Re: Question on release_one_tty

[Pavel Emelyanov <xemul@parallels.com>]

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
> .
>