linux-kernel - Question on release_one

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Date:	Thu, 7 Aug 2014 12:25:44 +0400
From:	Cyrill Gorcunov <gorcunov@...il.com>
To:	LKML <linux-kernel@...r.kernel.org>, Jiri Slaby <jslaby@...e.cz>,
	Greg Kroah-Hartman <gregkh@...uxfoundation.org>
Cc:	Pavel Emelyanov <xemul@...allels.com>
Subject: Question on release_one_tty

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?

	Cyrill
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