On 2/7/20 1:39 PM, Andy Shevchenko wrote:
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.
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.
+}
