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converting float to double

 
Greenhorn
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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

Jon
 
Ranch Hand
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Hi Jon,

How about this?


Joyce
 
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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...

http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigDecimal.html

For details on IEEE 754, follow the links here...

http://www.public.iastate.edu/~sarita/ieee754/homepage.html
[ October 04, 2004: Message edited by: marc weber ]
 
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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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