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

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Date:   Fri, 7 Feb 2020 13:28:40 -0600
From:   Eddie James <eajames@...ux.vnet.ibm.com>
To:     Andy Shevchenko <andy.shevchenko@...il.com>,
        Eddie James <eajames@...ux.ibm.com>
Cc:     linux-spi <linux-spi@...r.kernel.org>,
        Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
        Mark Brown <broonie@...nel.org>, Joel Stanley <joel@....id.au>,
        Andrew Jeffery <andrew@...id.au>
Subject: Re: [PATCH] spi: Add FSI-attached SPI controller driver


On 2/5/20 9:51 AM, Andy Shevchenko wrote:
> On Tue, Feb 4, 2020 at 6:06 PM Eddie James <eajames@...ux.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@...ux.vnet.ibm.com> wrote:
>>>> On 1/30/20 10:37 AM, Andy Shevchenko wrote:
>>>>> On Wed, Jan 29, 2020 at 10:09 PM Eddie James <eajames@...ux.ibm.com> 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.
>
>>>>>> +}