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java – Format Float to n decimal places

Posted by: admin March 10, 2020 Leave a comment


I need to format a float to “n”decimal places.

was trying to BigDecimal, but the return value is not correct…

public static float Redondear(float pNumero, int pCantidadDecimales) {
    // the function is call with the values Redondear(625.3f, 2)
    BigDecimal value = new BigDecimal(pNumero);
    value = value.setScale(pCantidadDecimales, RoundingMode.HALF_EVEN); // here the value is correct (625.30)
    return value.floatValue(); // but here the values is 625.3

I need to return a float value with the number of decimal places that I specify.

I need Float value return not Double


How to&Answers:

You may also pass the float value, and use:

String.format("%.2f", floatValue);



Take a look at DecimalFormat. You can easily use it to take a number and give it a set number of decimal places.

Edit: Example


Try this this helped me a lot

BigDecimal roundfinalPrice = new BigDecimal(5652.25622f).setScale(2,BigDecimal.ROUND_HALF_UP);

Result will be
roundfinalPrice –> 5652.26


Of note, use of DecimalFormat constructor is discouraged. The javadoc for this class states:

In general, do not call the DecimalFormat constructors directly, since the NumberFormat factory methods may return subclasses other than DecimalFormat.


So what you need to do is (for instance):

NumberFormat formatter = NumberFormat.getInstance(Locale.US);
Float formatedFloat = new Float(formatter.format(floatValue));


Here’s a quick sample using the DecimalFormat class mentioned by Nick.

float f = 12.345f;
DecimalFormat df = new DecimalFormat("#.00");

The output of the print statement will be 12.35. Notice that it will round it for you.


Kinda surprised nobody’s pointed out the direct way to do it, which is easy enough.

double roundToDecimalPlaces(double value, int decimalPlaces)
      double shift = Math.pow(10,decimalPlaces);
      return Math.round(value*shift)/shift;

Pretty sure this does not do half-even rounding though.

For what it’s worth, half-even rounding is going to be chaotic and unpredictable any time you mix binary-based floating-point values with base-10 arithmetic. I’m pretty sure it cannot be done. The basic problem is that a value like 1.105 cannot be represented exactly in floating point. The floating point value is going to be something like 1.105000000000001, or 1.104999999999999. So any attempt to perform half-even rounding is going trip up on representational encoding errors.

IEEE floating point implementations will do half-rounding, but they do binary half-rounding, not decimal half-rounding. So you’re probably ok


public static double roundToDouble(float d, int decimalPlace) {
        BigDecimal bd = new BigDecimal(Float.toString(d));
        bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
        return bd.doubleValue();


This is a much less professional and much more expensive way but it should be easier to understand and more helpful for beginners.

public static float roundFloat(float F, int roundTo){

    String num = "




."; for (int count = 0; count < roundTo; count++){ num += "0"; } DecimalFormat df = new DecimalFormat(num); df.setRoundingMode(RoundingMode.HALF_UP); String S = df.format(F); F = Float.parseFloat(S); return F; }


I think what you want ist

return value.toString();

and use the return value to display.


will always return 625.3 because its mainly used to calculate something.


I was looking for an answer to this question and later I developed a method! 🙂 A fair warning, it’s rounding up the value.

private float limitDigits(float number) {
    return Float.valueOf(String.format(Locale.getDefault(), "%.2f", number));