Re: Question on release_one_tty

[Cyrill Gorcunov]

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.
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