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converting float to double

Jon Wynett

Joined: Jun 28, 2004
Posts: 10
How can I convert a float to a double without have the number "adjusted". For example, if I create a Float with value 18.4 and then either get the doubleValue() or create a new Double with that Float, it comes out as 18.399999618530273.

I've tried this with many different numbers and all get the same type of problem. Also, casting a float as a double gets the same problem.

Here's some sample code:
public static void testFloatToDouble() {
Float f = new Float(18.4);
Double d = new Double(f.floatValue());
System.out.println("Float: " + f);
System.out.println("Float as double: " + f.doubleValue());
System.out.println("Float as Double: " + d);
And the results:
Float: 18.4
Float as double: 18.399999618530273
Float as Double: 18.399999618530273

Joyce Lee
Ranch Hand

Joined: Jul 11, 2003
Posts: 1392
Hi Jon,

How about this?

marc weber

Joined: Aug 31, 2004
Posts: 11343

This is a consequence of the way floating-point numbers are stored in computers (per IEEE 754 standards). If the decimal portion cannot be expressed as a sum of reciprocal powers of 2 within a particular "window," then the number loses precision.

If you want precision, then you should use the BigDecimal class in java.math. But be sure to use the String constructor, or you'll get similar errors. Compare the String and double constructor descriptions in the API...

For details on IEEE 754, follow the links here...
[ October 04, 2004: Message edited by: marc weber ]

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Ilja Preuss

Joined: Jul 11, 2001
Posts: 14112
With other words, you can't really store the value 18.4 in a float, because in binary doesn't have a finite number of decimal places (similar to 0.1 ternary = 1*3^(-1) = 1/3 doesn't have a finite number of decimal places in the decimal system). What gets stored is a close approximation.

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I agree. Here's the link:
subject: converting float to double
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