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

Trond Myklebust
2009-04-13T19:04:19Z

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@redhat.com>
>
> --- 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...

Cheers
Trond
--
Trond Myklebust
Linux NFS client maintainer

NetApp
Trond.Myklebust@netapp.com