Removing All But The First Two Numbers In A Decimal

- 1 answer

I'm trying to make a program that converts a number the user inputs into a percentage. Once converted, I want to keep the first two numbers after the decimal, but without rounding off the number. For example, inputting 1.23456 would result in 123.45.

My line of thinking was that I could take the input, separate the string into before and after the decimal place, and from there create a loop that removes the last digits until it has at most 2 decimal places. My issue is that whenever I create something that would have an output greater than 9 for the decimal, I get the error: Exception in thread "main" java.lang.StringIndexOutOfBoundsException: begin 0, end -1, length 0, so I can only get decimals to the tenths place at the moment.

My code:

import java.util.Scanner;
public class percentage {

public static void main(String[] args) {
    try (// TODO Auto-generated method stub
    Scanner x = new Scanner( {
        System.out.println("Please input a number with a decimal to be converted into a percentage: ");
        String numberAsString =;
        double number = Double.parseDouble(numberAsString);
        double percentage = (number * 100);         
        String toString = Double.toString(percentage);
        String[] parts = toString.split("[.]");
        String integer = parts[0];
        String decimal = parts[1];
        int length = decimal.length();

        while(length>2) {
            decimal = decimal.substring(0,decimal.length()-1);
        System.out.println("decimal is " + decimal);
        System.out.println("integer is " + integer);


Multiply by 10,000 (100 for percentage, 100 for the two decimal places), truncate to integer, divide by 100 (to get back to percentage).

Writing it out one step at a time, for clarity of exposition:

Double n = input.nextDouble();
int i = (int)(n * 10_000);
Double pc = i / 100.0;
System.out.printf("%.2f\n", pc);