Limit decimal digits in Java when using double

Question

I have a few variables, all int, and I make with them this operation:

percentage = ((double)100) - (((double)100)/(((double)64)*((double)dist.size()-1))*((double)bestDist));

forcing them to be double because I want to calculate a percentage. The problem is that (of course) I get results like 65.88841666666666, and I want to get a number with only 2 decimal digits, I don't mind about approximating it, I can also cut all the digits so I will get 65.88 instead of 65.89. How can I do this?

Do you actually want to change the *value* of `percentage` to only hold 2 digits past the decimal point, or do you want to *display* only the first 2 digits past the decimal point? — azurefrog, May 20 '15 at 18:28
possible duplicate of [round up to 2 decimal places in java?](http://stackoverflow.com/questions/11701399/round-up-to-2-decimal-places-in-java) — Jon, May 20 '15 at 18:29

score 5 · Answer 1 · answered May 20 '15 at 18:34

5

If this is for displaying result, you can use the formatter like:

String formatted = String.format("%.2f", 65.88888);

Here, .2 means display 2 digits after decimal. For more options, try BigDecimal

answered May 20 '15 at 18:34

Lumberjack

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5

score 3 · Accepted Answer · answered May 20 '15 at 18:29

3

You can use round for that

double roundedPercentage = (double) Math.round(percentage * 100) / 100;

answered May 20 '15 at 18:29

Karthik Cherukuri

603
2
9
15

score 0 · Answer 3 · answered May 20 '15 at 18:55

Although what @CHERUKURI suggested, is what I really use myself, but it will produce inaccurate result when it is MAX_DOUBLE.

Math.ceil(percentage) + Math.ceil((Math.ceil(percentage) - percentage) * 100)/100.0;

It has too many method calls, however, it will give your what it should.

IF the number is small, then the following is nice enough:

(double) Math.round(percentage * 100) / 100;

Limit decimal digits in Java when using double

3 Answers3