On Wed, 31 Jan 2007, Pavel Machek wrote:
> Hi!
>
> > > 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()
>
> "cease IO"? No, I believe it is enough not to start new I/O. Userspace
> is frozen at that point, it can't ask you to do I/O.
There may be I/O requests sitting in a queue, already submitted by
userspace. The suspend method should wait for existing I/O to complete
and stop processing new entries from the queue.
> > Isn't that a race until suspend() is called?
>
> I do not think so.
The part about not calling wake_up_interruptible() is indeed a race. We
have:
1. Task is frozen.
2. Driver must not call wake_up_interruptible().
3. Driver's suspend() method is called.
How is the driver supposed to satisfy (2) before (3) has occurred?
In fact this shouldn't matter. There shouldn't be anything wrong with
calling wake_up_interruptible() on a frozen task.
> > 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?
>
> We reuse signal handling code to do that for us. It is same situation
> as when someone signals task doing I/O.
Again you misunderstood the question. The driver must start queued I/O
when its resume() method is called. It should then be okay for the driver
to call wake_up_interruptible(), even before tasks are unfrozen.
Alan Stern
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