I'm trying to make my program throw an IllegalArgumentException if the input is not a number

Question

I'm trying to make it to were is will throw my exception if the input is not a number and i cant figure it out, Could someone help me get on the right track

import java.util.Scanner;

    class calculations
    {

        public static void main(String[] args)
        {
            Scanner scan = new Scanner(System.in);
            int number;
            int total = 0;
            try
            {

            } catch ( IllegalArgumentException e)
            {
                //error
            }
            while (true)
            {
                number = scan.nextInt();

                if (number == 0)
                {
                    break;
                }
                total += number;
            }
            System.out.println("Total is " + total);
        }

    }

You're catching `IllegalArgumentException ` instead of throwing it — Pankaj Gadge, Mar 22 '18 at 18:12
Why do you think that a `try`...`catch` block would do this? — EJoshuaS - Stand with Ukraine, Mar 22 '18 at 18:15
Possible duplicate of [Throw exception in Java](https://stackoverflow.com/questions/36706553/throw-exception-in-java) — EJoshuaS - Stand with Ukraine, Mar 22 '18 at 18:16

score 2 · Answer 1 · answered Mar 22 '18 at 18:17

2

You should use hasNextInt, which will allow you to check if the next token in the stream can be parsed as an int.

if (! scanner.hasNextInt()) throw new IllegalArgumentException()

answered Mar 22 '18 at 18:17

Floegipoky

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I haven't programmed in Java in a while, but would that work if the next item were a "." (period) ? – Phil N DeBlanc Mar 22 '18 at 18:26
1

Yes it should, it is checking if there is an int or not. A '.' is not an int, so hopefully there won't be a problem. – Sean Mar 22 '18 at 18:41

I'm trying to make my program throw an IllegalArgumentException if the input is not a number

1 Answers1