如何在不删除Java中尾随零的情况下找到双精度数字小数点后的长度？

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.

Answer 1

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

Answer 2

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);

Answer 3

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