2012/4/26 Alan Stern <stern@xxxxxxxxxxxxxxxxxxx>:
> On Thu, 26 Apr 2012, Greg KH wrote:
>
>> On Thu, Apr 26, 2012 at 10:25:41AM -0400, Alan Stern wrote:
>> > On Thu, 26 Apr 2012, Calin Demian wrote:
>> >
>> > > I don't know how, but adding the following line in function
>> > > quirk_usb_handoff_ohci makes the system work fine:
>> > >
>> > > /* reset requires max 10 us delay */
>> > > + udelay(1);
>> > > for (cnt = 30; cnt > 0; --cnt) { /* ... allow extra time */
>> > > if ((readl(base + OHCI_CMDSTATUS) & OHCI_HCR) == 0)
>> > >
>> > > I observed thad the condition in the for loop is allways true, so the
>> > > delay inside the loop is never executed.
>>
>> That implies that something is wrong with the complier here, and it
>> optimized out the loop? Are you sure that really happened?
>
> Here's the complete context:
>
> /* reset requires max 10 us delay */
> for (cnt = 30; cnt > 0; --cnt) { /* ... allow extra time */
> if ((readl(base + OHCI_CMDSTATUS) & OHCI_HCR) == 0)
> break;
> udelay(1);
> }
>
> I think Calim means that the "(readl(base + OHCI_CMDSTATUS) & OHCI_HCR)
> == 0" test succeeds on the first pass through the loop, so we break out
> early and the udelay(1) statement never gets executed.
>
> Alan Stern
>
Yes, that's what I was saying. Storing the readl result in a variable
and displaying it later shows that the (readl_result & OHCI_HCR) is 0
at first pass, but that doesn't imply reset is finished, I think. It
would be too fast. Maybe the readl doesn't work for some reason...
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