Why does Double.NaN==Double.NaN return false?
I was just studying OCPJP questions and I found this strange code:
public static void main(String a[]) {
System.out.println(Double.NaN==Double.NaN);
System.out.println(Double.NaN!=Double.NaN);
}
When I ran the code, I got:
false
true
How is the output false when we're comparing two things that look the same as each other? What does NaN mean?
NaN means "Not a Number".
Java Language Specification (JLS) Third Edition says:
An operation that overflows produces a signed infinity, an operation that underflows produces a denormalized value or a signed zero, and an operation that has no mathematically definite result produces NaN. All numeric operations with NaN as an operand produce NaN as a result. As has already been described, NaN is unordered, so a numeric comparison operation involving one or two NaNs returns false and any != comparison involving NaN returns true , including x!=x when x is NaN.
NaN is by definition not equal to any number including NaN. This is part of the IEEE 754 standard and implemented by the CPU/FPU. It is not something the JVM has to add any logic to support.
http://en.wikipedia.org/wiki/NaN
A comparison with a NaN always returns an unordered result even when comparing with itself. ... The equality and inequality predicates are non-signaling so x = x returning false can be used to test if x is a quiet NaN.
Java treats all NaN as quiet NaN.
Why that logic
NaN means Not a Number . What is not a number? Anything. You can have anything in one side and anything in the other side, so nothing guarantees that both are equals. NaN is calculated with Double.longBitsToDouble(0x7ff8000000000000L) and as you can see in the documentation of longBitsToDouble :
If the argument is any value in the range 0x7ff0000000000001L through 0x7fffffffffffffffL or in the range 0xfff0000000000001L through 0xffffffffffffffffL , the result is a NaN .
Also, NaN is logically treated inside the API.
Documentation
/**
* A constant holding a Not-a-Number (NaN) value of type
* {@code double}. It is equivalent to the value returned by
* {@code Double.longBitsToDouble(0x7ff8000000000000L)}.
*/
public static final double NaN = 0.0d / 0.0;
By the way, NaN is tested as your code sample:
/**
* Returns {@code true} if the specified number is a
* Not-a-Number (NaN) value, {@code false} otherwise.
*
* @param v the value to be tested.
* @return {@code true} if the value of the argument is NaN;
* {@code false} otherwise.
*/
static public boolean isNaN(double v) {
return (v != v);
}
Solution
What you can do is use compare / compareTo :
Double.NaN is considered by this method to be equal to itself and greater than all other double values (including Double.POSITIVE_INFINITY ).
Double.compare(Double.NaN, Double.NaN);
Double.NaN.compareTo(Double.NaN);
Or, equals :
If this and argument both represent Double.NaN , then the equals method returns true , even though Double.NaN==Double.NaN has the value false .
Double.NaN.equals(Double.NaN);
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