Is converting then shifting then bitwise-or'ing the only way to convert from two __m128d
to a single __m128i
?
This is perfectly acceptable to Xcode in an x64 build
m128d v2dHi = ....
m128d v2dLo = ....
__m128i v4i = _mm_set_epi64(_mm_cvtpd_pi32(v2dHi), _mm_cvtpd_pi32(v2dLo))
and the disassembly shows _mm_cvtpd_pi32
being used. However, Visual Studio cannot compile this, complaining about a linker error. This is supported in the VS docs, saying _mm_cvtpd_pi32
is not supported on x64.
I'm not too worried that it's not available, but is two conversions, a shift, then a bitwise-or the fastest way?
If you got a linker error, you're probably ignoring a warning about an undeclared intrinsic function.
Your current code has a high risk of compiling to terrible asm. If it compiled to a vector-shift and an OR, it already is compiling to sub-optimal code. (Update: that's not what it compiles to, IDK where you got that idea.)
Use 2x _mm_cvtpd_epi32 to get two __m128i
vectors with ints you want in the low 2 elements of each. Use _mm_unpacklo_epi64 to combine those two low halves into one vector with all 4 elements you want.
Compiler output from clang3.8.1 on the Godbolt compiler explorer. (Xcode uses clang by default, I think).
#include <immintrin.h>
// the good version
__m128i pack_double_to_int(__m128d a, __m128d b) {
return _mm_unpacklo_epi64(_mm_cvtpd_epi32(a), _mm_cvtpd_epi32(b));
}
cvtpd2dq xmm0, xmm0
cvtpd2dq xmm1, xmm1
punpcklqdq xmm0, xmm1 # xmm0 = xmm0[0],xmm1[0]
ret
// the original
__m128i pack_double_to_int_badMMX(__m128d a, __m128d b) {
return _mm_set_epi64(_mm_cvtpd_pi32(b), _mm_cvtpd_pi32(a));
}
cvtpd2pi mm0, xmm1
cvtpd2pi mm1, xmm0
movq2dq xmm1, mm0
movq2dq xmm0, mm1
punpcklqdq xmm0, xmm1 # xmm0 = xmm0[0],xmm1[0]
# note the lack of EMMS, because of not using the intrinsic for it
ret
MMX is almost totally useless when SSE2 and later is available; just avoid it. See the sse tag wiki for some guides.