CODEWITHSUNDEEP
javaexception
Matt
Matt

Reputation: 482

Java exception in loop

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?

Upvotes: 0

Views: 247

Answers (3)

Makoto
Makoto

Reputation: 106440

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.

Upvotes: 1

Matt
Matt

Reputation: 482

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

Upvotes: 0

Ankit Rustagi
Ankit Rustagi

Reputation: 5637

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");
}

Upvotes: 0

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