在Java中，如果在操作中使用Double.NaN会发生什么？

I have compiled code that erroneously tries to add a number and Double.NaN. I'm wondering if it's throwing an exception that's not getting caught? Does anyone know how that situation is handled?
Thanks.

我编译了错误地尝试添加数字和Double.NaN的代码。我想知道它是否会抛出一个未被捕获的异常?有谁知道这种情况是如何处理的?谢谢。

3 个解决方案

#1

Adding a number to NaN gives NaN. It isn't expected to cause an exception. I understand that this conforms to IEEE 754.

向NaN添加数字会得到NaN。预计不会导致异常。我知道这符合IEEE 754。

#2

To answer Steve B's question:

回答Steve B的问题:

POSITIVE_INFINITY is the largest postive number that you can store if you have unlimited storage space. Without this luxury we have to use a construction like 1.0 / 0.0 which does a fine job. Same goes for NEGATIVE_INFINITY but then the largest negative number.

如果您拥有无限的存储空间,POSITIVE_INFINITY是您可以存储的最大正号。如果没有这种奢侈品,我们必须使用像1.0 / 0.0这样的结构。 NEGATIVE_INFINITY也是如此,但是最大的负数。

NaN is normally defined as 0.0 / 0.0 because there is no such number as 0/0 so that perfectly qualifies for a NaN.

NaN通常定义为0.0 / 0.0,因为没有0/0这样的数字,因此完全符合NaN的要求。

#3

public static void main(String args[])
{
    Double d = Double.NaN + 1.0;
    System.out.println(d);
}

prints Double.Nan. Can anyone explain the source implementation?

打印Double.Nan。任何人都可以解释源实现吗?

  public static final double POSITIVE_INFINITY = 1.0 / 0.0;
  public static final double NEGATIVE_INFINITY = -1.0 / 0.0;
  public static final double NaN = 0.0d / 0.0;

#1