Is there any faster method to store two x86 32 bit registers in one 128 b开发者_开发问答it xmm register?
movd xmm0, edx
movd xmm1, eax
pshufd xmm0, xmm0, $1
por xmm0, xmm1
So if EAX is 0x12345678
and EDX is 0x87654321
, the result in xmm0 must be 0x8765432112345678
.
With SSE 4.1 you can use movd xmm0, eax
/ pinsrd xmm0, edx, 1
and do it in 2 instructions.
For older CPUs you can use 2 x movd
and then punpckldq
for a total of 3 instructions:
movd xmm0, edx
movd xmm1, eax
punpckldq xmm0, xmm1
I don't know much about MMX, but perhaps you want the PACKSSDW instruction.
The PACKSSDW instruction takes the two double words in the source operand and the two double words in the destination operand and converts these to four signed words via saturation. The instruction packs these four words together and stores the result in the destination MMX register.
(from http://webster.cs.ucr.edu/AoA/Windows/HTML/TheMMXInstructionSeta2.html)
Edit: I just realized that those were SSE registers. Oh well.
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