Try–catch block in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of try-catch with arithmetic exception

What is the output of this Java code snippet?

Java

public class Main {
    public static void main(String[] args) {
        try {
            int result = 10 / 0;
            System.out.println("Result: " + result);
        } catch (ArithmeticException e) {
            System.out.println("Cannot divide by zero");
        }
    }
}

A Compilation error

B Result: 0

C Cannot divide by zero

D Runtime error: NullPointerException

💡 Hint

Think about what happens when dividing by zero in Java.

Dividing by zero throws an ArithmeticException, which is caught and prints the message.

❓ Predict Output

intermediate

Output when no exception occurs in try block

What will be printed when this Java code runs?

Java

public class Main {
    public static void main(String[] args) {
        try {
            int result = 10 / 2;
            System.out.println("Result: " + result);
        } catch (ArithmeticException e) {
            System.out.println("Cannot divide by zero");
        }
    }
}

A No output

B Result: 5

C Cannot divide by zero

D Compilation error

💡 Hint

No exception happens here, so catch block is skipped.

Division 10/2 is valid, so try block prints the result and catch block is ignored.

🔧 Debug

advanced

Identify the error in try-catch syntax

Which option shows the code that will cause a compilation error due to incorrect try-catch syntax?

A

try {
    int x = 5;
} catch (Exception e) {
    System.out.println(e);
}

B

}
;)e(nltnirp.tuo.metsyS    
{ )e noitpecxE( hctac }
;5 = x tni    
{ yrt

C

try {
    int x = 5;
} catch (Exception e) {
    System.out.println(e);
} finally {
    System.out.println("Done");
}

D

try {
    int x = 5;
} catch Exception e {
    System.out.println(e);
}

💡 Hint

Check the syntax of the catch block parentheses.

Option D misses parentheses around Exception e in catch, causing a compilation error.

❓ Predict Output

advanced

Output with multiple catch blocks

What will this Java program print?

Java

public class Main {
    public static void main(String[] args) {
        try {
            int[] arr = new int[2];
            System.out.println(arr[5]);
        } catch (ArithmeticException e) {
            System.out.println("Arithmetic Exception caught");
        } catch (ArrayIndexOutOfBoundsException e) {
            System.out.println("Array index out of bounds caught");
        } finally {
            System.out.println("Finally block executed");
        }
    }
}

A

Array index out of bounds caught
Finally block executed

B

Arithmetic Exception caught
Finally block executed

C Runtime error: NullPointerException

D No output

💡 Hint

Which exception is thrown by accessing arr[5] when arr length is 2?

Accessing arr[5] throws ArrayIndexOutOfBoundsException, caught by second catch block, then finally runs.

🧠 Conceptual

expert

Behavior of try-catch-finally with return statements

What is the output of this Java code?

Java

public class Main {
    public static int test() {
        try {
            return 1;
        } catch (Exception e) {
            return 2;
        } finally {
            return 3;
        }
    }
    public static void main(String[] args) {
        System.out.println(test());
    }
}

A 3

B 2

C Compilation error

D 1

💡 Hint

Remember that finally block executes even after return statements.

The finally block's return overrides any previous return in try or catch, so 3 is returned.