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

[Eddie James <eajames@linux.vnet.ibm.com>]

On 2/7/20 4:04 PM, Andy Shevchenko wrote:
> On Fri, Feb 7, 2020 at 11:04 PM Eddie James <eajames@linux.vnet.ibm.com> wrote:
>> On 2/7/20 2:34 PM, Andy Shevchenko wrote:
>>> On Fri, Feb 7, 2020 at 10:04 PM Eddie James <eajames@linux.vnet.ibm.com> wrote:
>>>> 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.
>>>>>>> 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;
>>>>>>>>>>>> +}