Java Program: Method to check and handle odd numbers
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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