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Subject: Re: [PATCH] spi: Add FSI-attached SPI controller driver
To:     Andy Shevchenko <andy.shevchenko@gmail.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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From:   Eddie James <eajames@linux.vnet.ibm.com>
Message-ID: <cd8a7279-beb2-39e6-58e6-6ea505665547@linux.vnet.ibm.com>
Date:   Fri, 7 Feb 2020 14:04:22 -0600
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On 2/7/20 1:39 PM, Andy Shevchenko wrote:
> 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.


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.

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