How to set in Java a custom number of decimal digits in float or double with approximation

Question

Possible Duplicate:
How to round a number to n decimal places in Java

I want to set a specified number of decimal digits in a float (or double), with a method in this form

public float decimalDigits(int x, float n){
....
}

for example

->if I have

float n1=36.58529

the line

float n2=decimalDigits(2, n1);

should return

n2=36.59

->if n1 is:

float n1=36.58329

the line

float n2=decimalDigits(2, n1);

should return

n2=36.58

the line

float n2=decimalDigits(1, n2);

should return

n2=36.6

etc

Answer 1

Use this

public float decimalDigits(int decimaldigits, float x){
            final NumberFormat numFormat = NumberFormat.getNumberInstance();
            numFormat.setMaximumFractionDigits(decimaldigits);
            final String resultS = numFormat.format(x);
            String parsable=resultS.replace(".", "");
            parsable=resultS.replace(",", ".");
            float ris=Float.parseFloat(parsable);
            return ris;
        }

I have added the String replacement to the code to avoid Parsing issue caused by the dot convention (for example 1234.34 becomes 1.234,34 after the formatting causing error in reparsing in float)

If the your is simply a format visualization problem, you could also use the String and doesn't matter that floating point variables don't have decimal places, so this is another valid method:

public String decimalDigits(int decimaldigits, float x){
        final NumberFormat numFormat = NumberFormat.getNumberInstance();
        numFormat.setMaximumFractionDigits(decimaldigits);
        final String resultS = numFormat.format(x);
        return resultS;
    }

If someone has doubts about validity of this solution, should ask for details or try to compile the code and test it before downvote, thanks. The code is tested and works like a charm.

WARNING

Clearly you have to pass float because the method use Float.parseFloat, if you want to pass a double you have to use a cast to float before pass it in the method, otherwise you have to change all the method primitive and parse from float to double. Double and float are different.

Answer 2

It is impossible for a routine that returns a float or a double to return correctly rounded values such as 36.59 or 36.6 because these values are not representable in binary floating-point. Binary floating-point can only return values that are close, such as 36.60000000000000142108547152020037174224853515625. There is no bit pattern in binary floating-point that represents the value 36.6.

If you want decimal values, you must use a decimal radix, such as DecimalFormat or BigDecimal.

Answer 3

Possible pseudo code is below:

public class test {

    /**
     * @param args
     */
    public static void main(String[] args) {
        System.out.println(decimalDigits(10.09872, 4));
    }

     static double decimalDigits(double value, int n)
    {
        double decimal = value - ((int) value);
        System.out.println(decimal);

        double short_decimal = 0;
        for(int i = 0; i < n; i++)
        {
            /* current digit on decimal */
            decimal = decimal * 10;
            System.out.println(decimal);
            short_decimal += (Math.pow(10, n - i - 1) * (int)decimal);

            /* find further */
            decimal = decimal - (int)decimal;
        }

        return (int)value + (double)(short_decimal / Math.pow(10, n));
    }

}