Comparison with NaN

[leo donahue]

Why is this comparison false?

double x = 0;
double y = 0;
System.out.println((x/y));
System.out.println((x/y)==(java.lang.Double.NaN));

Just because?

[Tim LeMaster]

I would say just because that are both NaN doesn't make the equal. Lots and Lots of things are NaN (a, b, c, d) but are not equal.

[Ernest Friedman-Hill]

Yes, just because -- or as we computer sciency type fellas like to say, "by definition." The Java Lanaguage Specificiation explicitly states that NaN is not equal to anything, including itself.

[sven studde]

I think for the same reason the two strings in the following code are not equal:

String str1 = "hello";
String str2 = "world";
String str3 = "hello world";
 
str1 += str2;
 
if(str1 == str2)
System.out.println("they are equal");
else
System.out.println("not equal");

In other words, you have two different objects that are identical, but == compares whether the references are equal, and that's only true when the references refer to the same object.

[sven studde]

Specificiation explicitly states that NaN is not equal to anything, including itself.

That's the strange part.

[Keith Lynn]

Not really strange. It's specified in IEEE 754.

[leo donahue]

Originally posted by Ernest Friedman-Hill:
Yes, just because -- or as we computer sciency type fellas like to say, "by definition." The Java Lanaguage Specificiation explicitly states that NaN is not equal to anything, including itself.

If we compare with the inequality operator: != The comparison is true.

"By definition" is the only option.

If floating point division by zero is really not the same as whatever the value of java.lang.Double.NaN is, then what is the value of floating point division by zero? Did someone just choose to use the same "word"(NaN) to describe that value?

[Keith Lynn]

Originally posted by leo donahue:


If we compare with the inequality operator: != The comparison is true.

"By definition" is the only option.

If floating point division by zero is really not the same as whatever the value of java.lang.Double.NaN is, then what is the value of floating point division by zero? Did someone just choose to use the same "word"(NaN) to describe that value?

It depends on what the numerator is.

If the numerator and denominator are both 0, the result is NaN.

If the numerator is, for instance, 1.0, then 1.0/0 is Infinity.

As Tim pointed out, in the case of 0.0/0 != Double.NaN, just because two things are not numbers, they are not necessarily equal to each other.
[ October 16, 2006: Message edited by: Keith Lynn ]

[leo donahue]

Yes, I meant both numerator and denominator.

I keep forgetting the slightest inference really takes the converstation in different directions. I'll try harder.

[Robert Hill]

Mathematically speaking, x/0 is undefined, not infinity. Where x is a real number, not zero. Infinity is not a number, but can basically be described as if there exists a one-to-one correspondence between that set and a proper subset of itself then the set is infinite.

If you have 1/x, then its value approaches infinity as x goes to zero, but 1/0 is not an infinite number, since infinity is not a number.

0/0 is indeterminant.