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SubjectRe: [PATCH] net: phy: mdio_bus: add missing device_del() in mdiobus_register() error handling
Hello Andrew,

On Wed, 16 Jan 2019 16:44:39 +0100
Andrew Lunn <andrew@lunn.ch> wrote:

> > On Wed, 16 Jan 2019 15:48:29 +0100, Andrew Lunn wrote:
> >
> > > Reviewed-by: Andrew Lunn <andrew@lunn.ch>
> > >
> > > However, i wounder if it makes sense to add a label before the
> > > existing device_del() at the end of the function, and convert this,
> > > and the case above into a goto? That might scale better, avoiding the
> > > same issue in the future?
> >
> > That's another option indeed.
> >
> > Hmm, now that I looked at it, I think we should use device_unregister()
> > instead. device_unregister() does both device_del() and put_device().
>
> Hi Thomas
>
> device_unregister() does seem symmetrical with device_register() which
> is what we are trying to undo.

Even if DaveM already merged my simple fix, I had a further look at
whether we should be using device_unregister(), and in fact we should
not, but not really for a good reason: because the mdio API is not very
symmetrical.

The typical flow is:

probe() {
bus = mdiobus_alloc();
if (!bus)
return -ENOMEM;

ret = mdiobus_register(&bus);
if (ret) {
mdiobus_free(bus);

...
}

remove() {
mdiobus_unregister();
mdiobus_free();
}

mdiobus_alloc() only does memory allocation, i.e it has no side effects
on the device model data structures.

mdiobus_register() does a device_register(). If it fails, it only
cleans up with a device_del(), i.e it doesn't do the put_device() that
it should do to fully "undo" its effect.

mdiobus_unregister() does a device_del(), i.e it also doesn't do the
opposite of mdiobus_register(), which should be device_del() +
put_device() (device_unregister() is a shortcut for both).

mdiobus_free() does the put_device()

So:

* mdiobus_alloc() / mdiobus_free() are not symmetrical in terms of
their interaction with the device model data structures

* On error, mdiobus_register() leaves a non-zero reference count to the
bus->dev structure, which will be freed up by mdiobus_free()

* mdiobus_unregister() leaves a non-zero reference count to the
bus->dev structure, which will be freed up by mdiobus_free()

So, if we were to use device_unregister() in the error path of
mdiobus_register() and in mdiobus_unregister(), it would break how
mdiobus_free() works.

Best regards,

Thomas
--
Thomas Petazzoni, CTO, Bootlin
Embedded Linux and Kernel engineering
https://bootlin.com

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