On Wed, May 29, 2019 at 09:28:52PM -0700, David Miller wrote:
> From: Stefano Garzarella <sgarzare@xxxxxxxxxx>
> Date: Tue, 28 May 2019 12:56:20 +0200
>
> > @@ -68,7 +68,13 @@ struct virtio_vsock {
> >
> > static struct virtio_vsock *virtio_vsock_get(void)
> > {
> > - return the_virtio_vsock;
> > + struct virtio_vsock *vsock;
> > +
> > + mutex_lock(&the_virtio_vsock_mutex);
> > + vsock = the_virtio_vsock;
> > + mutex_unlock(&the_virtio_vsock_mutex);
> > +
> > + return vsock;
>
> This doesn't do anything as far as I can tell.
>
> No matter what, you will either get the value before it's changed or
> after it's changed.
>
> Since you should never publish the pointer by assigning it until the
> object is fully initialized, this can never be a problem even without
> the mutex being there.
>
> Even if you sampled the the_virtio_sock value right before it's being
> set to NULL by the remove function, that still can happen with the
> mutex held too.
Yes, I found out when I was answering Jason's question [1]. :(
I proposed to use RCU to solve this issue, do you agree?
Let me know if there is a better way.
>
> This function is also terribly named btw, it implies that a reference
> count is being taken. But that's not what this function does, it
> just returns the pointer value as-is.
What do you think if I remove the function, using directly the_virtio_vsock?
(should be easier to use with RCU API)
Thanks,
Stefano
[1] https://lore.kernel.org/netdev/20190529105832.oz3sagbne5teq3nt@steredhat
_______________________________________________
Virtualization mailing list
Virtualization@xxxxxxxxxxxxxxxxxxxxxxxxxx
https://lists.linuxfoundation.org/mailman/listinfo/virtualization
