On Fri, Feb 7, 2020 at 9:28 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> wrote: > On 2/5/20 9:51 AM, Andy Shevchenko wrote: > > On Tue, Feb 4, 2020 at 6:06 PM Eddie James <eajames@xxxxxxxxxxxxx> wrote: > >> On 2/4/20 5:02 AM, Andy Shevchenko wrote: > >>> On Mon, Feb 3, 2020 at 10:33 PM Eddie James <eajames@xxxxxxxxxxxxxxxxxx> 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