Select the correct answerWhat is the output of the below Java code with Exceptions?public class ExceptionTest2 { public static void main(String[] args) { try { int ary[] = { 10, 20, 30 }; int tempt = ary[4]; } catch (ArrayIndexOutOfBoundsException e1) { System.out.println(e1.getMessage()); } catch (Exception e2) { System.out.println("Some exception"); } }}OptionsIndex 4 out of bounds for length 3Some exceptionNo exception occursIndex 4 out of bounds for length 3 Some exception
Question
Select the correct answerWhat is the output of the below Java code with Exceptions?public class ExceptionTest2 { public static void main(String[] args) { try { int ary[] = { 10, 20, 30 }; int tempt = ary[4]; } catch (ArrayIndexOutOfBoundsException e1) { System.out.println(e1.getMessage()); } catch (Exception e2) { System.out.println("Some exception"); } }}OptionsIndex 4 out of bounds for length 3Some exceptionNo exception occursIndex 4 out of bounds for length 3 Some exception
Solution
The correct answer is "Index 4 out of bounds for length 3".
Here's the step-by-step explanation:
-
The code declares an array
ary[]of size 3 with elements 10, 20, and 30. -
Then it tries to access the element at index 4 of the array with
int tempt = ary[4];. -
However, arrays in Java are 0-indexed, meaning the first element is at index 0, the second at index 1, and so on. So, an array of size 3 will only have valid indices 0, 1, and 2.
-
Trying to access index 4 of the array
ary[]is out of bounds, which throws anArrayIndexOutOfBoundsException. -
This exception is caught by the first catch block, and the message of the exception is printed out with
System.out.println(e1.getMessage());. -
The message of an
ArrayIndexOutOfBoundsExceptionis typically in the format "Index X out of bounds for length Y", where X is the index you tried to access and Y is the size of the array. So in this case, the output will be "Index 4 out of bounds for length 3".
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