Multiple catch blocks in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of multiple catch blocks with specific exceptions

What is the output of the following Java code snippet?

Java

public class Test {
    public static void main(String[] args) {
        try {
            int[] arr = new int[2];
            System.out.println(arr[5]);
        } catch (ArithmeticException e) {
            System.out.println("ArithmeticException caught");
        } catch (ArrayIndexOutOfBoundsException e) {
            System.out.println("ArrayIndexOutOfBoundsException caught");
        } catch (Exception e) {
            System.out.println("General Exception caught");
        }
    }
}

A ArrayIndexOutOfBoundsException caught

B General Exception caught

C ArithmeticException caught

D Compilation error due to multiple catch blocks

💡 Hint

Think about which exception is actually thrown and which catch block matches it first.

The code throws an ArrayIndexOutOfBoundsException because it tries to access index 5 in an array of size 2. The catch block for ArrayIndexOutOfBoundsException matches this exception and executes, printing "ArrayIndexOutOfBoundsException caught".

❓ Predict Output

intermediate

Which catch block executes for NullPointerException?

Given the code below, which catch block will handle the exception?

Java

public class Test {
    public static void main(String[] args) {
        try {
            String s = null;
            System.out.println(s.length());
        } catch (IllegalArgumentException e) {
            System.out.println("IllegalArgumentException caught");
        } catch (NullPointerException e) {
            System.out.println("NullPointerException caught");
        } catch (Exception e) {
            System.out.println("General Exception caught");
        }
    }
}

A IllegalArgumentException caught

B No exception caught, program runs normally

C NullPointerException caught

D General Exception caught

💡 Hint

What happens when you call a method on a null reference?

Calling length() on a null String throws a NullPointerException. The catch block for NullPointerException matches and executes, printing "NullPointerException caught".

🔧 Debug

advanced

Identify the compilation error in multiple catch blocks

Why does the following code cause a compilation error?

Java

public class Test {
    public static void main(String[] args) {
        try {
            int a = 5 / 0;
        } catch (Exception e) {
            System.out.println("Exception caught");
        } catch (ArithmeticException e) {
            System.out.println("ArithmeticException caught");
        }
    }
}

A No compilation error, code runs and prints "Exception caught"

B Compilation error because ArithmeticException catch block is unreachable

C Compilation error because try block has no code

D Compilation error because catch blocks must be in alphabetical order

💡 Hint

Remember the order of catch blocks matters when exceptions have inheritance relationships.

ArithmeticException is a subclass of Exception. The first catch block catches all Exceptions, so the second catch block for ArithmeticException is unreachable, causing a compilation error.

❓ Predict Output

advanced

Output when multiple exceptions are possible

What will be printed when running this code?

Java

public class Test {
    public static void main(String[] args) {
        try {
            int[] arr = null;
            System.out.println(arr.length);
            int a = 5 / 0;
        } catch (NullPointerException e) {
            System.out.println("NullPointerException caught");
        } catch (ArithmeticException e) {
            System.out.println("ArithmeticException caught");
        } catch (Exception e) {
            System.out.println("General Exception caught");
        }
    }
}

A NullPointerException caught

B ArithmeticException caught

C General Exception caught

D No output, program crashes

💡 Hint

Which exception happens first in the try block?

The first statement in try throws NullPointerException because arr is null. The catch block for NullPointerException executes and prints "NullPointerException caught". The division by zero is never reached.

🧠 Conceptual

expert

Why must catch blocks be ordered from specific to general?

In Java, why should catch blocks for exceptions be ordered from the most specific to the most general exception type?

A Because catch blocks are executed in parallel and order does not matter

B Because Java requires catch blocks to be alphabetically ordered by exception name

C Because specific exceptions cannot be caught unless they are last

D Because a general exception catch block before a specific one makes the specific catch block unreachable, causing a compilation error

💡 Hint

Think about how Java matches exceptions to catch blocks and what happens if a parent exception is caught first.

If a general exception catch block (like Exception) is placed before a specific one (like ArithmeticException), the specific catch block becomes unreachable because the general one catches all exceptions of that type and its subclasses. This causes a compilation error.