Java Program: Method to check and handle odd numbers

Java Exception Handling: Exercise-2 with Solution

Write a Java program to create a method that takes an integer as a parameter and throws an exception if the number is odd.

Sample Solution:

Java Code:

public class Exception_OddNumber {
  public static void main(String[] args) {
    int n = 18;
    trynumber(n);
    n = 7;
    trynumber(n);
  }

  public static void trynumber(int n) {
    try {
      checkEvenNumber(n);
      System.out.println(n + " is even.");
    } catch (IllegalArgumentException e) {
      System.out.println("Error: " + e.getMessage());
    }
  }

  public static void checkEvenNumber(int number) {
    if (number % 2 != 0) {
      throw new IllegalArgumentException(number + " is odd.");
    }
  }
}

Sample Output:

18 is even.
Error: 7 is odd.

Explanation:

In the above exercise,

• The Exception_OddNumber class is the main class.
• In the main method, an integer n is declared and assigned 18. The trynumber method is then called with n as an argument.
• The trynumber method handles the exception. It contains a try-catch block. Inside the try block, the method checkEvenNumber is called, passing n as an argument. If the number is even, the message "[number] is even." is printed.
• If an exception occurs in the try block, it is caught by the catch block, which handles IllegalArgumentException. In this case, the error message "Error: [exception message]" is printed.
• After the first call to trynumber(n), the value of n is updated to 7, and the trynumber method is called again. This time, since 7 is an odd number, an exception is thrown.
• The checkEvenNumber method checks if a given number is even or odd. If the number is odd, it throws an IllegalArgumentException with the message "[number] is odd."

Flowchart:

Java Code Editor:

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