Re: Decimal rounding of a float data type in Java.

["James Donkin" <James@xxxxxxxxxxxxxxx>]

> I expect you didn't notice the original source of this thread, where the
> questioner was asking why he or she got something like 8.23999999999993755
> instead of 8.24, after a simple calculation.

This is not the original question - the Mangesh wants to round the floating
point to two decimal places for use in both display and later in validation.
I think that Mangesh fully understands what floating point numbers are.

Original Question

> I need to display a float value on the screen with 2 decimal points and
later use that value to perform validations. The
> problem is that the float value which calculated has 6plus decimal
precision which is totally unwarranted. for e.g.
>   3.00 - 1.70 - 0.20 - 0.20 = 0.89999999... instead of 0.90
> Since my validations are based on 0.90 there is a mismatch.
>
> Is there any way in which I can calculate the float value without
unnecessary precision or at least display the right value of >
> 0.90
 > Mangesh

Jesus.Bautista@chase.com  showed a good way for formating to String for
display using DecimalFormat, e.g.

>
> DecimalFormat   twoDigits = new  DecimalFormat( " 0.00" );
>

All I was suggesting was two ways to get a float with two decimal places...
e.g.

>
>    float roundedInt = Float.parseFloat(twoDigits.format( average ));
>

To reuse the string, However as Sting and DecimalFormat are classes it might
be more efficient to use primitives - e.g.

>
>    int temp = (int)(average * 100);
>    average = (float)temp / 100;
>

Which converts without creating a string. If you multiply any float by one
hundred and turn in into an integer it will be a whole number (by
definition). If you divide any whole number by one hundred and cast it into
a float it will have at most two decimal places.

> Floating point numbers
> work in binary, not in decimal.  So dividing by 100 is not "just a
two-digit
> shift" in floating point, that would be dividing by 4.

Java data types are decimal not binary. The underlying implementation might
be binary but the float is a decimal number - dividing it by one hundred is
a two digit shift.

> And by the way,
> 323/34 is no more irrational than 422/100.  Both are rational numbers.

true - but is 8.23999999999993753  ?



----- Original Message -----
From: Clapham, Paul <pclapham@core-mark.com>
To: <JAVA400-L@midrange.com>
Sent: Friday, March 02, 2001 4:41 PM
Subject: RE: Decimal rounding of a float data type in Java.


> I expect you didn't notice the original source of this thread, where the
> questioner was asking why he or she got something like 8.23999999999993755
> instead of 8.24, after a simple calculation.  This IS how floating point
> numbers work, regardless of your innumerate denial.  Floating point
numbers
> work in binary, not in decimal.  So dividing by 100 is not "just a
two-digit
> shift" in floating point, that would be dividing by 4.  And by the way,
> 323/34 is no more irrational than 422/100.  Both are rational numbers.
>
> -----Original Message-----
> From: James Donkin [mailto:James@DuMaurier.co.uk]
> Sent: March 2, 2001 01:25
> To: JAVA400-L@midrange.com
> Subject: Re: Decimal rounding of a float data type in Java.
>
>
>
> > Well, not exactly.  Your "average" doesn't become 4.22, it becomes
> something
> > that's extremely close to 4.22.  Floating point numbers can't express
> *any*
> > fraction exactly unless it's a power of 0.5.  So don't be surprised if
you
> > do "System.out.println(average)" and you find something like
> > "4.2200000000033775" appearing.
> e.g. average = 422 / 100;
>
> This will always give 4.22. The time you get problems is with 323 / 34 or
> some other fraction which generates an irrational number. Saying that
>
>     float average = 422 / 100;
>
>  will give 4.22000032 is like saying that
>
>     float average = 4.22;
>
>  will leave average = 4.2200042. Dividing by 100 just generates a two
digit
> shift.
>
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