我决定继续快速角落优化并坚持下去
_mm_movemask_epi8
SSE指令。如何使用uint8x16_t
输入为ARM Neon重写它?
答案 0 :(得分:6)
我知道这篇文章已经过时但我觉得提供我的(经验证的)解决方案很有用。它假定Input参数的每个通道中的所有1 /全零。
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,无论如何。)
与Michael类似,诀窍是形成非空条目的索引的权力,并将它们成对地加起来三次。这必须通过增加数据大小来完成,以使每次添加的步幅加倍。您将2 x 8 8位条目减少到2 x 4 16位,然后是2 x 2 32位和2 x 1 64位。这两个数字的低字节给出了解决方案。我不认为有一种简单的方法可以将它们打包在一起,形成一个使用NEON的短值。
如果输入采用合适的形式并且可以预加载功率,则需要6条NEON指令。
答案 1 :(得分:2)
明显的解决方案似乎在这里完全被忽略了。
// 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);
}
答案 2 :(得分:1)
经过一些测试后,看起来下面的代码是正确的:
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));
}
答案 3 :(得分:1)
这个问题值得为 aarch64 提供更新的答案。向 Armv8 添加新功能允许用更少的指令实现相同的功能。这是我的版本:
uint32_t _mm_movemask_aarch64(uint8x16_t input)
{
const uint8_t __attribute__ ((aligned (16))) ucShift[] = {-7,-6,-5,-4,-3,-2,-1,0,-7,-6,-5,-4,-3,-2,-1,0};
uint8x16_t vshift = vld1q_u8(ucShift);
uint8x16_t vmask = vandq_u8(input, vdupq_n_u8(0x80));
uint32_t out;
vmask = vshlq_u8(vmask, vshift);
out = vaddv_u8(vget_low_u8(vmask));
out += (vaddv_u8(vget_high_u8(vmask)) << 8);
return out;
}
答案 4 :(得分:0)
请注意,我还没有对此进行测试,但是这样的事情可能有用:
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
这需要重复一次以处理128位向量,因为vpadd
仅适用于64位向量。