Re: [PATCH] slow_work_thread() should do the exclusive wait

[Oleg Nesterov]

On 04/13, Trond Myklebust wrote:
>
> On Mon, 2009-04-13 at 20:17 +0200, Oleg Nesterov wrote:
> > slow_work_thread() sleeps on slow_work_thread_wq without WQ_FLAG_EXCLUSIVE,
> > this means that slow_work_enqueue()->__wake_up(nr_exclusive => 1) wakes up
> > all kslowd threads. Afaics this is not what we want, change slow_work_thread()
> > to use prepare_to_wait_exclusive().
> > 
> > Signed-off-by: Oleg Nesterov <oleg@xxxxxxxxxx>
> > 
> > --- 6.30/kernel/slow-work.c~1_SW_EXCLUSIVE	2009-04-06 00:03:42.000000000 +0200
> > +++ 6.30/kernel/slow-work.c	2009-04-13 19:40:20.000000000 +0200
> > @@ -372,8 +372,8 @@ static int slow_work_thread(void *_data)
> >  		vsmax *= atomic_read(&slow_work_thread_count);
> >  		vsmax /= 100;
> >  
> > -		prepare_to_wait(&slow_work_thread_wq, &wait,
> > -				TASK_INTERRUPTIBLE);
> > +		prepare_to_wait_exclusive(&slow_work_thread_wq, &wait,
> > +						TASK_INTERRUPTIBLE);
> >  		if (!freezing(current) &&
> >  		    !slow_work_threads_should_exit &&
> >  		    !slow_work_available(vsmax) &&
> > 
> 
> Should that really be TASK_INTERRUPTIBLE? I don't see anything obvious
> in the enclosing for(;;) loop that checks for or handles signals...

I guess TASK_INTERRUPTIBLE was chosen to not contribute to calc_load(),
nr_active() returns nr_running + nr_uninterruptible.

Oleg.

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