Re: [PATCH] gpss: core: no waiters left behind on deregister

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



On Wed, Jun 26, 2019 at 01:04:07PM +0200, Oliver Neukum wrote:
> Am Dienstag, den 25.06.2019, 09:04 +0200 schrieb Johan Hovold:
> > On Mon, Jun 24, 2019 at 10:33:23AM +0200, Oliver Neukum wrote:
> > > If you deregister a device you need to wake up all waiters
> > > as there will be no further wakeups.
> > > 
> > > Signed-off-by: Oliver Neukum <oneukum@xxxxxxxx>
> > > ---
> > >  drivers/gnss/core.c | 2 +-
> > >  1 file changed, 1 insertion(+), 1 deletion(-)
> > > 
> > > diff --git a/drivers/gnss/core.c b/drivers/gnss/core.c
> > > index e6f94501cb28..0d13bd2cefd5 100644
> > > --- a/drivers/gnss/core.c
> > > +++ b/drivers/gnss/core.c
> > > @@ -303,7 +303,7 @@ void gnss_deregister_device(struct gnss_device *gdev)
> > >  	down_write(&gdev->rwsem);
> > >  	gdev->disconnected = true;
> > >  	if (gdev->count) {
> > > -		wake_up_interruptible(&gdev->read_queue);
> > > +		wake_up_interruptible_all(&gdev->read_queue);
> > 
> > GNSS core doesn't have any exclusive waiters, so no need to use use the
> > exclusive wake-up (all) interface.
> 
> Well, yes, but that is the problem. In gnss_read() you drop the lock:

> That means that an arbitrary number of tasks can get here.
> 
>                 ret = wait_event_interruptible(gdev->read_queue,
>                                 gdev->disconnected ||
>                                 !kfifo_is_empty(&gdev->read_fifo));
> 
> Meaning that an arbitrary number can be sleeping here.

I understand wait you're getting at, but I think your mistaken regarding
exclusive wait. Note that wait_event_interruptible() uses nonexclusive
wait.

> Yet in gnss_deregister_device() you use a simple wake_up:
> 
> void gnss_deregister_device(struct gnss_device *gdev)
> 
> {
> 
>         down_write(&gdev->rwsem);
>         gdev->disconnected = true;
>         if (gdev->count) {
>                 wake_up_interruptible(&gdev->read_queue);
> 
> 
> wake_up_interruptible() will wake up one waiting task. But after that
> the device is gone. There will be no further events. The other tasks
> will sleep forever.

No, wake_up_interruptible() will wake up all nonexclusive waiters,
which is all we care about here.

Johan



[Index of Archives]     [Linux Media]     [Linux Input]     [Linux Audio Users]     [Yosemite News]     [Linux Kernel]     [Linux SCSI]     [Old Linux USB Devel Archive]

  Powered by Linux