比较两个双精度浮点数或两个浮点数最有效的方法是什么?
简单地这样做是不正确的:
bool CompareDoubles1 (double A, double B)
{
return A == B;
}
比如:
bool CompareDoubles2 (double A, double B)
{
diff = A - B;
return (diff < EPSILON) && (-diff < EPSILON);
}
似乎是浪费加工。
有人知道更聪明的浮点比较器吗?
你写的代码有bug:
return (diff < EPSILON) && (-diff > EPSILON);
正确的代码应该是:
return (diff < EPSILON) && (diff > -EPSILON);
(…是的,这是不同的)
我想知道晶圆厂是否会让你在某些情况下失去懒惰的评价。我会说这取决于编译器。你可能想两种都试试。如果它们在平均水平上是相等的,则采用晶圆厂实现。
如果你有一些关于两个浮点数中哪一个比另一个更大的信息,你可以根据比较的顺序来更好地利用惰性求值。
最后,通过内联这个函数可能会得到更好的结果。不过不太可能有太大改善……
编辑:OJ,谢谢你纠正你的代码。我相应地删除了我的评论
比较浮点数取决于上下文。因为即使改变操作的顺序也会产生不同的结果,所以知道你希望这些数字有多“相等”是很重要的。
在研究浮点数比较时,比较Bruce Dawson编写的浮点数是一个很好的开始。
以下定义来自Knuth的《The art of computer programming》:
bool approximatelyEqual(float a, float b, float epsilon)
{
return fabs(a - b) <= ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool essentiallyEqual(float a, float b, float epsilon)
{
return fabs(a - b) <= ( (fabs(a) > fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyGreaterThan(float a, float b, float epsilon)
{
return (a - b) > ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyLessThan(float a, float b, float epsilon)
{
return (b - a) > ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
当然,选择取决于上下文,并决定你想要的数字有多相等。
比较浮点数的另一种方法是查看数字的ULP(最后位置的单位)。虽然没有专门处理比较,但“每个计算机科学家都应该知道浮点数”这篇论文是了解浮点数如何工作以及陷阱是什么,包括什么是ULP的很好的资源。
使用任何其他建议都要非常小心。这完全取决于上下文。
我花了很长时间在一个系统中追踪错误,该系统假设|a-b|<epsilon,则a==b。潜在的问题是:
The implicit presumption in an algorithm that if a==b and b==c then a==c.
Using the same epsilon for lines measured in inches and lines measured in mils (.001 inch). That is a==b but 1000a!=1000b. (This is why AlmostEqual2sComplement asks for the epsilon or max ULPS).
The use of the same epsilon for both the cosine of angles and the length of lines!
Using such a compare function to sort items in a collection. (In this case using the builtin C++ operator == for doubles produced correct results.)
就像我说的,这完全取决于上下文和a和b的预期大小。
顺便说一下,std::numeric_limits<double>::epsilon()是“机器epsilon”。它是1.0和下一个用double表示的值之间的差值。我猜它可以用在比较函数中,但只有当期望值小于1时。(这是对@cdv的回答的回应…)
同样,如果你的int算术是双精度的(这里我们在某些情况下使用双精度来保存int值),你的算术是正确的。例如,4.0/2.0将等同于1.0+1.0。只要你不做导致分数(4.0/3.0)的事情,或者不超出int的大小。
General-purpose comparison of floating-point numbers is generally meaningless. How to compare really depends on a problem at hand. In many problems, numbers are sufficiently discretized to allow comparing them within a given tolerance. Unfortunately, there are just as many problems, where such trick doesn't really work. For one example, consider working with a Heaviside (step) function of a number in question (digital stock options come to mind) when your observations are very close to the barrier. Performing tolerance-based comparison wouldn't do much good, as it would effectively shift the issue from the original barrier to two new ones. Again, there is no general-purpose solution for such problems and the particular solution might require going as far as changing the numerical method in order to achieve stability.