisbecker wrote:
> On Sun, Dec 18, 2022 at 07:13:09PM +0000, Joel Fernandes (Google) wrote:
> > During a flip, we have a full memory barrier before idx is incremented.
> >
> > The effect of this seems to be to guarantee that, if a READER sees srcu_idx
> > updates (srcu_flip), then prior scans would not see its updates to counters on
> > that index.
> >
> > That does not matter because of the following reason: If a prior scan did see
> > counter updates on the new index, that means the prior scan would would wait
> > for the reader when it probably did not need to.
>
> I'm confused, isn't it actually what we want to prevent from?
> The point of the barrier here is to make sure that the inactive index that
> we just scanned is guaranteed to remain seen as inactive during the whole scan
> (minus the possible twice residual increments from a given task that we debated
> on Paul's patch, but we want the guarantee that the inactive index won't be
> incremented thrice by a given task or any further while we are scanning it).

I believe you are talking about the memory barrier after the flip, that's the
one that guarantees what you are talking about it, I feel. That is, readers
see the newly inactivated index eventually, so that we are not scanning
indefinitely.

For that, we need smp_mb() after the flip but before the second scan which is
a much needed memory barrier IMHO, and not what this patch is talking about.

> If some readers see the new index and increments the lock and we see that while
> we are scanning it, there is a risk that the GP is going to be delayed indefinetly.

The "new" index is the index after the flip, do you mean the "old" index?
i.e. the index before the flip? That is what barrier E is talking about, not
the index after the flip.

>
> > @@ -982,14 +982,6 @@ static bool try_check_zero(struct srcu_struct *ssp, int idx, int trycount)
> > */
> > static void srcu_flip(struct srcu_struct *ssp)
> > {
> > - /*
> > - * Ensure that if a given reader sees the new value of ->srcu_idx, this
> > - * updater's earlier scans cannot have seen that reader's increments
> > - * (which is OK, because this grace period need not wait on that
> > - * reader).
> > - */
> > - smp_mb(); /* E */ /* Pairs with B and C. */
>
> That said, I've been starring at this very barrier for the whole day, and I'm
> wondering what does it match exactly on the other end?
>
> UPDATER READER
> ------- ------
> idx = ssp->srcu_idx; idx = srcu_idx;
> READ srcu_unlock_count[srcu_idx ^ 1] srcu_lock_count[idx]++
> smp_mb(); smp_mb();
> READ srcu_lock_count[srcu_idx ^ 1] srcu_unlock_count[old_idx]++
> smp_mb()
> srcu_idx++;
>
> For a true match, I would expect a barrier between srcu_idx read and
> srcu_lock_count write. I'm not used to ordering writes after reads.
> So what is the pattern here? I would expect something like the below
> but that doesn't match the above:

IMHO, it is matching updates to index and the lock count of a reader.

>
> C rwrw
>
> {}
>
>
> P0(int *X, int *Y)
> {
> int x;
>
> x = READ_ONCE(*X);
> smp_mb();
> WRITE_ONCE(*Y, 1);
> }
>
> P1(int *X, int *Y)
> {
>
> int y;
>
> y = READ_ONCE(*Y);
> smp_mb();
> WRITE_ONCE(*X, 1);
> }
>
> exists (0:x=1 /\ 1:y=1)

Hmm, I guess first lets degenerate the real code to an access pattern:


READER UPDATER

scanner() {
count_all_unlocks();
smp_mb();
count_all_locks(); (Y)
}

rcu_read_lock() {
idx = READ(idx); (X)
lock_count[idx]++;

smp_mb(); // mb B
}

rcu_read_unlock() {
smp_mb(); // mb C
unlock_count[idx]++;
}
srcu_flip() {
smp_mb(); //E
idx++; (X)
rcu_read_lock() {
idx = READ(idx);
lock_count[idx]++; (Y)

smp_mb(); // mb B
smp_mb();
}
}


That becomes:

// READER
P0(int *X, int *Y)
{
int r0;

r0 = READ_ONCE(*X); // PP
smp_mb(); // B+C // QQ
WRITE_ONCE(*Y, 1); // RR
}

// UPDATER
P1(int *X, int *Y)
{
int r1;

r1 = READ_ONCE(*Y); // SS
smp_mb(); // E // TT
WRITE_ONCE(*X, 1); // UU
}

Impossible that:
exists (0:r0=1 /\ 1: r1:1)

Because if r0=1, there is PP ->rf UU relation. So because of the smp_mb(), it
is impossible that r1=1.

So "E" is saying, if a reader saw new idx, that is the "X" in the litmus
test, then previous scan where it count all the locks (SS) cannot see the
lock count updates made at the new index.

However, that does not matter IMHO because due to preemption after current
index is sampled, we have no control anyway over which lock counts are
incremented anyway, so this cannot effect correctness.

And if forward progress is a problem, we are doing a full memory barrier
after the flip anyway so I am not seeing the point of "E".

thanks,

- Joel