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SubjectRe: [linux-usb-devel] Re: [OOPS, usbcore, releaseintf] 2.6.0-test10-mm1
On Wed, 10 Dec 2003, Greg KH wrote:

> Sorry, the pci driver has a remove() function, that's what I ment.
> But anyway, yes, it does get called. It looks like it is time to help
> explain the driver core...
> When pci_unregister_driver() gets called, here is what happens:
> - pci_unregister_driver calls driver_unregister() with a pointer
> to the pci_driver->driver field.
> - That pci_driver->driver field is set up when
> pci_register_driver() is called, and contains pointers to the
> pci_device_remove function.
> - driver_unregister() calls bus_remove_driver()
> - bus_remove_driver() locks some locks and calls
> driver_detach().
> - driver_detach() walks the list of all devices that this driver
> is attached to, and calls device_release_driver() on every
> device.

That's about as far as I had followed it. I should have kept going... but
it wouldn't have made any difference.

> - device_release_driver() unlinks some sysfs files, does some
> power management stuff, and then calls the remove() function
> that is associated with that driver. That remove function for
> a pci driver is pci_device_remove()
> - pci_device_remove() then calls down to the pci_driver's remove
> function, which in our case for a USB PCI Host controller
> driver would be usb_hcd_pci_remove()
> - I think you can follow what usb_hcd_pci_remove() does. After
> it is finished, the call stack is unwound, and eventually
> returns back to the caller of pci_unregister_driver().
> Now grasshopper, are you wise in the ways of the driver core or are you
> wishing you never asked? :)

Both, I think. I still don't see where pci_unregister_driver() ends up
waiting for the reference count to drop to 0. In fact, I think maybe you
agree that it _doesn't_ wait. Which was my earlier point.

> > Or sleeping until the actual release function (struct hc_driver->hcd_free)
> > is called. But you have to make sure it was called for the host you are
> > trying to deregister, not some other host.
> That is done by the following logic (yeah, it's a maze of twisty
> paths...)
> - in usb_hcd_pci_remove() we call usb_deregister_bus() to
> unregister the bus structure.
> - usb_deregister_bus() calls class_device_unregister() with a
> pointer to the bus's class device structure. That class
> device structure was previously reigstered to the
> usb_host_class. The usb_host_class's release function is the
> usb_host_release() call. That release function will be called
> when the last reference on the class device is release.
> - usb_host_release() calls the release() function of the usb_bus
> structure, which points back to how to clean up the memory for
> that specific usb bus driver. Now for all host controllers
> that use the hcd framework, that points to the
> hcd_pci_release() function. Which will then call the
> hcd_free() function for that specific hcd driver.
> Does that help? Or does your head hurt even more now?

I had already figured that much out for myself. So
pci_unregister_driver() will follow this all the way down to
class_device_unregister(), which will decrement a reference count and
return immediately without calling usb_host_release() if the count isn't
0, which it wasn't in this case.

As a result pci_unregister_driver() returns immediately and the module is
unloaded. Later on when usb_host_release() does get called -- BOOM!

> So, if we wait for the class_device_unregister() function to actually
> free the memory (wait for the usb_host_release() function to complete)
> then we know it is absolutely safe for the driver to be removed from the
> system.
> > Of course, if all you want to do is unload the module then it doesn't
> > matter which host is which. You just have to wait until they are all
> > gone.
> Exactly, and that will happen, if we wait on that
> class_device_unregister() call. An example of how to do that can be
> seen in the i2c_del_adapter() function in drivers/i2c/i2c-core.c.

In the absence of the class_device_unregister_wait() function, the patch
you have created appears to be necessary.

As Pat LaVarre would say, I think we're agreeing violently.

Alan Stern

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