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Message-ID: <20150530192531.GB13956@redhat.com>
Date: Sat, 30 May 2015 21:25:31 +0200
From: Oleg Nesterov <oleg@...hat.com>
To: "Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Cc: Peter Zijlstra <peterz@...radead.org>, tj@...nel.org,
mingo@...hat.com, linux-kernel@...r.kernel.org, der.herr@...r.at,
dave@...olabs.net, torvalds@...ux-foundation.org
Subject: Re: [RFC][PATCH 1/5] rcu: Create rcu_sync infrastructure
On 05/30, Oleg Nesterov wrote:
>
> On 05/30, Paul E. McKenney wrote:
> >
> > On Tue, May 26, 2015 at 01:43:57PM +0200, Peter Zijlstra wrote:
> > > From: Oleg Nesterov <oleg@...hat.com>
> > >
> > > It is functionally equivalent to
> > >
> > > struct rcu_sync_struct {
> > > atomic_t counter;
> > > };
> > >
> > > static inline bool rcu_sync_is_idle(struct rcu_sync_struct *rss)
> > > {
> > > return atomic_read(&rss->counter) == 0;
> > > }
> > >
> > > static inline void rcu_sync_enter(struct rcu_sync_struct *rss)
> > > {
> > > atomic_inc(&rss->counter);
> > > synchronize_sched();
> > > }
> >
> > For vanilla RCU, this is called get_state_synchronize_rcu().
> >
> > > static inline void rcu_sync_exit(struct rcu_sync_struct *rss)
> > > {
> > > synchronize_sched();
> > > atomic_dec(&rss->counter);
> > > }
> > >
> > > except: it records the state and synchronize_sched() is only called by
> > > rcu_sync_enter() and only if necessary.
> >
> > Again for vanilla RCU, this is called cond_synchronize_rcu().
>
> Hmm. I do not understand... I think rcu_sync doesn't need
> get_state/cond_synchronize.
>
> The first caller of rcu_sync_enter() always needs sync(). The next one
> could use cond_synchronize_rcu(), but for what? The 2nd one will wait
> for the end of gp started by the first caller, and this is more optimal?
>
> Note that rcu_sync_enter/rcu_sync_func never call sync() unless strictly
> necessary.
If you meant that rcu_sync_exit() could use cond_synchronize_rcu(), this
doesn't look right too... We always need another synchronize_rcu() after
the last writer does rcu_sync_exit(). Except rcu_sync_exit() uses call_rcu()
and thus it never blocks.
> Or I misunderstood you?
Yes...
Oleg.
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