From: "Rafael J. Wysocki" <rjw@sisk.pl>
To: Oliver Neukum <oliver@neukum.name>
Subject: Re: question on resume()
Date: Wed, 31 Jan 2007 09:33:28 +0100
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On Tuesday, 30 January 2007 23:32, Rafael J. Wysocki wrote:
> [Added linux-pm to the Cc list, because I'm going to talk about things that
> I know only from reading the code.]
> 
> On Tuesday, 30 January 2007 17:50, Oliver Neukum wrote:
> > Am Dienstag, 30. Januar 2007 17:32 schrieb Rafael J. Wysocki:
> > > However, you can always inspect the PF_FROZEN flag of the tasks in question
> > > if that's practicable.
> > 
> > What would I do with that information? Ignore completion of IO?
> 
> I probably should say "that depends", but that wouldn't be very helpful.
> 
> Getting back to your initial question, which is if wake_up() may be called
> from a driver's .resume() routine, I think the answer is no, it may not,
> because in that case the "notified" tasks would be removed from the wait
> queue, but the refrigerator() would (wrongly) restore their states as
> TASK_UNINTERRUPTIBLE (or TASK_INTERRUPTIBLE for wake_up_interruptible()).
> 
> Generally, you are safe if your driver only calls wake_up() from a process
> context, but not from .resume() or .suspend() routines (or from an
> unfreezeable kernel thread).

Ah, sorry, I've just realized I was wrong.  Processes in TASK_UNINTERRUPTIBLE
cannot be frozen!  So, the above only applies to wake_up_interruptible().

You don't need to call wake_up() from .resume(), because there are no tasks
to be notified this way and you shouldn't call wake_up_interruptible() from
there.

Greetings,
Rafael


-- 
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