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Date:	Tue, 11 Dec 2012 12:00:55 +0530
From:	Preeti U Murthy <preeti@...ux.vnet.ibm.com>
To:	Alex Shi <alex.shi@...el.com>
CC:	rob@...dley.net, mingo@...hat.com, peterz@...radead.org,
	gregkh@...uxfoundation.org, andre.przywara@....com, rjw@...k.pl,
	paul.gortmaker@...driver.com, akpm@...ux-foundation.org,
	paulmck@...ux.vnet.ibm.com, linux-kernel@...r.kernel.org,
	pjt@...gle.com, vincent.guittot@...aro.org
Subject: Re: [PATCH 01/18] sched: select_task_rq_fair clean up

On 12/11/2012 10:58 AM, Alex Shi wrote:
> On 12/11/2012 12:23 PM, Preeti U Murthy wrote:
>> Hi Alex,
>>
>> On 12/10/2012 01:52 PM, Alex Shi wrote:
>>> It is impossible to miss a task allowed cpu in a eligible group.
>>
>> The one thing I am concerned with here is if there is a possibility of
>> the task changing its tsk_cpus_allowed() while this code is running.
>>
>> i.e find_idlest_group() finds an idle group,then the tsk_cpus_allowed()
>> for the task changes,perhaps by the user himself,which might not include
>> the cpus in the idle group.After this find_idlest_cpu() is called.I mean
>> a race condition in short.Then we might not have an eligible cpu in that
>> group right?
> 
> your worry make sense, but the code handle the situation, in
> select_task_rq(), it will check the cpu allowed again. if the answer is
> no, it will fallback to old cpu.
>>
>>> And since find_idlest_group only return a different group which
>>> excludes old cpu, it's also imporissible to find a new cpu same as old
>>> cpu.

I doubt this will work correctly.Consider the following situation:sched
domain begins with sd that encloses both socket1 and socket2

cpu0 cpu1  | cpu2 cpu3
-----------|-------------
 socket1   |  socket2

old cpu = cpu1

Iteration1:
1.find_idlest_group() returns socket2 to be idlest.
2.task changes tsk_allowed_cpus to 0,1
3.find_idlest_cpu() returns cpu2

* without your patch
   1.the condition after find_idlest_cpu() returns -1,and sd->child is
chosen which happens to be socket1
   2.in the next iteration, find_idlest_group() and find_idlest_cpu()
will probably choose cpu0 which happens to be idler than cpu1,which is
in tsk_allowed_cpu.

* with your patch
   1.the condition after find_idlest_cpu() does not exist,therefore
a sched domain to which cpu2 belongs to is chosen.this is socket2.(under
the for_each_domain() loop).
   2.in the next iteration, find_idlest_group() return NULL,because
there is no cpu which intersects with tsk_allowed_cpus.
   3.in select task rq,the fallback cpu is chosen even when an idle cpu
existed.

So my concern is though select_task_rq() checks the
tsk_allowed_cpus(),you might end up choosing a different path of
sched_domains compared to without this patch as shown above.

In short without the "if(new_cpu==-1)" condition we might get misled
doing unnecessary iterations over the wrong sched domains in
select_task_rq_fair().(Think about situations when not all the cpus of
socket2 are disallowed by the task,then there will more iterations in
the wrong path of sched_domains before exit,compared to what is shown
above.)

Regards
Preeti U Murthy


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