CODEWITHSUNDEEP
javaexceptionswitch-statement
Brian
Brian

Reputation: 115

Exception handling in a switch

The following code is in a while loop which then houses a switch as such:

  System.out.println("Enter in a selection.");
  System.out.println("Enter \"1\" for a default selection of die");
  System.out.println("Enter \"2\" for a custom number of sides.");
  //try the input to see if its an integer
  try {
  selection = sc.nextInt();
  } catch (NumberFormatException e){
      System.out.print("Your selection can only be an integer!");
   } 
 switch (selection){
   case1:
   ...
   break;

   case2:
   ...
   break;

   default:
   ...
   //yell at them
   continue;
  }

I already have a default selection in the switch so that if a user enters in an invalid number like 4 (since there are only 2 cases) that it brings them back to the beginning of the loop. So, the issue is handling the exception. The exception does not get handled with the following above code and I don't know why. The try is housing the offending code.

As always, please ask for clarification if needed. Thanks.

Upvotes: 1

Views: 16497

Answers (6)

Bhesh Gurung
Bhesh Gurung

Reputation: 51030

You should continue on the exception as there is no point in going ahead do the switch on the selection if the exception occurred.

} catch (InputMismatchException e){
    System.out.print("Your selection can only be an integer!");
    sc.nextLine();
    continue;
}

The way you have it right now, if the exception were to occur you would be switching the old stale value held by selection.

But the main problem in your code is that the nextInt() method doesn't throw the NumberFormatException, instead it throws the InputMismatchException.

So you are trying to catch the wrong exception.

Upvotes: 1

Yogendra Singh
Yogendra Singh

Reputation: 34367

Scanner#nextInt() doesn't throw NumberFormatException

Scans the next token of the input as an int. An invocation of this method of the form nextInt() behaves in exactly the same way as the invocation nextInt(radix) , where radix< is the default radix of this scanner.

 return the int scanned from the input
 throws InputMismatchException
         if the next token does not match the <i>Integer</i>
          regular expression, or is out of range
 throws NoSuchElementException if input is exhausted
 throws IllegalStateException if this scanner is closed

Your exception handling seems right, with only issue that you are catching wrong exception. Please catch InputMismatchException and other two.

e.g. below:

    try {
          selection = sc.nextInt();
      } catch (InputMismatchException e){
          System.out.print("Your selection can only be an integer!");
      }

Also you may want to put the above code in a while loop as below:

       boolean validInput = false;
       while(!validInput){
     try {
          selection = sc.nextInt();
              validInput = true;
      } catch (InputMismatchException e){
          System.out.print("Your selection can only be an integer!");
      }
       }

This will repeatedly ask for input until a number is entered.

Upvotes: 1

Zach
Zach

Reputation: 897

If you want to really handle the exception in the 'switch' statement, you need to expand the scope of the try-catch block:

  System.out.println("Enter in a selection.");
  System.out.println("Enter \"1\" for a default selection of die");
  System.out.println("Enter \"2\" for a custom number of sides.");
  //try the input to see if its an integer
  try {
  selection = sc.nextInt();

  switch (selection){
   case1:
   ...
   break;

   case2:
   ...
   break;

   default:
   ...
   //yell at them
   throw new NumberFormatException("Yelling message");
   continue;
  }
 } catch (NumberFormatException e){
      System.out.print("Your selection can only be an integer!");
 }

Upvotes: 0

paulsm4
paulsm4

Reputation: 121669

Personally, I would put your input (and the try/catch block that goes with it) in its own, separate method. Return a boolean (true = valid integer) or a value that's out of range (perhaps "-1", depending on your program).

IMHO...

Upvotes: 1

Platinum Azure
Platinum Azure

Reputation: 46193

In the case of a Scanner, you can use the hasNextInt() method (and equivalents for other data types) rather than crashing and burning via exception:

while (some_condition) {
  System.out.println("Enter in a selection.");
  System.out.println("Enter \"1\" for a default selection of die");
  System.out.println("Enter \"2\" for a custom number of sides.");
  //try the input to see if its an integer
  if (!sc.hasNextInt()) {
    System.out.println("You must enter an integer!");
    continue;
  }
  selection = sc.nextInt();

  switch (selection){
   case 1:
   ...
   break;

   case 2:
   ...
   break;

   default:
   ...
   //yell at them
   continue;
  }
}

As Bhesh Gurung said, if you're in a loop, you should just use continue to go back to the beginning, like you do in the default case.

Upvotes: 0

Summer_More_More_Tea
Summer_More_More_Tea

Reputation: 13356

The reason exception is not handled when you enter 4 is that 4 is a valid integer and nextInt() will not raise NumberFormatException exception.

You'd better enclose the switch in the try block.

Upvotes: 1

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