On Wed, Feb 05, 2020 at 10:23:00AM -0800, Dan Williams wrote:
> > > > 506 if (device_add(dev) != 0) {
> > > > 507 dev_err(dev, "%s: failed\n", __func__);
> > > > 508 put_device(dev);
> > > > ^^^^^^^^^^^^^^^
> > > > 509 }
> > > > 510 put_device(dev);
> > > > ^^^^^^^^^^^^^^
> > > > 511 if (dev->parent)
> > > > 512 put_device(dev->parent);
> > > > 513 }
> > > >
> > > > We call get_device() from __nd_device_register(), I guess. It seems
> > > > buggy to call put device twice on error.
> > >
> > > The registration path does:
> > >
> > > get_device(dev);
> > >
> > > async_schedule_dev_domain(nd_async_device_register, dev,
> > > &nd_async_domain);
> > >
> > > ...and device_add() does its own get_device().
> >
> > device_add() does its own put_device() at the end so it's a net zero.
> >
>
> It does it's own, yes, but the put_device() after device_add() failure
> is there to drop the reference taken by device_initialize().
> Otherwise, device_add() has always documented:
>
> * NOTE: _Never_ directly free @dev after calling this function, even
> * if it returned an error! Always use put_device() to give up your
> * reference instead.
>
> ...so what am I missing?
The "never call kfree" is hopefully straight forward because the kobject
needs to do its own cleanup.
__nvdimm_create() allocates the dev.
nd_device_register() calls device_initialize() which call kobject_init()
so the refcount is 1.
__nd_device_register() call get_device() so the refcount is now two.
nd_async_device_register() decrements the refcount once on success.
But if device_add() fails then it decrements it twice. Now the refcount
is zero so we call nvdimm_release(). This leads to a use after free on
the next line:
put_device(dev);
if (dev->parent)
There is a trick here because depending on the debug options it
might free immediately or it might call nvdimm_release() after 4
seconds. See kobject_release() for details.
Either way if device_add() fails we return back to __nvdimm_create()
and return the zero reference count "nvdimm" pointer, which is going
to be a problem.
regards,
dan carpenter
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