On Fri, Jul 23, 2021 at 02:29:03PM +0200, Christoph Hellwig wrote:
> On Fri, Jul 23, 2021 at 09:22:27AM -0300, Jason Gunthorpe wrote:
> > > But do we even need the else part? Assingning &mc_dev->dev is
> > > equivalent to the default per-device set anyway, isn't it?
> >
> > Not quite, the default is this:
> >
> > if (!device->dev_set)
> > vfio_assign_device_set(device, device);
> >
> > Where 'device' is the vfio_device itself, the above is connecting to
> > the struct fsl_mc_device.
>
> Isn't there a 1:1 relation?
Yes, but one is a struct device * and the other is a struct
vfio_device *. They don't have the same pointer value.
The above default code has the goal of creating a singleton dev_set
because no other driver can reasonably obtain the 'struct vfio_device
*'.
FSL is using a 'struct device *' as the key and the interesting case
is when two drivers are loaded such that:
mc_dev->dev.parent == &mc_dev->dev
Then they will have the same dev_set and the locking works out. The
vfio_device * can never be == &mc_dev->dev because it is a different
allocation.
Jason
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