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Message-ID: <53E338CA.9070503@parallels.com>
Date: Thu, 7 Aug 2014 12:28:58 +0400
From: Pavel Emelyanov <xemul@...allels.com>
To: Cyrill Gorcunov <gorcunov@...il.com>,
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 12:25 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.
> Cyrill
> .
>
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