8 bit shift operation in AVX2 with shifting in zeros

Question

Is there any way to rebuild the _mm_slli_si128 instruction in AVX2 to shift an __mm256i register by x bytes?

The _mm256_slli_si256 seems just to execute two _mm_slli_si128 on a[127:0] and a[255:128].

The left shift should work on a __m256i like this:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ..., 32] -> [2, 3, 4, 5, 6, 7, 8, 9, ..., 0]

I saw in thread that it is possible to create a shift with _mm256_permutevar8x32_ps for 32bit. But I need a more generic solution to shift by x bytes. Has anybody already a solution for this problem?

If you find yourself needing to do this a lot, you might want to think of an alternate approach. AVX and beyond breaks up the vectors into "lanes" of 128-bit. Cross-lane operations are very expensive. Looking at Agner Fog's docs, it looks like cross-lane operations are more expensive than misaligned memory access. — Mysticial
thanks a lot for the answer. I will checkout his docs. But I don't have to use the command extensive. But it would be good if I could use SIMD commands. — martin s
@Mysticial: It's a latency penalty, not throughput. VPERMD y,y,y, VPERMQ y,y,i, and VPERM2I128 y,y,y,i are all 1uop, lat=3c, throughput=1/cycle. (And all run on port5 only in Haswell.) I agree, if you can structure things to work without crossing lanes all the time, that's best. But if your algo inherently benefits, and the extra latency isn't killer, then it could be a win. — Peter Cordes

martin s martin s · Accepted Answer · 2013-12-29T16:35:33

okay I implemented a function that can shift left up to 16 byte.

template  <unsigned int N> __m256i _mm256_shift_left(__m256i a)
{
  __m256i mask =  _mm256_srli_si256(
          _mm256_permute2x128_si256(a, a, _MM_SHUFFLE(0,0,3,0))
          , 16-N);
  return _mm256_or_si256(_mm256_slli_si256(a,N),mask);
}

Example:

int main(int argc, char* argv[]) {
   __m256i reg =  _mm256_set_epi8(32,31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,
                                  14,13,12,11,10,9,8,7,6,5,4,3,2,1);

   __m256i result = _mm256_shift_left<1>(reg);
   for(int i = 0; i < 32; i++)
     printf("%2d ",((unsigned char *)&result)[i]);
   printf("\n");
}

The output is
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

Edit: New version with new alignr instruction. Thanks for the hint @Evgney Kluev

template  <unsigned int N> __m256i _mm256_shift_left(__m256i a)
{
  __m256i mask = _mm256_permute2x128_si256(a, a, _MM_SHUFFLE(0,0,3,0) );
  return _mm256_alignr_epi8(a,mask,16-N);
}

8 bit shift operation in AVX2 with shifting in zeros

1 Answers