On 06/25/2012 05:21 PM, Jason Garrett-Glaser wrote:
>> +    mova       m0, [srcq         ]  ; m0 =  0,  1,  2,  3,  4,  5,  6,  7
>> > +    mova       m1, [srcq+  mmsize]  ; m1 =  8,  9, 10, 11, 12, 13, 14, 15
>> > +    mova       m2, [srcq+2*mmsize]  ; m2 = 16, 17, 18, 19, 20, 21, 22, 23
>> > +    movhlps    m3, m1
>> > +    movlhps    m3, m2               ; m3 = 12, 13, 14, 15, 16, 17, 18, 19
>> > +    movlhps    m1, m1
>> > +    movhlps    m1, m0               ; m1 =  4,  5,  6,  7,  8,  9, 10, 11
> Using more loads (and fewer shuffles) might be better here, e.g:
> 
> mova  m0, [srcq         ]
> mova  m2, [srcq+2*mmsize]
> movq  m3, [srcq + mmsize + mmsize/2]
> movlhps m3, m2
> movq m1, [srcq+  mmsize]
> movhlps m1, m0
> 
> One more load, two less shuffles, the load is probably free.
> 

That is not equivalent for m1. I think it would have to use movhpd
instead of movq for the load. But I'll try that and see how it compares.

-Justin
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