Storing two x86 32 bit registers into 128 bit xmm register

Question

Is there any faster method to store two x86 32 bit registers in one 128 bit 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.

Thanks

Answer 1

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

Answer 2

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.

Edit: I'm going to shut up now.