Java try catch not handling IndexOutOfBoundsException

Question

I was having some problem when try to try catch the IndexOutOfBoundsException for a List in Java. So I declared my list with 2 elements as:

List<String> list = new ArrayList<>(Arrays.asList("item1", "item2"));

Then I tried to do a try catch:

do {
    for (int i = 0; i < list.size(); i++) {
        System.out.print("(" + (i + 1) + ")" + list.get(i));
    }
    System.out.println(" ");

    try{
        option = sc.nextInt();
    } catch (IndexOutOfBoundsException e){
        System.out.println("Invalid option");
        sc.next();
        continue;
    } catch (InputMismatchException e) {
        System.out.println("Option input mismatch.");
        sc.next();
        continue;
    } 
    sc.nextLine();
    if (option == 1) {
        System.out.print("Enter name: ");
        // scanner takes in input
    } else if (option == 2) {
        System.out.print("Enter desc: ");
        // scanner takes in input
    }
type = list.get((option - 1));
} while (option <= 0 || option >= 3);

However, when I entered anything larger than 2 for option, it threw me IndexOutOfBounds exception but I thought I did a try catch for it already?

Thanks in advance.

Solution 1

    do {
        for (int i = 0; i < list.size(); i++) {
            System.out.print("(" + (i + 1) + ")" + list.get(i));
        }
        System.out.println(" ");

        try {
            option = sc.nextInt();
        } catch (IndexOutOfBoundsException e) {
            System.out.println("Invalid option");
            sc.next();
            continue;
        } catch (InputMismatchException e) {
            System.out.println("Option input mismatch.");
            sc.next();
            continue;
        }
        sc.nextLine();
        if (option == 1) {
            System.out.print("Enter name: ");
            // scanner takes in input
        } else if (option == 2) {
            System.out.print("Enter desc: ");
            // scanner takes in input
        }
        try {
            type = list.get((option - 1));
        } catch (IndexOutOfBoundsException e) {
            System.out.println("Invalid option");
            option=3;
        }
    } while (option <= 0 || option >= 3);

I have added new try-catch at type = list.get((option - 1)); To force user re-input option, I will set option to 3 at the catch cause

Solution 2

You are not going to catch the exception if you don't use an invalid value to call the list.

    ArrayList<String> list = new ArrayList<>(Arrays.asList("item1", "item2"));
    Scanner sc = new Scanner(System.in);
    int option;
    try {
        option = sc.nextInt();
        System.out.println(list.get(option));
    } catch (IndexOutOfBoundsException e) {
        System.out.println("Invalid option");
    } catch (InputMismatchException e) {
        System.out.println("Option input mismatch.");
    }

    sc.close();

Comments

[Admin]

I figured out that the problem came from this line: type = list.get((option - 1));. After I surround it with a try catch, the error is gone but the problem is, when user entered anything larger than 2, it does not prompt for input again

[Admin]

Hello, that's what I thought but it does not prompt user to enter input again thou. As in in I entered larger than 2, the first for loop that part wont print out again and ask for input

[VinhNT]

What do you got in console?

[Admin]

just an empty space and I have to enter any value then the for loop that part will come out after that. Ah alright I knew already because of the sc.next() in last try catch and that 's why

[VinhNT]

You need to put some promt into console, for example System.out.print("Re-enter the option: ");