Count digits after decimal including last zero

Question

I have this,

import java.util.Scanner;

public class StringDecimalPartLength {

    public static void main(String[] args){
       Scanner input = new Scanner(System.in);
       System.out.print("Enter a decimal number: ");
       Double string_Temp = Double.parseDouble(input.nextLine().replace(',', '.'));
       String string_temp = Double.toString(string_Temp);
       String[] result = string_temp.split("\\.");
       System.out.print(result[1].length() + " decimal place(s)");
    }
}

it works until I enter a number with trailing zero, such as 4,90. It ignores the zero and returns 1.

How to fix this? Thank you!

Have you calculating the length before converting to a double? If you convert `"4.90"` into a double, you get `4.9`. Then if you convert back to string you get `"4.9"`. — Robert Vunabandi, Feb 27 '19 at 19:02

Omari Celestine · Accepted Answer · 2019-02-27T19:32:20.550

1

Since you are already reading the input as a string, you can save that value and then test to see if it is a valid decimal number:

import java.util.Scanner;

public class StringDecimalPartLength {
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        System.out.print("Enter a decimal number: ");
        String value = input.nextLine().replace(',', '.');

        try {
            Double.parseDouble(value);

            String[] result = value.split("\\.");

            System.out.print(result[1].length() + " decimal place(s)");
        } catch (NumberFormatException e) {
            System.out.println("The entered value is not a decimal number.");
        }
    }
}

edited Feb 27 '19 at 19:32

answered Feb 27 '19 at 19:09

Omari Celestine

1,272
1
9
18

thanks Omari, it worked, yet it didn't catch all the errors. I had to adjust it and it worked. – RoiboisTx Feb 27 '19 at 20:09
when something like 1. was entered, it didn't catch the exception, now it does – RoiboisTx Feb 28 '19 at 12:36
@RoiboisTx That would be because in parsing the value `1` can be converted to a decimal number since it is equivalent to `1.0`. – Omari Celestine Feb 28 '19 at 12:40

score 0 · Answer 2 · answered Feb 27 '19 at 19:21

4.9 or 4.90 as a double are both represented by the same approximation. There is no difference between them.

You can, of course, do what you want using String processing on the input. However, you may find BigDecimal a better data type for what you are doing. It distinguishes between 4.9 and 4.90, and can represent each of them exactly.

import java.math.BigDecimal;

public class Test {
  public static void main(String[] args) {
    System.out.println(new BigDecimal(4.9));
    System.out.println(new BigDecimal(4.90));
    BigDecimal x1 = new BigDecimal("4.9");
    System.out.println(x1);
    System.out.println(x1.scale());
    BigDecimal x2 = new BigDecimal("4.90");
    System.out.println(x2);
    System.out.println(x2.scale());
  }
}

Output:

4.9000000000000003552713678800500929355621337890625
4.9000000000000003552713678800500929355621337890625
4.9
1
4.90
2

I will look into this approach, appreciate it. – RoiboisTx Feb 27 '19 at 20:16 — RoiboisTx, Feb 27 '19 at 20:16

score 0 · Answer 3 · answered Feb 27 '19 at 20:14

Omari, I adjusted your code like this and it worked, thanks!

if(result[0].length() == 0 ){

            System.out.print("The entered value is not a decimal number.");
        }
        else {
            System.out.print(result[1].length() + " decimal place(s)");
        }

    } catch (NumberFormatException e) {

        System.out.println("The entered value is not a decimal number.");

    }catch (ArrayIndexOutOfBoundsException e) {
System.out.print("Error Message" );     
      }

Count digits after decimal including last zero

3 Answers3