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

[Eddie James]

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:
> > > > > On Wed, Jan 29, 2020 at 10:09 PM Eddie James <eajames@xxxxxxxxxxxxx> wrote:
> ...
>
> > > > > > +       struct device *dev;
> > > > > Isn't fsl->dev the same?
> > > > > Perhaps kernel doc to explain the difference?
> > > > No, it's not the same, as dev here is the SPI controller. I'll add a
> > > > comment.
> > > Why to have duplication then?
> >
> > Nothing is being duplicated, the two variables are storing entirely
> > different information, both of which are necessary for each SPI
> > controller that this driver is driving.
> Oh, I see now, thanks!
>
> ...
>
> > > > > > +       for (i = 0; i < num_bytes; ++i)
> > > > > > +               rx[i] = (u8)((in >> (8 * ((num_bytes - 1) - i))) & 0xffULL);
> > > > > Redundant & 0xffULL part.
> > > > >
> > > > > Isn't it NIH of get_unalinged_be64 / le64 or something similar?
> > > > No, these are shift in/out operations. The read register will also have
> > > > previous operations data in them and must be extracted with only the
> > > > correct number of bytes.
> > > Why not to call put_unaligned() how the tail in this case (it's 0 or
> > > can be easily made to be 0) will affect the result?
> >
> > The shift-in is not the same as any byte-swap or unaligned operation.
> > For however many bytes we've read, we start at that many bytes
> > left-shifted in the register and copy out to our buffer, moving right
> > for each next byte... I don't think there is an existing function for
> > this operation.
> 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.


Thanks,

Eddie


> > > > > > +       return num_bytes;
> > > > > > +}
> > > > > > +static int fsi_spi_data_out(u64 *out, const u8 *tx, int len)
> > > > > > +{
> > > > > Ditto as for above function. (put_unaligned ...)
> > > Ditto.
> >
> > I don't understand how this could work for transfers of less than 8
> > bytes, any put_unaligned would access memory that it doesn't own.
> Ditto.
>
> > > > > > +}