How can I catch multiple exceptions at the same time?

Question

I was trying to catch multiple exceptions, but I could not. I was only able to catch one exception.

class Main {
  public static void main(String[] args) {
    try {
      int array[] = new int[10];
      array[10] = 30 / 0;
    } catch (ArithmeticException e) {
      System.out.println(e.getMessage());
    } catch (ArrayIndexOutOfBoundsException e) {
      System.out.println(e.getMessage());
    }
  }
}

https://docs.oracle.com/javase/7/docs/technotes/guides/language/catch-multiple.html This has existed for some time; searching the web works too. — Dave Newton, Jun 27 '21 at 02:38
What multiple exceptions? When the division-by-zero error occurs, the code in the `try` block stops executing, so it never reaches the code that would otherwise cause index-out-of-bounds error. — Andreas, Jun 27 '21 at 02:39
There can only be one exception at a time not more than that, as when exception occurs it moves to the catch block — Ashish Mishra, Jun 27 '21 at 02:43

score 1 · Answer 1 · answered Jun 27 '21 at 02:53

As in the statements array[10] = 30 / 0; you can see what the compiler will do at the first hand is evaluation of 30 divide by zero that will surely throw a arithmetic exception. As a statement before assigning operation throws exception thus it will break from the try block and skip to catch (ArithmeticException e) block , array[invalidIndex] is never executed so it will never throw ArrayIndexOutOfBoundsException.

How can I catch multiple exceptions at the same time?

1 Answers1