Java exception in loop

Question

Scanner b = new Scanner(System.in);

public void menu() {
    while (true) {
        try {
            System.out.println("Your options is:");
            int option = 0;
            option = b.nextInt();
            b.nextLine();

            if (option == 1)
                add();
            else if (option == 2)
                delete();
            else if (option == 3)
                print();
            else if (option == 4)
                nr();
            else if (option == 0)
                break;
        } catch (Exception e) {
            System.out.println("Not a valid option!");
            // break;
        }
    }
    b.close();
}

Thing is that whenever I'm giving something else than the selected options(0,1,2,3,4) the function prints the message from the catch and then asks for a new option but this thing is done repeatedly. If the break is there, the whole execution is stopped and all I want is to ask again for an option but without the infinite loop. For example, if I type a "r" instead of a number, it should catch the exception and get again through the while. What am I missing?

Makoto · Accepted Answer · 2013-11-18 07:49:01Z

1

The fundamental purpose of try...catch is to try to execute code that could throw an exception, and if it does throw the exception, catch it and deal with it.

Invalid input in this case won't throw an exception, unless they managed to sneak a character in there, which would result in an infinite loop.

Instead of assuming that the user will enter in a digit - which is what Scanner#nextInt generally assumes - read the whole line instead, and use Integer.parseInt() to construct the int.

option = Integer.parseInt(b.nextLine());

If this blows up because someone decided to put "foo" instead of "1", then you'll hit your exceptional case, and be able to re-enter values again.

answered Nov 18, 2013 at 7:49

Makoto

107k29 gold badges199 silver badges236 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Ankit Rustagi · Accepted Answer · 2013-11-18 07:46:33Z

0

do this

try
{
  do
  {
   option=b.nextInt();
   // continue

   switch(option){


        case 0:  
                 break;
        case 1:  add();
                 break;
        case 2:  delete();
                 break;
        case 3:  print();
                 break;
        case 4:  nr();
                 break;
        default: System.out.println("Not a valid option!");
                 break;

   }

  }while(!(option==1||option==2||option==3||option==4))
}
catch(Exception e){
    System.out.println("Please enter a number");
}

edited Nov 18, 2013 at 7:46

answered Nov 18, 2013 at 7:40

Ankit Rustagi

5,63712 gold badges44 silver badges71 bronze badges

1 Comment

Makoto Over a year ago

What happens if I enter in a series of letters?

Matt · Accepted Answer · 2013-11-18 07:48:10Z

0

solved it.I added a b.nextLine(); in the catch and the loop stops.

answered Nov 18, 2013 at 7:48

Matt

4825 silver badges15 bronze badges

Collectives™ on Stack Overflow

Java exception in loop

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related