Re: question on resume()

[Oliver Neukum <oneukum@suse.de>]

Am Mittwoch, 31. Januar 2007 09:49 schrieb Rafael J. Wysocki:
> On Wednesday, 31 January 2007 09:40, Oliver Neukum wrote:
> > Am Mittwoch, 31. Januar 2007 09:33 schrieb Rafael J. Wysocki:
> > > On Tuesday, 30 January 2007 23:32, Rafael J. Wysocki wrote:
> > 
> > > > 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().
> > 
> > So the kernel will wait for tasks in TASK_UNINTERRUPTIBLE to finish IO
> > before it calls suspend()? I am confused.
> 
> Yes, it will.  The process freezer can only return success if there are no more
> TASK_UNINTERRUPTIBLE tasks.  Otherwise it fails (after a timeout).

So, this means, on suspend():

1. Don't worry about TASK_UNINTERRUPTIBLE
2. Do worry about TASK_INTERRUPTIBLE
We have to cease IO and must not call wake_up_interruptible()

Isn't that a race until suspend() is called?

On resume():

1. Don't worry about TASK_UNINTERRUPTIBLE
2. Do not restart IO that may call wake_up_interruptible()

When do we restart such IO?

	Regards
		Oliver