编写代码来检查两个双精度值是否足够不同

simd code to check if two double values are sufficiently different

本文关键字：是否 双精度 两个 代码 检查      更新时间：2023-10-16

假设我有两个双精度值old和new。我想实现一个向量化函数返回旧的if abs(x-y) <</P>

下面是代码(test.cpp):
#include <emmintrin.h>
#include <iostream>
#define ARRAY_LENGTH 2
int main(void) {
    // x = old value, y = new value, res = result
    double *x, *y, *res;
    posix_memalign((void **)&x, 16, sizeof(double) * ARRAY_LENGTH);
    posix_memalign((void **)&y, 16, sizeof(double) * ARRAY_LENGTH);
    posix_memalign((void **)&res, 16, sizeof(double) * ARRAY_LENGTH);
    double p = 1e-4; // precision
    __m128d sp = _mm_set1_pd(p);
    x[0] = 1.5; y[0] = 1.50011; // x - old value, y - new value
    x[1] = 2.; y[1] = 2.0000001;
    __m128d sx = _mm_load_pd(x);
    __m128d sy = _mm_load_pd(y);
    // sign mask to compute fabs()
    __m128d sign_mask = _mm_set1_pd(-0.);
    // |x-y|
    __m128d absval = _mm_andnot_pd(sign_mask, _mm_sub_pd(sx, sy) );
    // mask of |x-y| < p
    __m128d mask = _mm_cmplt_pd(absval, sp);
    // sres = |x-y| < p ? x : y;
    __m128d sres = _mm_or_pd(
            _mm_and_pd(mask, sx), _mm_andnot_pd(mask, sy) );
    _mm_store_pd(res, sres);
    std::cerr << "res=" << res[0] << "," << res[1] << std::endl;
    return 0;
}
构建

: 

g++ -std=c++11 -msse4 test.cpp

我们首先计算fab (x-y)，比较p，并使用获得面具。

有人看到一个更有效的方法来编码这个吗?谢谢。

有一种方法可以使该算法更快一些，但它可能会降低准确性:

// d = x - y;
__m128d diff = _mm_sub_pd(sx, sy);
// mask of |y - x| < p
__m128d mask = _mm_cmplt_pd(_mm_andnot_pd(sign_mask, diff), sp);
// sres = y + (|y - x| < p) ? (x - y) : 0;
__m128d sres = _mm_add_pd(sy, _mm_and_pd(mask, diff));

另一种方法-使用AVX或/和单精度

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