Re: [PATCH 01/18] sched: select_task_rq_fair clean up

From: Namhyung Kim
Date: Thu Dec 20 2012 - 23:28:49 EST


Hi,

On Tue, 11 Dec 2012 12:00:55 +0530, Preeti U. Murthy wrote:
> 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

AFAIK The tsk->cpus_allowed cannot be changed during the operation since
it's protected by tsk->pi_lock. I can see the following comment:

kernel/sched/core.c:
/*
* The caller (fork, wakeup) owns p->pi_lock, ->cpus_allowed is stable.
*/
static inline
int select_task_rq(struct task_struct *p, int sd_flags, int wake_flags)
{
int cpu = p->sched_class->select_task_rq(p, sd_flags, wake_flags);



Thanks,
Namhyung
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