Question

I know how to format a double to keep only the available decimal places (DP), up to a certain number of DPs. This example keeps up to 4 DPs.

    double d = 1.0;
    DecimalFormat df = new DecimalFormat("#.####");
    System.out.print(df.format(d)); // returns "1"

    double d = 1.23;
    DecimalFormat df = new DecimalFormat("#.####");
    System.out.print(df.format(d)); // returns "1.23"

    double d = 1.2345678;
    DecimalFormat df = new DecimalFormat("#.####");
    System.out.print(df.format(d)); // returns "1.2346", rounding off: bad!

Now I want whole numbers e.g. 1.0 to return "1" without the unnecessary .0, and the # format character does provide that functionality. But how do I make sure that the number never gets rounded off? Is there any other way other than an arbitrarily long chain of # such as "#.###########################################"?

Or should I just use the default conversion of double to string, and truncate the ".0" if it appears at the end:

    String s = "" + d;
    if(s.substring(s.length()-2).equals(".0")) {
        s=s.substring(0, s.length()-2);
    }

Both ways seems terribly clumsy.

Answer 1

I use the following

double d =
String s = (long) d == d ? "" + (long) d : "" + d; 

if you need Double instead for double. (Personally I would avoid using the wrapper if you can)

Double d =
String s = d.longValue() == d ? "" + d.longValue() : "" + d;