将一个小范围除以另一个小范围时出现浮点错误

double min, max, min2, max2, mid, result;
// min2 < max2 is a subrange of min < max with min2 >= min and max2 <= max
// where max-min may be very small, which means max2-min2 will be very small as well.
mid = (min2+max2)/2.0;
result = (mid - min)/(max-min)

我有一个问题,我得到的值大于1 min, max min2, max2 是双打。 double 变量,但我仍然不确定浮点错误是否会产生错误的结果。

稍后,我想使用结果通过使用

const int MAX=63;
double array[64];
int index = (int)(result*MAX);
array[index];

NUM_BINS = 64
min = -5.00958252
max = -5.00958014
min2 = -5.0095801960014716
max2 = -5.0095799398356107
mid = (max2+min2) / 2.0
mid - min = 2.4516127083984429e-06
(mid - min) / (max - min)) * (NUM_BINS - 1) = 64.781696632504463