Format Float to n decimal places

Question

You may also pass the float value, and use:

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

Documentation

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.

https://docs.oracle.com/javase/8/docs/api/java/text/DecimalFormat.html

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

NumberFormat formatter = NumberFormat.getInstance(Locale.US);
formatter.setMaximumFractionDigits(2);
formatter.setMinimumFractionDigits(2);
formatter.setRoundingMode(RoundingMode.HALF_UP); 
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");
System.out.println(df.format(f));

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