[patch] input: function call order in serio_exit()

[patch] input: function call order in serio_exit()

[Németh Márton]

[-- Attachment #1: Type: text/plain, Size: 385 bytes --]

Hi,

the order of the bus registration and the kthread start was changed
between linux kernel 2.6.17.11 and 2.6.18. The order is now first
register the bus and then start the kthread. The serio_exit() left
unchanged.

I think that the order of the function calls in serio_exit() should also
be changed: first stop the kthread and then unregister the bus.

What do you think?

	NMarci


[-- Attachment #2: serio-exit.diff --]
[-- Type: text/plain, Size: 395 bytes --]

--- linux-2.6.19-rc2.orig/drivers/input/serio/serio.c	2006-10-13 18:25:04.000000000 +0200
+++ linux-2.6.19-rc2/drivers/input/serio/serio.c	2006-10-17 08:17:30.000000000 +0200
@@ -958,8 +958,8 @@ static int __init serio_init(void)
 
 static void __exit serio_exit(void)
 {
-	bus_unregister(&serio_bus);
 	kthread_stop(serio_task);
+	bus_unregister(&serio_bus);
 }
 
 subsys_initcall(serio_init);

Re: [patch] input: function call order in serio_exit()

[Dmitry Torokhov]

On Sunday 22 October 2006 11:45, Németh Márton wrote:
> Hi,
> 
> the order of the bus registration and the kthread start was changed
> between linux kernel 2.6.17.11 and 2.6.18. The order is now first
> register the bus and then start the kthread. The serio_exit() left
> unchanged.
> 
> I think that the order of the function calls in serio_exit() should also
> be changed: first stop the kthread and then unregister the bus.
> 
> What do you think?
> 

Hi,

It really does not matter - if bus is unregistered that means the there
are no drivers using it and so noone can submit new requests to kseriod.

I'd leave it as is.

-- 
Dmitry