[Solved] What will be the output of this code?try { int[] arr = {1, 2, 3}; System.out.println(arr[3]); } catch (ArrayIndexOutOfBoundsException e) { System.out.println("Index error"); } finally { System.out.... | Java

Java - Exception Handling

What will be the output of this code?

try {
  int[] arr = {1, 2, 3};
  System.out.println(arr[3]);
} catch (ArrayIndexOutOfBoundsException e) {
  System.out.println("Index error");
} finally {
  System.out.println("Done");
}

AArrayIndexOutOfBoundsException\nDone

B3\nDone

CIndex error\nDone

DDone

Step-by-Step Solution

Solution:

Step 1: Identify exception thrown
Accessing arr[3] causes ArrayIndexOutOfBoundsException because valid indexes are 0-2.
Step 2: Catch block and finally execution
The exception is caught, printing "Index error". Then finally prints "Done".
Final Answer:
Index error\nDone -> Option C
Quick Check:
Exception caught + finally runs = Index error\nDone [OK]

Quick Trick: Catch runs on exception; finally always runs [OK]

Common Mistakes:

Assuming arr[3] prints 3
Thinking exception message prints automatically
Ignoring finally block output

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What will be the output of this code?

Step 1: Identify exception thrown

Step 2: Catch block and finally execution

Final Answer:

Quick Check:

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