Messages in this thread |  | | Date | Fri, 20 Oct 2017 16:57:31 +0200 | From | Greg KH <> | Subject | Re: [PATCH v10 13/15] platform/x86: wmi: create userspace interface for drivers |
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On Fri, Oct 20, 2017 at 01:54:36PM +0000, Mario.Limonciello@dell.com wrote: > > -----Original Message----- > > From: Greg KH [mailto:greg@kroah.com] > > Sent: Friday, October 20, 2017 8:22 AM > > To: Limonciello, Mario <Mario_Limonciello@Dell.com> > > Cc: dvhart@infradead.org; Andy Shevchenko <andy.shevchenko@gmail.com>; > > LKML <linux-kernel@vger.kernel.org>; platform-driver-x86@vger.kernel.org; Andy > > Lutomirski <luto@kernel.org>; quasisec@google.com; pali.rohar@gmail.com; > > rjw@rjwysocki.net; mjg59@google.com; hch@lst.de; Alan Cox > > <gnomes@lxorguk.ukuu.org.uk> > > Subject: Re: [PATCH v10 13/15] platform/x86: wmi: create userspace interface for > > drivers > > > > On Thu, Oct 19, 2017 at 12:50:16PM -0500, Mario Limonciello wrote: > > > + wblock = container_of(wdev, struct wmi_block, dev); > > > + if (!wblock) > > > + return -ENODEV; > > > > How can container_of() ever return NULL? If so, you have a very odd > > memory layout... > > > > I'm assuming this is from set_required_buffer_size right? > > The symbol is exported out for other drivers to use. It's possible for another > driver to allocate a wmi_device structure that's not part of a wblock.
Fine, but your test does not do anything at all.
> > > + list_for_each_entry_safe(wblock, next, &wmi_block_list, list) { > > > + wdriver = container_of(wblock->dev.dev.driver, > > > + struct wmi_driver, driver); > > > + if (!wdriver) > > > + continue; > > > > Same here. And other places in this file. > > > > This one it's possible that a driver isn't bound to a device, and when > that happens wdriver is NULL.
Again, that's not what you are testing at all.
container_of() is just pointer math. If you pass in NULL, you will get a non-NULL value (incremented or decremented). If you pass in a very tiny number, you might get NULL, but that's still really wrong.
In other words, these tests will _NEVER_ fail. Go ahead, try it :)
thanks,
greg k-h
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