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

[Eddie James]

On 2/7/20 4:04 PM, Andy Shevchenko wrote:
> 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.


I just tried to implement as you suggested but I realized something: The 
value is already swapped from BE to CPU when the register is read in 
fsi_spi_read_reg. It happens to work out correctly to use 
put_unaligned_be64 on a LE CPU to flip the bytes here. But on a BE CPU, 
this wouldn't be correct I think. Now I don't anticipate this driver 
running on a BE CPU, but I think it is weird to flip it twice, and 
better to do it manually here.

What do you think Andy?

Thanks,

Eddie


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