Re: about split the udev

[Kay Sievers <kay.sievers@vrfy.org>]

On Sun, Jan 25, 2004 at 12:06:52AM +0800, Ling, Xiaofeng wrote:
> Thank you for continuing my work. (I'm in a long vocation for two weeks)
> I'm already working on making udev to a daemon but I like your idea to let udevd 
> invoke udev. This way there will be less effect to current udev.

Yes, but now we wait for the exec of udev , before we do the next.
This is much too slow if you have a lot of callouts in the rules.
I think we need something that can execute without blocking, but we
need to care that "remove" is much much faster than "add".

Greg, do we need to care about the order of execution of udev for
different device pathes?
If not, we may fork all udev's in the background, unless we have one
with the same path already running.
We may maintain the state of the fork in our sequence list and delay
all events for device pathes udev is already working on.

If we receive the SIGCHLD, then we execute the delayed one.
Hmm, if this may work?


> I've some comments about the timeout value, see below.
> > From: Kay Sievers [mailto:kay.sievers@vrfy.org] 
> > So it's now possible to use the test script at any time, 
> > cause it resets the daemon, if real hotplug event coming in 
> > later all missing nimbers will be skipped after a timeout of 
> > 10 seconds and the queued events are applied.

Yes, but we need this to follow the log while testing.
If we have a working codebase, we can set it to something useful.


> Is 10 seconds a little too longer? 
> If there is only one event lost and one user have to wait 10 seconds when plugging in 
> a device, is this acceptable for he/she?
> Initial, my idea is making the timeout value related with the number of lost events, so
> if happened there is one event lost, we just need to wait for 3 seconds.If two, wait for 6 seconds.
> Of couse, maybe a maximum value is also needed.

I think this is already solved.
On timeout we skip all missing events and go ahead with the next
event in the queue. So if 100 events in a row missing, there will
be one timeout only.


thanks,
Kay


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