Am Mittwoch, 31. Januar 2007 12:37 schrieb Jiri Kosina:
> On Wed, 31 Jan 2007, Oliver Neukum wrote:
>
> > in hiddev_ioctl() IO is submitted with usbhid_submit_report() and waited
> > for with usbhid_wait_io(), which uses the following condition:
> > if (!wait_event_timeout(hid->wait, (!test_bit(HID_CTRL_RUNNING,
> > &usbhid->iofl) &&
> > !test_bit(HID_OUT_RUNNING,
> > &usbhid->iofl)),
> > 10*HZ)) {
> > It seems to me that in the case of concurrent IO this condition will swallow
> > the wakeup and timeout. Or am I dense today?
>
> Hi Oliver,
>
> I don't currently see the scenario you are talking about, could you
> elaborate a little bit more, please?
>
> Every URB submission (both ctrl and out) is paired with wake_up() in its
> completion handler. Before the wake_up() is issued, the HID_CTRL_RUNNING
> or HID_OUT_RUNNING flag is cleared. As this is a "1:1 pairing", it's
> guaranteed that there will once be a wake_up() called on the waitqueue
But it is not guaranteed that you will already be on the waitqueue.
The URB may complete before you call usbhid_wait_io(), then it'll
dependent on the flags.
> with both the flags unset (if a timeout doesn't happen sooner, which is
> OK). If this "final" wake_up() is called before the usbhid_wait_io() is
> called, this is also OK, because then the condition will evaluate to true
> and the task won't event sleep on the waitqueue.
Who tells you that the flags will still be cleared? Somebody else may
have called usbhid_submit_report() by the time you check the flags.
Regards
Oliver
>
> Thanks,
>
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