Re: [RFC 1/1] workqueue: Add mod_delayed_work()

[Gustavo F. Padovan]

Hi Tejun,

* Tejun Heo <tj@xxxxxxxxxx> [2011-06-25 13:43:31 +0200]:

> Hello,
> 
> On Fri, Jun 24, 2011 at 07:27:37PM -0300, Gustavo F. Padovan wrote:
> > mod_delayed_work() updates a timer if the work is pending otherwise calls
> > queue_delayed_work_on() to queue the work with the specified delay.
> > 
> > Call cancel_delayed_work_sync() and then queue_delayed_work() again to
> > change a timer's delays is too expensive (and requires process context).
> > Istead we call mod_delayed_work() to only modify the timer's timeout.
> 
> Yes, this part of the interface is lacking.  It might be best to
> modify queue_delayed_work() to adjust the timer according to the new
> timeout but we would need to audit the current users to make sure
> nothing breaks and I agree introducing a new function probably makes
> sense.
> 
> > +int mod_delayed_work(struct workqueue_struct *wq,
> > +			struct delayed_work *dwork, unsigned long delay)
> > +{
> > +	struct timer_list *timer = &dwork->timer;
> > +	struct work_struct *work = &dwork->work;
> > +
> > +	if (!test_bit(WORK_STRUCT_PENDING_BIT, work_data_bits(work)))
> > +		return queue_delayed_work_on(-1, wq, dwork, delay);
> > +
> > +	BUG_ON(!timer_pending(timer));
> > +
> > +	mod_timer(timer, jiffies + delay);
> > +
> > +	return 0;
> > +}
> 
> But I think the current implementation is as it is because modifying
> delayed work safely wasn't very simple.  The above code is broken in
> multiple ways - a delayed work could be pending without timer pending,
> and timer may expire after test_bit() but before the rest of the code.
> 
> I haven't thought about it too hard but think it would require the
> timer sync part of __cancel_work_timer() (sans wait_on_work()) to get
> it correctly.  Care to delve into it?

Sure, I can do that. I'm kind of stuck in the Bluetooth changes I need to do
due to this workqueue patch. I'll send a -v2 soon.

	Gustavo
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