In Java, what happens if you use Double.NaN in an operation?

Question

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.

Answer 1

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

Answer 2

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?

  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;

Answer 3

To answer Steve B's question:

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.

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.