1: Double a = new Double(Double.NaN); 2: Double b = new Double(Double.NaN); 3: 4: if( Double.NaN == Double.NaN ) 5: System.out.println("True"); 6: else 7: System.out.println("False"); 8: 9: if( a.equals(b) ) 10: System.out.println("True"); 11: else 12: System.out.println("False"); I have not read the stuff about NaN. The above code is printing False True. What gives? Is there anything else that I need to know about NaN?
NAN (not a number) is unordered. Comparing it to any other number, including itself (using the == comparison operator), always yields false. If you use the equals() method overridden by double from the object class it returns true because NAN is a constant. I'm not sure if my logic is accurate...but I was looking over the same subject this evening. Also, I was thinking that constants are stored on the heap. If that's the case...can someone confirm. Thanks.
Marilyn de Queiroz
Joined: Jul 22, 2000
NaN is unordered, so the numerical comparison operators <, <=, >, and >= return false if either or both operands are NaN (�15.20.1). The equality operator == returns false if either operand is NaN, and the inequality operator != returns true if either operand is NaN (�15.21.1). In particular, x!=x is true if and only if x is NaN, and (x<y) == !(x>=y) will be false if x or y is NaN.JLS Floating-point operators produce no exceptions (�11). 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. JLS