Messages in this thread | | | Subject | Re: [PATCH] spi: Add FSI-attached SPI controller driver | From | Eddie James <> | Date | Mon, 10 Feb 2020 14:50:17 -0600 |
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On 2/10/20 2:33 PM, Andy Shevchenko wrote: > On Mon, Feb 10, 2020 at 10:05 PM Eddie James <eajames@linux.vnet.ibm.com> wrote: >> 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. > Hmm... Any BE conversion op on BE architecture is no-op. > Same for LE on LE.
Right. So regardless of architecture, by the time we get to fsi_spi_data_in, the data is in the correct endianness. But on a BE architecture, it would still need to get flipped because that's what the specification indicates. So doing it manually seems correct to me.
> >> 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? > > >>>>>>>>>>>>>> + return num_bytes; >>>>>>>>>>>>>> +} > >
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