Re: [PATCH] spi: Add FSI-attached SPI controller driver

[Andy Shevchenko]

On Fri, Feb 7, 2020 at 11:04 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> wrote:
> On 2/7/20 2:34 PM, Andy Shevchenko wrote:
> > On Fri, Feb 7, 2020 at 10:04 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> wrote:
> >> On 2/7/20 1:39 PM, Andy Shevchenko wrote:
> >>> On Fri, Feb 7, 2020 at 9:28 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> wrote:
> >>>> On 2/5/20 9:51 AM, Andy Shevchenko wrote:
> >>>>> On Tue, Feb 4, 2020 at 6:06 PM Eddie James <eajames@xxxxxxxxxxxxx> wrote:
> >>>>>> On 2/4/20 5:02 AM, Andy Shevchenko wrote:
> >>>>>>> On Mon, Feb 3, 2020 at 10:33 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> wrote:
> >>>>>>>> On 1/30/20 10:37 AM, Andy Shevchenko wrote:
> >
> >>>>>>>>>> +       for (i = 0; i < num_bytes; ++i)
> >>>>>>>>>> +               rx[i] = (u8)((in >> (8 * ((num_bytes - 1) - i))) & 0xffULL);
> >>>>>>>>> Redundant & 0xffULL part.
> >>>>> For me it looks like
> >>>>>
> >>>>>      u8 tmp[8];
> >>>>>
> >>>>>      put_unaligned_be64(in, tmp);
> >>>>>      memcpy(rx, tmp, num_bytes);
> >>>>>
> >>>>> put_unaligned*() is just a method to unroll the value to the u8 buffer.
> >>>>> See, for example, linux/unaligned/be_byteshift.h implementation.
> >>>> Unforunately it is not the same. put_unaligned_be64 will take the
> >>>> highest 8 bits (0xff00000000000000) and move it into tmp[0]. Then
> >>>> 0x00ff000000000000 into tmp[1], etc. This is only correct for this
> >>>> driver IF my transfer is 8 bytes. If, for example, I transfer 5 bytes,
> >>>> then I need 0x000000ff00000000 into tmp[0], 0x00000000ff000000 into
> >>>> tmp[1], etc. So I think my current implementation is correct.
> >>> Yes, I missed correction of the start address in memcpy(). Otherwise
> >>> it's still the same what I was talking about.
> >>
> >> I see now, yes, thanks.
> >>
> >> Do you think this is worth a v3? Perhaps put_unaligned is slightly more
> >> optimized than the loop but there is more memory copy with that way too.
> > I already forgot the entire context when this has been called. Can you
> > summarize what the sequence(s) of num_bytes are expected usually.
> >
> > IIUC if packets small, less than 8 bytes, than num_bytes will be that value.
> > Otherwise it will be something like 8 + 8 + 8 ... + tail. Is it
> > correct assumption?
>
>
> Yes, it will typically be 8 + 8 +... remainder. Basically, on any RX,
> the driver polls for the rx register full. Once full, it will read
> however much data is left to be transferred. Since we use min(len, 8)
> then we read 8 usually, until we get to the end.

I asked that because we might have a better optimization, i.e, call
directly put_unaligned_be64() when we know that length is 8 bytes. For
the rest your approach might be simpler. Similar for the TX case.

> >>>>>>>>>> +       return num_bytes;
> >>>>>>>>>> +}

-- 
With Best Regards,
Andy Shevchenko