[Solved] What will be the output of this Java program?public class ArrayTest { public static void main(String[] args) { try { int[] numbers = new int[3]; numbers[4] = 50;... | Java

Java - Exception Handling

What will be the output of this Java program?

public class ArrayTest {
  public static void main(String[] args) {
    try {
      int[] numbers = new int[3];
      numbers[4] = 50;
    } catch (ArrayIndexOutOfBoundsException e) {
      System.out.println("Array index is out of bounds");
    }
  }
}

A0

B50

CArray index is out of bounds

DCompilation error

Step-by-Step Solution

Solution:

Step 1: Understand array indexing
The array 'numbers' has size 3, valid indices are 0 to 2.
Step 2: Accessing invalid index
Accessing index 4 causes ArrayIndexOutOfBoundsException.
Step 3: Catch block
The exception is caught and prints "Array index is out of bounds".
Final Answer:
Array index is out of bounds -> Option C
Quick Check:
Invalid array index triggers exception catch [OK]

Quick Trick: Invalid array index triggers ArrayIndexOutOfBoundsException [OK]

Common Mistakes:

Expecting the program to print 50
Assuming index 4 is valid for size 3 array
Thinking this causes a compile-time error

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What will be the output of this Java program?

Step 1: Understand array indexing

Step 2: Accessing invalid index

Step 3: Catch block

Final Answer:

Quick Check:

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