On 10/10/2017 07:48 AM, Kishon Vijay Abraham I wrote:
> Hi,

Hi,

[...]

>>>>>>>>>>> -   dwc3_flush_cache((uintptr_t)trb, sizeof(*trb));
>>>>>>>>>>> +   dwc3_flush_cache((uintptr_t)dwc->ep0_trb_addr, sizeof(*trb) * 
>>>>>>>>>>> 2);
>>>>>>>>>>
>>>>>>>>>> Why *2 ?
>>>>>>>>>
>>>>>>>>> Because its allocated as sizeof(*dwc->ep0_trb) * 2 below. This is not
>>>>>>>>> strictly required as dwc3_flush_cache() rounds up the size to
>>>>>>>>> CACHELINE_SIZE but from a caller POV, flush everything we allocated.
>>>>>>>>
>>>>>>>> Can the other TRB be in use ? Maybe aligning the TRBs to cacheline size
>>>>>>>> would be better ?
>>>>>>>>
>>>>>>> A single trb is 16 bytes in size and two of them are allocated
>>>>>>> contiguously.
>>>>>>
>>>>>> Why are two allocated continuously ? (I am not dwc3 expert)
> 
> The TRB's should be allocated contiguously for dwc3 and only the base of the
> entire TRB table is programmed in the HW.
>  ________________ <------------------ TRB table base address
> |     TRB0       |
> |________________|
> |     TRB1       |
> |________________|
> |     TRB2       |
> |________________|
> |     TRBn       |
> |________________|
> 
> 
>>>>>
>>>>> Neither am I. I did try to pad to the dwc_trb structure such that each
>>>>> trb is 64 bytes in size but this leads to failures when testing. I
>>>>> didn't get a chance to debug this though. I suspect its because the code
>>>>> expects the trbs to be contiguous and/or 16 bytes in size.
> 
> It's not the code but it's the HW.

That'd imply we need either some sort of flushing scheme or non-cachable
memory allocation. What does Linux do ?

-- 
Best regards,
Marek Vasut
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