No pointation on a Double value

Question

The following Code

private double roundTheReading(double toRoundValue) { double roundetValue = 0; if (formatter == null){ DecimalFormatSymbols dfs = new DecimalFormatSymbols(); dfs.setDecimalSeparator('.'); formatter = new DecimalFormat(("#0." + zerosAfterDot),dfs); } roundetValue = Double.parseDouble(formatter.format(toRoundValue)); return roundetValue; }

makes a shorting of the double "toRoundValue" because it haves 9 after dot characters and i need only 0, 1 or 2 of them (is set by zerosAfterDot by an Input Method).

Problem is: if zerosAfterDot is "0" the answer should be 5 or 3421. I get the . and 1 Zero everytime (5.0 3421.0).

Dose this have something to do with the Line:

formatter = new DecimalFormat(("#0." + zerosAfterDot),dfs);

Is the dot in this line the dot i get in my output? If so i could make an If() what checks if the value fter the dot is zero. But i would rather not do that because it would also kill the Dot if the zerosAfterDot > 0. And i want to get around to many if statements.

TL,DR How do i get rid of the "." when i dont have numbers following it that are significant?

possible duplicate of How to nicely format floating numbers to String without unnecessary decimal 0? — icza
– icza, Commented Sep 5, 2014 at 10:11

Reimeus · Accepted Answer · 2014-09-05 10:14:04Z

0

You're probably displaying the double value returned from roundTheReading which will always include non-significant digits. Instead you can return a String and use

formatter = new DecimalFormat(("0.#");

answered Sep 5, 2014 at 10:14

Reimeus

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1 Comment

Fulli Over a year ago

The value is used in other ways to, so an String return is not the best way because i need to typecast it back everytime i want to use it. But i can give the UI the string in the getter Method. its a simple answer i didnt think of until i saw yours. Thanks for that!

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