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How to round a number to n decimal places in Java

What I would like is a method to convert a double to a string which rounds using the half-up method - i.e. if the decimal to be rounded is 5, it always rounds up to the next number. This is the standard method of rounding most people expect in most situations.

I also would like only significant digits to be displayed - i.e. there should not be any trailing zeroes.

I know one method of doing this is to use the String.format method:

String.format("%.5g%n", 0.912385);

returns:

0.91239

which is great, however it always displays numbers with 5 decimal places even if they are not significant:

String.format("%.5g%n", 0.912300);

returns:

0.91230

Another method is to use the DecimalFormatter:

DecimalFormat df = new DecimalFormat("#.#####");
df.format(0.912385);

returns:

0.91238

However as you can see this uses half-even rounding. That is it will round down if the previous digit is even. What I'd like is this:

0.912385 -> 0.91239
0.912300 -> 0.9123

What is the best way to achieve this in Java?

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Any reason why Double.valueOf() was chosen over Double.parseDouble()? The valueOf() method returns a Double object, while parseDouble() will return a double primitive. With the way the current code is written, you also apply auto-unboxing to the return to cast it to the primitive that your twoDouble variable expects, an extra bytecode operation. I'd change the answer to use parseDouble() instead.

ecbrodie
– ecbrodie

2015-12-03 06:25:05 +00:00
Commented Dec 3, 2015 at 6:25
Double.parseDouble() needs String input.

Ryde
– Ryde

2016-05-05 23:40:46 +00:00
Commented May 5, 2016 at 23:40

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How to round a number to n decimal places in Java

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