Hi Tim,
Actually, I left out another very good reason why you may want to use
vec_vsx_ld/st. Sorry for forgetting this.
As you saw, vec_ld translates into the lvx instruction. This
instruction loads a sequence of 16 bytes into a vector register. For
big endian, the first byte in memory is loaded into the high order byte
of the register. For little endian, the first byte in memory is loaded
into the low order byte of the register.
This is fine if the data you are loading is arrays of characters, but is
not so fine if you are loading arrays of larger items. Suppose you are
loading four integers {1, 2, 3, 4} into a register with lvx. In big
endian you will see:
00 00 00 01 00 00 00 02 00 00 00 03 00 00 00 04
In little endian you will see:
04 00 00 00 03 00 00 00 02 00 00 00 01 00 00 00
But for this to be interpreted as a vector of integers ordered for
little endian, what you really want is:
00 00 00 04 00 00 00 03 00 00 00 02 00 00 00 01
If you use vec_vsx_ld, the compiler will generate a lxvw2x instruction
followed by an xxpermdi that swaps the doublewords. After the lxvw2x
you will have:
00 00 00 02 00 00 00 01 00 00 00 04 00 00 00 03
because the two LE doublewords are loaded in BE (reversed) order.
Swapping the two doublewords restores sanity:
00 00 00 04 00 00 00 03 00 00 00 02 00 00 00 01
So, even if your data is properly aligned, the use of vec_ld = lvx is
only correct if you are loading arrays of bytes. Arrays of anything
larger must use vec_vsx_ld to avoid errors.
Again, sorry for my previous omission!
Thanks,
Bill Schmidt, Ph.D.
IBM Linux Technology Center
On Fri, 2015-03-13 at 15:42 +0000, Ewart Timothée wrote:
> thank you very much for this answer.
> I know my memory is aligned so I will use vec_ld/st only.
>
> best
>
> Tim
>
>
>
>
>
