The storage requirements for float and double are defined by Java, however
precision during calculation is not. This means that a program that uses
floating point arithmetic can produce different answers on different
systems, with the degree of difference increasing with the number of
calculations a particular value goes through. This is true of floating point
in general, not just in Java.
Justy
----- Original Message -----
> The reason is that floating point numbers only approximate real numbers.
> But this should work if all you want is two decimal places:
> (*Chris*)
>
> public class Junk {
>
> public static void main (String[] args) {
> float d1 = 99.44f;
> float d2 = 99.40f;
> float d3 = (float)(Math.rint((d1 - d2) * 100.00) / 100.00);
> System.out.println(d3);
> } file://main
>
> }
>
> ----- Original Message -----
> From: "lancelot" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Friday, June 29, 2001 1:41 AM
> Subject: [JSP-INTEREST] Need your help one float and double calc.
>
>
> > Dear all,
> > Who can tell me why the result is 0.040000916? and why it's not =
> > 0.04, if I want to get 0.04 how could I do? thanks
> >
> > Lancelot
> >
> > D:\test>type testD.java
> > public class testD
> > {
> > public static void main(String args[])
> > {
> > float d1=3D99.44f;
> > float d2=3D99.40f ;
> > float d3 =3D d1-d2 ;
> > System.out.println(d3);
> > }
> > }
> >
> >
> > D:\test>java testD
> > 0.040000916
> >
> >
>
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