问题
I decided to continue Fast corners optimisation and stucked at
_mm_movemask_epi8
SSE instruction. How can i rewrite it for ARM Neon with uint8x16_t
input?
回答1:
I know this post is quite outdated but I found it useful to give my (validated) solution. It assumes all ones/all zeroes in every lane of the Input argument.
const uint8_t __attribute__ ((aligned (16))) _Powers[16]=
{ 1, 2, 4, 8, 16, 32, 64, 128, 1, 2, 4, 8, 16, 32, 64, 128 };
// Set the powers of 2 (do it once for all, if applicable)
uint8x16_t Powers= vld1q_u8(_Powers);
// Compute the mask from the input
uint64x2_t Mask= vpaddlq_u32(vpaddlq_u16(vpaddlq_u8(vandq_u8(Input, Powers))));
// Get the resulting bytes
uint16_t Output;
vst1q_lane_u8((uint8_t*)&Output + 0, (uint8x16_t)Mask, 0);
vst1q_lane_u8((uint8_t*)&Output + 1, (uint8x16_t)Mask, 8);
(Mind http://gcc.gnu.org/bugzilla/show_bug.cgi?id=47553, anyway.)
Similarly to Michael, the trick is to form the powers of the indexes of the non-null entries, and to sum them pairwise three times. This must be done with increasing data size to double the stride on every addition. You reduce from 2 x 8 8-bit entries to 2 x 4 16-bit, then 2 x 2 32-bit and 2 x 1 64-bit. The low byte of these two numbers gives the solution. I don't think there is an easy way to pack them together to form a single short value using NEON.
Takes 6 NEON instructions if the input is in the suitable form and the powers can be preloaded.
回答2:
The obvious solution seems to be completely missed here.
// Use shifts to collect all of the sign bits.
// I'm not sure if this works on big endian, but big endian NEON is very
// rare.
int vmovmaskq_u8(uint8x16_t input)
{
// Example input (half scale):
// 0x89 FF 1D C0 00 10 99 33
// Shift out everything but the sign bits
// 0x01 01 00 01 00 00 01 00
uint16x8_t high_bits = vreinterpretq_u16_u8(vshrq_n_u8(input, 7));
// Merge the even lanes together with vsra. The '??' bytes are garbage.
// vsri could also be used, but it is slightly slower on aarch64.
// 0x??03 ??02 ??00 ??01
uint32x4_t paired16 = vreinterpretq_u32_u16(
vsraq_n_u16(high_bits, high_bits, 7));
// Repeat with wider lanes.
// 0x??????0B ??????04
uint64x2_t paired32 = vreinterpretq_u64_u32(
vsraq_n_u32(paired16, paired16, 14));
// 0x??????????????4B
uint8x16_t paired64 = vreinterpretq_u8_u64(
vsraq_n_u64(paired32, paired32, 28));
// Extract the low 8 bits from each lane and join.
// 0x4B
return vgetq_lane_u8(paired64, 0) | ((int)vgetq_lane_u8(paired64, 8) << 8);
}
回答3:
after some tests it looks like following code works correct:
int32_t _mm_movemask_epi8_neon(uint8x16_t input)
{
const int8_t __attribute__ ((aligned (16))) xr[8] = {-7,-6,-5,-4,-3,-2,-1,0};
uint8x8_t mask_and = vdup_n_u8(0x80);
int8x8_t mask_shift = vld1_s8(xr);
uint8x8_t lo = vget_low_u8(input);
uint8x8_t hi = vget_high_u8(input);
lo = vand_u8(lo, mask_and);
lo = vshl_u8(lo, mask_shift);
hi = vand_u8(hi, mask_and);
hi = vshl_u8(hi, mask_shift);
lo = vpadd_u8(lo,lo);
lo = vpadd_u8(lo,lo);
lo = vpadd_u8(lo,lo);
hi = vpadd_u8(hi,hi);
hi = vpadd_u8(hi,hi);
hi = vpadd_u8(hi,hi);
return ((hi[0] << 8) | (lo[0] & 0xFF));
}
回答4:
Note that I haven't tested any of this, but something like this might work:
X := the vector that you want to create the mask from
A := 0x808080808080...
B := 0x00FFFEFDFCFB... (i.e. 0,-1,-2,-3,...)
X = vand_u8(X, A); // Keep d7 of each byte in X
X = vshl_u8(X, B); // X[7]>>=0; X[6]>>=1; X[5]>>=2; ...
// Each byte of X now contains its msb shifted 7-N bits to the right, where N
// is the byte index.
// Do 3 pairwise adds in order to pack all these into X[0]
X = vpadd_u8(X, X);
X = vpadd_u8(X, X);
X = vpadd_u8(X, X);
// X[0] should now contain the mask. Clear the remaining bytes if necessary
This would need to be repeated once to process a 128-bit vector, since vpadd
only works on 64-bit vectors.
来源:https://stackoverflow.com/questions/11870910/sse-mm-movemask-epi8-equivalent-method-for-arm-neon