用于计算欧几里德距离的函数

Question

我编写了计算欧几里得距离的函数。函数的返回值不符合要求的精度范围。我将所有变量都设为最准确类型的double，并使用了sqrt()函数，该函数用于double而不是sqrtf，但返回的值仍然不合适。这是我的功能：

float distance(const struct color_t* p1, const struct color_t* p2, int* err_code) {
        if (p1 == NULL || p2 == NULL) {
            return -1;
        }
     double r1 = 1, r2 = 1, r3 = 1, r4=1;
    double p11 = (double)p1->r - (double)p2->r;
    double p22 = (double)p1->g - (double)p2->g;
    double p33 = (double)p1->b - (double)p2->b;
    double p44 = (double)p1->a - (double)p2->a;
    for (int exponent = 2; exponent > 0; exponent--)
    {
        r1 = r1 * p11;
        r2 = r2 * p22;
        r3 = r3 * p33;
        r4 = r4 * p44;
    }
    double b = r1 + r2 + r3 + r4;
    double a = sqrt(b);
    if (a < 0) {
        return -1;
    }   
    return a;
}

提示是：函数distance（）返回的值367.106262不符合要求的精度范围。

和测试：

struct color_t c1 = {.r = 14, .g = 253, .b = 23, .a = 76}, c2 = {.r = 253, .g = 14, .b = 148, .a = 6};

            printf("#####START#####");
            float dist = distance(&c1, &c2, NULL);
            printf("#####END#####\n");

            test_error(360.4206425334894 > dist && 360.3206425334894 < dist, "Value %f returned by function distance() doesn't fit in required accuracy range", dist);

Answer 1

丢弃透明术语

@abelenky

OP正在寻找颜色的“距离”。

[int数学除最后一步外均可使用

dist2的最大值是255 * 255 * 3 = 195,075，正好在float完全可编码的范围内。

float distance(const struct color_t *p1, const struct color_t *p2, int *err_code /* unused */ ) {
  if (p1 == NULL || p2 == NULL) {
    return -1;
  }
  int p11 = p1->r - p2->r;
  int p22 = p1->g - p2->g;
  int p33 = p1->b - p2->b;
  int dist2 = p11*p11 + p22*p22 + p33*p33;
  return sqrtf(dist2);
}