C 在_mm256_rsqrt_ps（）中处理零
考虑到C 在_mm256_rsqrt_ps（）中处理零,c,x86,sse,intrinsics,avx,C,X86,Sse,Intrinsics,Avx,考虑到\u mm256\u sqrt\u ps（）相对较慢，并且我生成的值会立即被\u mm256\u floor\u ps（）截断，环顾四周，似乎执行了以下操作： _mm256_mul_ps(_mm256_rsqrt_ps(eightFloats), eightFloats); 是获得额外性能和避免管道停滞的方法 不幸的是，对于零值，我当然会在计算1/sqrt（0）时遇到崩溃。最好的解决办法是什么？我已经尝试过这个方法（它有效而且更快），但是有没有更好的方法，或者
\u mm256\u sqrt\u ps（）\u mm256\u floor\u ps（）_mm256_mul_ps(_mm256_rsqrt_ps(eightFloats),
eightFloats);
1/sqrt（0）_mm256_mul_ps(_mm256_rsqrt_ps(_mm256_max_ps(eightFloats,
_mm256_set1_ps(0.1))),
eightFloats);
float vals[MAX];
int sum = 0;
for (int i = 0; i < MAX; i++)
{
int thisSqrt = (int) floor(sqrt(vals[i]));
sum += min(thisSqrt, 0x3F);
}
float vals[MAX]；
整数和=0；
对于（int i=0；ivalsint为了解决
0.00.00.00*+无穷大insqrt（x）=x*rsqrt（x）
）中产生的任何NaN归零。自动矢量化时，Clang会执行此操作。（但它将blendps
与归零向量一起使用，而不是将比较掩码与和nps
一起直接用于归零或保留元素。）
也可以使用sqrt（x）~=recip（rsqrt（x））
<代码>rsqrt（0）=+Inf
<代码>往复（+Inf）=0你错过的东西是： 当输入为完全平方时，截断为整数（而不是四舍五入）需要精确的
sqrt45rsqrtps5.0-1ulp- 。请注意，我们关心的是吞吐量而不是延迟，因为我们在循环中使用它，没有做很多其他事情。sqrt不是循环携带的dep链的一部分，因此不同的迭代可以同时运行它们的sqrt计算
在执行
（x+offset）*近似值（x+offset）sqrt_约（63*63-1+偏移量）63n=36sqrt（n*n-1）+errorsqrt（n*n）define f(n) { local x=sqrt(n*n-1); local e=x*1.5*2^(-12); print x; print e, x+e; }
; f(36)
35.98610843089316319413
~0.01317850650545403926 ~35.99928693739861723339
; f(37)
36.9864840178138587015
~0.01354485498699237990 ~37.00002887280085108140
isqrt=static\u cast（（x+0.5）*大约rsqrt（x+0.5））20*2020*20-1; define test_approx_sqrt(x, off) { local s=x*x+off; local sq=s/sqrt(s); local sq_1=(s-1)/sqrt(s-1); local e=1.5*2^(-12); print sq, sq_1; print sq*e, sq_1*e; }
; test_approx_sqrt(20, 0.5)
~20.01249609618950056874 ~19.98749609130668473087 # (x+0.5)/sqrt(x+0.5)
~0.00732879495710064718 ~0.00731963968187500662 # relative error
val*（x+/-err）=val*x+/-val*err无论如何，我想你需要一次牛顿-拉斐逊迭代。 最好的办法是在使用
rsqrt0.5编译器输出：一个不错的起点 我从clang 3.7.1中获得了非常好的自动矢量化结果，它具有
sum\u int-fno math errno-funsafe math optimizations-ffinite math only-ffast mathrsqrtpssqrt（0）=NaNsum\u fp-ffast mathclangsqrtps-ffast math; define test_approx_sqrt(x, off) { local s=x*x+off; local sq=s/sqrt(s); local sq_1=(s-1)/sqrt(s-1); local e=1.5*2^(-12); print sq, sq_1; print sq*e, sq_1*e; }
; test_approx_sqrt(20, 0.5)
~20.01249609618950056874 ~19.98749609130668473087 # (x+0.5)/sqrt(x+0.5)
~0.00732879495710064718 ~0.00731963968187500662 # relative error
// autovectorizes with clang, but has rounding problems.
// Note the use of sqrtf, and that floorf before truncating to int is redundant. (removed because clang doesn't optimize away the roundps)
int sum_int(float vals[]){
int sum = 0;
for (int i = 0; i < MAX; i++) {
int thisSqrt = (int) sqrtf(vals[i]);
sum += std::min(thisSqrt, 0x3F);
}
return sum;
}
#include <immintrin.h>
#define MAX 4096
// 2*sqrt(x) ~= 2*x*approx_rsqrt(x), with a Newton-Raphson iteration
// dividing by 2 is faster in the integer domain, so we don't do it
__m256 approx_2sqrt_ps256(__m256 x) {
// clang / gcc usually use -3.0 and -0.5. We could do the same by using fnmsub_ps (add 3 = subtract -3), so we can share constants
__m256 three = _mm256_set1_ps(3.0f);
//__m256 half = _mm256_set1_ps(0.5f); // we omit the *0.5 step
__m256 nr = _mm256_rsqrt_ps( x ); // initial approximation for Newton-Raphson
// 1/sqrt(x) ~= nr * (3 - x*nr * nr) * 0.5 = nr*(1.5 - x*0.5*nr*nr)
// sqrt(x) = x/sqrt(x) ~= (x*nr) * (3 - x*nr * nr) * 0.5
// 2*sqrt(x) ~= (x*nr) * (3 - x*nr * nr)
__m256 xnr = _mm256_mul_ps( x, nr );
__m256 three_minus_muls = _mm256_fnmadd_ps( xnr, nr, three ); // -(xnr*nr) + 3
return _mm256_mul_ps( xnr, three_minus_muls );
}
// packed int32_t: correct results for inputs from 0 to well above 63*63
__m256i isqrt256_ps(__m256 x) {
__m256 offset = _mm256_set1_ps(0.5f); // or subtract -0.5, to maybe share constants with compiler-generated Newton iterations.
__m256 xoff = _mm256_add_ps(x, offset); // avoids 0*Inf = NaN, and rounding error before truncation
__m256 approx_2sqrt_xoff = approx_2sqrt_ps256(xoff);
__m256i i2sqrtx = _mm256_cvttps_epi32(approx_2sqrt_xoff);
return _mm256_srli_epi32(i2sqrtx, 1); // divide by 2 with truncation
// alternatively, we could mask the low bit to zero and divide by two outside the loop, but that has no advantage unless port0 turns out to be the bottleneck
}
__m256i isqrt256_ps_simple_exact(__m256 x) {
__m256 sqrt_x = _mm256_sqrt_ps(x);
__m256i isqrtx = _mm256_cvttps_epi32(sqrt_x);
return isqrtx;
}
int hsum_epi32_avx(__m256i x256){
__m128i xhi = _mm256_extracti128_si256(x256, 1);
__m128i xlo = _mm256_castsi256_si128(x256);
__m128i x = _mm_add_epi32(xlo, xhi);
__m128i hl = _mm_shuffle_epi32(x, _MM_SHUFFLE(1, 0, 3, 2));
hl = _mm_add_epi32(hl, x);
x = _mm_shuffle_epi32(hl, _MM_SHUFFLE(2, 3, 0, 1));
hl = _mm_add_epi32(hl, x);
return _mm_cvtsi128_si32(hl);
}
int sum_int_avx(float vals[]){
__m256i sum = _mm256_setzero_si256();
__m256i upperlimit = _mm256_set1_epi32(0x3F);
for (int i = 0; i < MAX; i+=8) {
__m256 v = _mm256_loadu_ps(vals+i);
__m256i visqrt = isqrt256_ps(v);
// assert visqrt == isqrt256_ps_simple_exact(v) or something
visqrt = _mm256_min_epi32(visqrt, upperlimit);
sum = _mm256_add_epi32(sum, visqrt);
}
return hsum_epi32_avx(sum);
}
sum_int_avx(float*):
vpxor ymm3, ymm3, ymm3
xor eax, eax
vbroadcastss ymm0, dword ptr [rip + .LCPI4_0] ; set1(0.5)
vbroadcastss ymm1, dword ptr [rip + .LCPI4_1] ; set1(3.0)
vpbroadcastd ymm2, dword ptr [rip + .LCPI4_2] ; set1(63)
LBB4_1: ; latencies
vaddps ymm4, ymm0, ymmword ptr [rdi + 4*rax] ; 3c
vrsqrtps ymm5, ymm4 ; 7c
vmulps ymm4, ymm4, ymm5 ; x*nr ; 5c
vfnmadd213ps ymm5, ymm4, ymm1 ; 5c
vmulps ymm4, ymm4, ymm5 ; 5c
vcvttps2dq ymm4, ymm4 ; 3c
vpsrld ymm4, ymm4, 1 ; 1c this would be a mulps (but not on the critical path) if we did this in the FP domain
vpminsd ymm4, ymm4, ymm2 ; 1c
vpaddd ymm3, ymm4, ymm3 ; 1c
; ... (those 9 insns repeated: loop unrolling)
add rax, 16
cmp rax, 4096
jl .LBB4_1
;... horizontal sum