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Date:	Thu, 7 Aug 2014 13:05:28 +0400
From:	Pavel Emelyanov <xemul@...allels.com>
To:	Cyrill Gorcunov <gorcunov@...il.com>
CC:	LKML <linux-kernel@...r.kernel.org>, Jiri Slaby <jslaby@...e.cz>,
	Greg Kroah-Hartman <gregkh@...uxfoundation.org>
Subject: Re: Question on release_one_tty

On 08/07/2014 12:34 PM, Cyrill Gorcunov wrote:
> On Thu, Aug 07, 2014 at 12:28:58PM +0400, Pavel Emelyanov wrote:
>> On 08/07/2014 12:25 PM, Cyrill Gorcunov wrote:
>>> Hi guys, could you please explain me the sequence
>>>
>>> static void release_one_tty(struct work_struct *work)
>>> {
>>> 	struct tty_struct *tty =
>>> 		container_of(work, struct tty_struct, hangup_work);
>>> 	struct tty_driver *driver = tty->driver;
>>>
>>> 	if (tty->ops->cleanup)
>>> 		tty->ops->cleanup(tty);
>>>
>>> 	tty->magic = 0;
>>> -->	tty_driver_kref_put(driver);
>>> -->	module_put(driver->owner);
>>>
>>> why tty_driver_kref_put is called before module_put? As far as I understand
>>> tty_driver_kref_put may call the destruct_tty_driver which eventually does
>>>
>>> static void destruct_tty_driver(struct kref *kref)
>>> {
>>> 	struct tty_driver *driver = container_of(kref, struct tty_driver, kref);
>>> 	...
>>> 	kfree(driver->cdevs);
>>> 	kfree(driver->ports);
>>> 	kfree(driver->termios);
>>> 	kfree(driver->ttys);
>>> -->	kfree(driver);
>>> }
>>>
>>> so that the module_put(driver->owner) would access freed memory. Should not we
>>> call the reverse module_put and then tty_driver_kref_put, or I miss something
>>> obvious?
>>
>> If you put the module it can be unloaded at any time killing the code that would
>> be potentially required by kref_put.
> 
> So how this code supposed to work then? I mean tty_driver_kref_put must never call
> for destruct_tty_driver, otherwise we're accessing freed memory.

mod = driver->owner;
tty_driver_kref_put(driver);
module_put(mod);

Check the upstream whether the same issue exists there.

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