🚀 go-pugleaf

On 12 Mar 2001 14:11:43 -0800, Ron Reynolds wrote:

> this i don't get, especially since the numbers are so simple.
> the following code:
>   double d1 = 2.05, d2 = 0.05;
>   System.out.println("d1/d2 = " + d1/d2);
> produces:
>   d1/d2 = 40.99999999999999
> if i cast the result to "float" i can get 41.0, which is nice,
> but my question is why this sort of rounding error pops up with
> such low-precision numbers and what's the accepted way of
> handling this sort of problem?

That is how floating point numbers work, not only in Java but in most 
other languages.  While your numbers divide nicely in decimal, that is 
not the case with floating point values, which are encoded as binary 
numbers.

The issue is how you format the numbers.  You are implicitly using the 
toString() method, which displays all of the significant digits of the 
number.  There are other methods for formatting numbers that should 
give you the results you are looking for.  Take a look at the java.util 
DecimalFormat class.

-- 
Regards,

John McGrath [TeamB]

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