Scanner inside a loop - Why does it not exit loop when incorrect input is entered?

Question

The code -

for(int i = 0; i < count; i++)
    {
     System.out.println("Please enter a number!");
    try{
         numbers[i] = keyboardScanner.nextInt();
    }catch(InputMismatchException ex)
    {
      System.out.println("You did not enter a number!");
    }
}

If you enter a String, it loops till 'count'. Why does that happen ?

nneonneo · Accepted Answer · 2012-10-26 23:39:07Z

4

Sure, it'll loop until count: you caught the exception and continued the loop anyway.

You want to break out of your loop when you get that exception:

System.out.println("You did not enter a number!");
break;

answered Oct 26, 2012 at 23:39

nneonneo

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3 Comments

Apple Grinder Over a year ago

Yes, the break will help. But, i have another question. Why not design the Java language such that the loop will not execute fully if some exception is thrown ?

Matthias Braun Over a year ago

@Apple Grinder: Because there might be good reasons to stay in the loop after an exception was thrown.

le3th4x0rbot Over a year ago

It is because the exception is caught, period. Caught exceptions are not thrown from try-catch!!!!!

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