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# Doubt with NaN

Ranch Hand
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source SK Majji Q6

The following code will print

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");

A) True
True

B) True
False

C) False
True

D) False
False
Ans: C

Doubt: why are we getting true for line 10?

pls help,
regards,
Gitesh

Greenhorn
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Hi Gitesh

The Ans is correct.
The reason is the Double class overrides the equals() method.
so it will compare the values.

since both the objects containing Double.NaN it returns true.
like
Double d1 = new Double(12.2);
Double d2 = new Double(12.2);
if( d1.equals(d2)) // true.

the question why if( Double.NaN == Double.NaN ) returns false is unknow, you have to check with JLS.

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Wrapper and Strings implements method equals for standard.

Gitesh Ramchandani
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My doubt is both a and b contain NaN, and if NaN!=NaN then how
( a.equals(b) )gives true??

Pls help,

regards,
Gitesh

Java Cowboy
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Have a look at the API documentation of the equals() method in class Double. It explains why it works like this:

Note that in most cases, for two instances of class Double, d1 and d2, the value of d1.equals(d2) is true if and only if

d1.doubleValue() == d2.doubleValue()

also has the value true. However, there are two exceptions:

• If d1 and d2 both represent Double.NaN, then the equals method returns true, even though Double.NaN==Double.NaN has the value false.
• If d1 represents +0.0 while d2 represents -0.0, or vice versa, the equal test has the value false, even though +0.0==-0.0 has the value true.

• This definition allows hash tables to operate properly.

[ September 04, 2007: Message edited by: Jesper Young ]

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