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From:   Andy Shevchenko <andy.shevchenko@gmail.com>
Date:   Fri, 7 Feb 2020 21:39:17 +0200
Message-ID: <CAHp75Vf6HJw=SpK9_HUgcMaaabs5pZEybP4SS-gc1wz5GRcqeg@mail.gmail.com>
Subject: Re: [PATCH] spi: Add FSI-attached SPI controller driver
To:     Eddie James <eajames@linux.vnet.ibm.com>
Cc:     Eddie James <eajames@linux.ibm.com>,
        linux-spi <linux-spi@vger.kernel.org>,
        Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
        Mark Brown <broonie@kernel.org>, Joel Stanley <joel@jms.id.au>,
        Andrew Jeffery <andrew@aj.id.au>
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On Fri, Feb 7, 2020 at 9:28 PM Eddie James <eajames@linux.vnet.ibm.com> wrote:
> On 2/5/20 9:51 AM, Andy Shevchenko wrote:
> > On Tue, Feb 4, 2020 at 6:06 PM Eddie James <eajames@linux.ibm.com> wrote:
> >> On 2/4/20 5:02 AM, Andy Shevchenko wrote:
> >>> On Mon, Feb 3, 2020 at 10:33 PM Eddie James <eajames@linux.vnet.ibm.com> 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.
> >>>>>
> >>>>> 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.

Yes, I missed correction of the start address in memcpy(). Otherwise
it's still the same what I was talking about.

> >>>>>> +       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.
> >
> >>>>>> +}

-- 
With Best Regards,
Andy Shevchenko