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

[Tejun Heo]

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?

Thanks.

-- 
tejun
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