how to find length after decimal point of a double number without removing trailing zeroes in java?

Question

Suppose you have a double number say Double d = 12.123223800000. How can we find length of number from decimal point till the end including all trailing zeroes in it.

Double d = 12.123223800000;
String t = d.toString();

Here it's removing all the trailing zeroes from double number.

Possible duplicate of [counting the number of zeros](https://stackoverflow.com/questions/13853921/counting-the-number-of-zeros) — yassadi, Jul 10 '18 at 11:18
What is the difference between 12.123223800000 and 12.1232238? — TheJavaGuy-Ivan Milosavljević, Jul 10 '18 at 11:22
The trailing zeros removed while you initialize it to variable — Venkataraghavan Yanamandram, Jul 10 '18 at 11:23
`Double d = 12.123223800000;` will just store `12.1232238`, so you've already lost the zeros. — Steve Smith, Jul 10 '18 at 11:36
A floating point value doesn't have a number of trailing or leading zeros. It just has a value which is the closest representation to the one you have. i.e. it's not even exactly `12.1232238` — Peter Lawrey, Jul 10 '18 at 11:41

score 0 · Answer 1 · answered Jul 10 '18 at 11:35

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Are you able to assign the string directly? Java primitives don't keep trailing zeros because of their binary representation, yet in Mathematics trailing zeros are meaningless .. and so on. This sample would kinda work

Double d = 12.123223800000;
String str = "12.123223800000"; // d.toString() won't work here
int length = str.length() - str.indexOf(".");
System.out.println(String.format("%."+ length +"f", d)); 
//prints out - 12.1232238000000

answered Jul 10 '18 at 11:35

noobed

1,329
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I am using Double.toString(d) here. – Aayushi Tayal Jul 10 '18 at 12:48
@AayushiTayal Then I guess... there is no option to do so, unless you specify as a hard-coded constant how many digits after the decimal point you want. – noobed Jul 10 '18 at 13:54
if we'll take double value as a String input (String str = "12.123223800000"), then there are other options also, but don't know how we can convert double value to string without removing zeroes. – Aayushi Tayal Jul 11 '18 at 08:14
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1. str = str.substring(str.indexOf(".")+1); int len = str.length()+1; .........2. There is another option also,we can parse the string to double value.. it'll give the number in exponential form.String str = "12.123223800000"; int length = str.length() - str.indexOf(".");Double d2 =Double.parseDouble(str.substring(str.indexOf('.')+1));String ch = d2.toString();int o = ch.indexOf('E'); String v=ch.substring(o+1); (The value of v will give you length after decimal point-1) – Aayushi Tayal Jul 11 '18 at 08:25

score 0 · Answer 2 · answered Jul 10 '18 at 11:35

0

There's no difference between 12.123223800000 and 12.1232238. The trailing zeros are removed while you initialize it to Double. If you still need such solution

String d = "12.123223800000";
System.out.println((d.length()-d.indexOf("."))-1);

answered Jul 10 '18 at 11:35

Venkataraghavan Yanamandram

431
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score 0 · Answer 3 · answered Jul 10 '18 at 11:38

This should work:

String text = Double.toString(Math.abs(d))
int decimalPlaces = text.length()  - text.indexOf('.') - 1;

There is no way to get the trailing zeros counted because they don't affect the mathematical value of your double figure. Edit: Except you assign your value directly to a string. With Javas primitve types, it is not possible

how to find length after decimal point of a double number without removing trailing zeroes in java?

3 Answers3