AP Computer Science Exam Question

You are given the following code snippet:

public class TryCatchExample {
    public static void main(String[] args) {
        try {
            int num1 = 10;
            int num2 = 0;
            int result = num1 / num2;
            System.out.println("Result: " + result);
        } catch (ArithmeticException e) {
            System.out.println("Error: Cannot divide by zero!");
        }
    }
}

Explain what will happen when the above code is run. Identify the purpose of the try-catch block and provide a detailed explanation of the output that will be produced.

Answer with Explanation

The purpose of the try-catch block in the given code is to catch and handle an ArithmeticException that may occur when dividing num1 by num2.

When this code is run, an ArithmeticException will be thrown because dividing any number by zero is not allowed in mathematics.

The try block contains the code that may throw an exception, in this case, the division operation. If an exception occurs within the try block, the execution immediately transfers to the catch block.

In the catch block, the ArithmeticException is caught, and the message "Error: Cannot divide by zero!" is printed to the console using System.out.println(). This message informs the user that attempting to divide by zero has occurred and handles the exception gracefully.

Therefore, the output of running this code will be:

Error: Cannot divide by zero!

The program will exit gracefully after the exception is caught in the catch block, preventing the program from crashing due to the division by zero error.