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Message-ID: <CAKfTPtCCF6Et7gu_GKTW6oP50OtBa4=PXpfW-CCkCDdGS4wXOQ@mail.gmail.com>
Date:	Wed, 10 Apr 2013 17:23:37 +0200
From:	Vincent Guittot <vincent.guittot@...aro.org>
To:	Frederic Weisbecker <fweisbec@...il.com>
Cc:	LKML <linux-kernel@...r.kernel.org>,
	"linaro-kernel@...ts.linaro.org" <linaro-kernel@...ts.linaro.org>,
	Peter Zijlstra <peterz@...radead.org>,
	Ingo Molnar <mingo@...nel.org>,
	Steven Rostedt <rostedt@...dmis.org>,
	Mike Galbraith <efault@....de>
Subject: Re: [PATCH Resend v5] sched: fix init NOHZ_IDLE flag

On 9 April 2013 14:45, Frederic Weisbecker <fweisbec@...il.com> wrote:
> 2013/4/4 Vincent Guittot <vincent.guittot@...aro.org>:
>> On 4 April 2013 19:07, Frederic Weisbecker <fweisbec@...il.com> wrote:
>>> Is it possible that we can be dealing here with a
>>> sched_group/sched_group_power that is used on another CPU (from that
>>> CPU's rq->rq_sd->sd) concurrently?
>>> When we call build_sched_groups(), we might reuse an exisiting struct
>>> sched_group used elsewhere right? If so, is there a race with the
>>> above initialization?
>>
>> No we are not reusing an existing struct, the
>> sched_group/sched_group_power that is initialized here, has just been
>> created by __visit_domain_allocation_hell in build_sched_domains. The
>> sched_group/sched_group_power is not already attached to any CPU
>
> I see. Yeah the group allocations/initialization is done per domain
> found in ndoms_new[]. And there is no further reuse of these groups
> once these are attached.
>
> Looking at the code it seems we can make some symetric conclusion with
> group release? When we destroy a per cpu domain hierarchy and then put
> our references to the struct sched_group, all the other per cpu
> domains that reference these sched_group are about to put their
> reference soon too, right? Because IIUC we always destroy these per
> cpu domains per highest level partition (those found in doms_cur[])?

Yes

>
> I'm just asking to make sure we don't need some
> atomic_dec(nr_busy_cpus) on per cpu domain release, which is not
> necessary the sched group is getting released soon.

yes, it's not needed

>
> Thanks for your patience :)

That's fine.
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