I don't quite get your meaning, but maybe this is clearer:

Jython 2.1 on java1.3.1_01 (JIT: null)
Type "copyright", "credits" or "license" for more information.
> >> from java.lang.Float import NaN
> >> NaN == 1
1
> >>

From the JLS:

"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."

http://java.sun.com/docs/books/jls/second_edition/html/typesValues.doc.html#16083

a.k.a. Bug, IMHO.

-Clark

-----Original Message-----
From: On Behalf Of Widhalm, Eric
Sent: Monday, November 15, 2004 10:32 AM
Subject: RE: [Jython-users] (NaN == 111.0 ) returns 1 !

>  could you tell me if  the following  code
>  
>   >>from java.lang import Math
>   >>g = Math.log(-1.0)
>   >>print (g==111.0)
>  1
>  
>  highlights a bug or is an (un)expected behaviour of the rich
>  comparison?

Nicola, the return value will either be 0 or 1, representing boolean values true or false. I
n this case, -1 is != 111.0, so you get back 1, representing false.

-Eric


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