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SubjectRe: [PATCH] Remove GP_REPLAY state from rcu_sync
Hi Oleg,

On Fri, Oct 04, 2019 at 05:41:03PM +0200, Oleg Nesterov wrote:
> On 10/04, Joel Fernandes (Google) wrote:
> >
> > But this is not always true if you consider the following events:
>
> I'm afraid I missed your point, but...
>
> > ---------------------->
> > GP num 111111 22222222222222222222222222222222233333333
> > GP state i e p x r rx i
> > CPU0 : rse rsx
> > CPU1 : rse rsx
> > CPU2 : rse rsx
> >
> > Here, we had 3 grace periods that elapsed, 1 for the rcu_sync_enter(),
> > and 2 for the rcu_sync_exit(s).
>
> But this is fine?
>
> We only need to ensure that we have a full GP pass between the "last"
> rcu_sync_exit() and GP_XXX -> GP_IDLE transition.
>
> > However, we had 3 rcu_sync_exit()s, not 2. In other words, the
> > rcu_sync_exit() got batched.
> >
> > So my point here is, rcu_sync_exit() does not really always cause a new
> > GP to happen
>
> See above, it should not.

Ok, I understand now. The point is to wait for a full GP, not necessarily
start a new one on each exit.

> > Then what is the point of the GP_REPLAY state at all if it does not
> > always wait for a new GP?
>
> Again, I don't understand... GP_REPLAY ensures that we will have a full GP
> before rcu_sync_func() sets GP_IDLE, note that it does another "recursive"
> call_rcu() if it sees GP_REPLAY.

Ok, got it.

> > Taking a step back, why did we intend to have
> > to wait for a new GP if another rcu_sync_exit() comes while one is still
> > in progress?
>
> To ensure that if another CPU sees rcu_sync_is_idle() (GP_IDLE) after you
> do rcu_sync_exit(), then it must also see all memory changes you did before
> rcu_sync_exit().

Would this not be better implemented using memory barriers, than starting new
grace periods just for memory ordering? A memory barrier is lighter than
having to go through a grace period. So something like: if the state is
already GP_EXIT, then rcu_sync_exit() issues a memory barrier instead of
replaying. But if state is GP_PASSED, then wait for a grace period. Or, do
you see a situation where this will not work?

thanks,

- Joel

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