On Tue, 17 Aug 2010, John Youn wrote:
> > John, I'm concerned that you're blindly assigning the USB 3.0 hub
> > descriptor to a usb_hub_descriptor. The current structure is:
> >
> > struct usb_hub_descriptor {
> > __u8 bDescLength;
> > __u8 bDescriptorType;
> > __u8 bNbrPorts;
> > __le16 wHubCharacteristics;
> > __u8 bPwrOn2PwrGood;
> > __u8 bHubContrCurrent;
> > /* add 1 bit for hub status change; round to bytes */
> > __u8 DeviceRemovable[(USB_MAXCHILDREN + 1 + 7) / 8];
> > __u8 PortPwrCtrlMask[(USB_MAXCHILDREN + 1 + 7) / 8];
> > } __attribute__ ((packed));
> >
> > The USB 3.0 hub descriptor should look something like this:
> >
> > struct usb_hub_usb3_descriptor {
> > __u8 bDescLength;
> > __u8 bDescriptorType;
> > __u8 bNbrPorts;
> > __le16 wHubCharacteristics;
> > __u8 bPwrOn2PwrGood;
> > __u8 bHubContrCurrent;
> > /* add 1 bit for hub status change; round to bytes */
This comment doesn't belong here. It's present in the USB-2.0
structure because the DeviceRemovable and PortPwrCtrlMask fields have
variable length, depending on the number of ports. In the USB-3.0
structure the fields are all fixed length.
> > __u8 bHubHdrDecLat;
> > __u16 wHubDelay;
> > __u8 DeviceRemovable[2];
> > } __attribute__ ((packed));
> >
> > The descriptor is longer, and the fields after bHubContrCurrent don't
> > align properly. So I don't think your current approach of just
> > assigning the new hub descriptor to a pointer of the old hub descriptor
> > type is going to work very well.
> >
>
> Actually the 3.0 descriptor is smaller than the largest 2.0 descriptor, so it fits into the same buffer. And most of the structure is the same so I left it like that so the core can access the same structure and fields.
>
> The differences are only at the end, so for 3.0 hubs I skipped where the hub accesses the 2.0-only fields DeviceRemovable and PortPwrCtrlMask, and put the FIXME comment. For the new 3.0 fields, there is no code to use them yet.
>
> Not a complete solution obviously but I did somewhat account for the difference.
Perhaps the best answer is to use a union.
Alan Stern
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