exception handling java user input

Question

I'am trying to put an exception in my code. But it doesnt work fully. Im trying for user to enter integers and then adding it to a list.

how come it only works for the first input? say if the first I enter 1.0 then it will raise the error. but if I put say 10, 1.0 it doesnt produce an error. and when i print the list. there is only the 10 well this is actually correct. but i want to make it so that if anything is entered other than integer it raise an error, and does not terminate. instead ask the user to try again.

this is what my code looks like

package basic.functions;
import java.util.*;
import java.text.DecimalFormat;

public class Percent {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        reader.useDelimiter(System.getProperty("line.separator"));
        List<Integer> list = new ArrayList<>();
        System.out.println("Enter Integer: ");

        do {
            try {
                int n = reader.nextInt();
                list.add(Integer.valueOf(n));
            } catch (InputMismatchException exception) {
                System.out.println("Not an integer, please try again");
            }
        }

        //When user press enter empty
        while (reader.hasNextInt());
        reader.close();

I still need some upvotes today; so I would love to have a real answer for you, but: I added System.out.println("got: " + list); after that close() statement ... and everything works fine. I can enter: 10 20 30 (with ENTER inbetween!) ... and 10, 20 , 30 is printed. Thus: unable to reproduce! — GhostCat
– GhostCat, Commented May 3, 2017 at 13:04

Sridhar · Accepted Answer · 2017-05-03 14:19:29Z

2

What you are running is a do-while which will execute once before checking the condition.

So, your first input is tried to be parsed into an integer to which you have handled the exception, thus printing the error message.

If you give a non-integer anywhere other than first time, your code will just terminate the loop before trying to parse the input.

Something like the following will give you the error message properly as you expect.

public static void main(String[] args) {
    Scanner reader = new Scanner(System.in);
    reader.useDelimiter(System.getProperty("line.separator"));
    List<Integer> list = new ArrayList<>();
    System.out.println("Enter Integer: ");

    while (true) {
        try {
            int n = reader.nextInt();
            list.add(Integer.valueOf(n));
        } catch (InputMismatchException exception) {
            System.out.println("Not an integer, please try again. Press enter key to exit");
            if (reader.next().isEmpty()) {
                break;
            }
        }
    }

    System.out.println(list);
    reader.close();
}

edited May 3, 2017 at 14:19

answered May 3, 2017 at 13:11

Sridhar

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19 Comments

Kutam Over a year ago

that works. Oh im just wondering, how do i make the code robust so that, instead of break the code ask the user for another try?

Kutam Over a year ago

and why does that also print the exception?

Kutam Over a year ago

it works, but its also printing the exception. why is that?

Sridhar Over a year ago

What exception is it printing?

Kutam Over a year ago

the ("Not an integer, please try again") from the exception handling

|

dumbPotato21 · Accepted Answer · 2017-05-03 13:03:47Z

2

Whenever you input a String,reader.hasNextInt() returns false, and the loop terminates. Therefore, not printing anything.

However, since it is a do-while loop, it must execute once, and therefore, if you input a String first, then it prints out the error

This again, only happens if you press Enter after each input

answered May 3, 2017 at 13:03

dumbPotato21

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