Separating double according to "." in java

Question

i have a double number which is "547.123456"

i just want to use this double as "547.1" like only 1 number after "."

How can i do that?

You will find this question and answer interesting:- [Round a double in Java]: [http://stackoverflow.com/questions/22036885/round-a-double-in-java/22037280#22037280] — Rahul Tripathi, Mar 05 '14 at 12:09
Do you want to round it or to represent it (print it) that way? You can continue treating it as a double, and use a DecimalFormat to show only one decimal. — helderdarocha, Mar 05 '14 at 12:12
If you want to cut away everything after the first decimal (*not* round it) - trick a little: `d = ((int) (d * 10)) / 10.0` — Michael Rose, Mar 05 '14 at 12:13

score 5 · Accepted Answer · answered Mar 05 '14 at 12:13

5

Use BigDecimal

double f=547.123456;
BigDecimal d=new BigDecimal(f);
System.out.print( d.setScale(1, RoundingMode.FLOOR));

answered Mar 05 '14 at 12:13

Nambi

1

+1 this is the only reliable way of ensuring that you are **always** working with `547.1`. – skiwi Mar 05 '14 at 12:15
@skiwi except that the snippet above use a double to create the BigDecimal, in which case the precision is already lost. Only the constructors with String should be used. – gizmo Mar 05 '14 at 12:23
@gizmo the precision is not lost - it is as good as the precision of the original double... – assylias Mar 05 '14 at 14:43

score 2 · Answer 2 · answered Mar 05 '14 at 12:20

If you want to trancate a value e.g.

  47.12345 -> 47.1
  47.56789 -> 47.5 // <- not 47.6!

you can do it via floor

  double value = 47.123456;
  double result = Math.floor(value * 10.0) / 10.0;

If you want to round a value e.g.

  47.12345 -> 47.1
  47.56789 -> 47.6 // <- not 47.5!

you can do it via round

  double value = 47.123456;
  double result = Math.round(value * 10.0) / 10.0;

score 2 · Answer 3 · answered Mar 05 '14 at 12:21

2

Use a String.format

String.format("%.1f",547.123456);

This an easiest way i think.

answered Mar 05 '14 at 12:21

Sergey Shustikov

libik · Answer 4 · 2014-03-05T12:30:04.447

1

Very easily :)

double x = 47.123456;
x = (long)(x*10)/(double)10

edited Mar 05 '14 at 12:30

answered Mar 05 '14 at 12:10

libik

Even x+1 can overflow... – libik Mar 05 '14 at 12:13
Yes and? Your code can't handle any double larger than Integer.MAX_VALUE, which is a lot less than the largest possible double. Even without the `int` cast, the statement could overflow for very large doubles. – assylias Mar 05 '14 at 12:16
You should at least cast to long here. – Radiodef Mar 05 '14 at 12:17
Now it's in the wrong place. ; P – Radiodef Mar 05 '14 at 12:23
@Radiodef - damn me today :) – libik Mar 05 '14 at 12:30

score 1 · Answer 5 · answered Mar 05 '14 at 12:11

1

There is a general solution for every precision, you can specify your own like so:

DecimalFormat decimalFormat = new DecimalFormat("#.#");
System.out.println(decimalFormat.format(<your double>));

answered Mar 05 '14 at 12:11

nikis

5 Answers5