floating point - Precision lost in float value using java

Question

Given below the test code and its output. When I get float value from number value, precision is lost.. Can anyone tell me why this behaviour and also how to handle this?

public static void main(String[] args)
    {
        try
        {
            java.lang.Number numberVal = 676543.21;
            float floatVal = numberVal.floatValue();
            System.out.println("Number value    : " + numberVal);
            System.out.println("float value     : " + floatVal);
            System.out.println("Float.MAX_VALUE : " + Float.MAX_VALUE);
            System.out.println("Is floatVal > Float.MAX_VALUE ? " + ( floatVal > Float.MAX_VALUE));
        }catch(Exception e)
        {
            e.printStackTrace();
        }
    }

output:

    Number value    : 676543.21
    float value     : 676543.2
    Float.MAX_VALUE : 3.4028235E38
    Is floatVal > Float.MAX_VALUE ? false

also why the float value lesser than Float.MAX_VALUE?

Answer 1

floats are usually good up to 6 significant digits. Your number is 676543.21 which has 8 significant digits. You will see errors past 6 digits, and those errors will easily propagate to more significant digits the more calculations you perform. If you value your sanity (or precision), use doubles. Floats can't even count past 10 million accurately.

Now, you have 2 significant digits, which suggests to me that there is a chance you want to represent currency - DO NOT . Use your own class that internally represents values using fixed-point arithmetic.

Now, as to Float.MAX_VAL which is 3.4028235E38, meaning 3.4028235*10^38 which is about 10^32 times larger than your value. Notice the 'E' in there? That's the exponent.