CODEWITHSUNDEEP
javajava.util.scanner
Hussain
Hussain

Reputation: 663

Java.util.scanner error handling

I'm helping a friend with a java problem. However, we've hit a snag. We're using Java.Util.Scanner.nextInt() to get a number from the user, asking continiously if the user gives anything else. Only problem is, we can't figure out how to do the error handeling.

What we've tried:

do {
  int reloop = 0;
  try {
    number = nextInt(); 
  } catch (Exception e) {
    System.out.println ("Please enter a number!");
    reloop ++; 
  }
} while(reloop != 0);

Only problem is, this loops indefinatly if you enter in something not a number.

Any help?

Upvotes: 7

Views: 33670

Answers (2)

Nick Z.
Nick Z.

Reputation: 39

if the number is non-int , exception will pop, if not reloop will become 1 , and loop will exit 

  int reloop = 0;
  do {
   try {
        number = nextInt();
        reloop ++; 
  } catch (Exception e) {
       System.out.println ("Please enter a number!");
  }}
 while(reloop == 0);

Upvotes: 2

polygenelubricants
polygenelubricants

Reputation: 383686

You can use hasNextInt() to verify that the Scanner will succeed if you do a nextInt(). You can also call and discard nextLine() if you want to skip the "garbage".

So, something like this:

Scanner sc = new Scanner(System.in);
while (!sc.hasNextInt()) {
   System.out.println("int, please!");
   sc.nextLine();
}
int num = sc.nextInt();
System.out.println("Thank you! (" + num + ")");

See also:

The problem with your code, in addition to the unnecessarily verbose error handling because you let nextInt() throw an InputMismatchException instead of checking for hasNextInt(), is that when it does throw an exception, you don't advance the Scanner past the problematic input! That's why you get an infinite loop!

You can call and discard the nextLine() to fix this, but even better is if you use the exception-free hasNextInt() pre-check technique presented above instead.

Upvotes: 16

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