Java Code Example: calculate cross sum (iterative and recursive)

In this Java code example, the checksum of a number is calculated – both iterative and recursive.

Iterative version

This code calculates the cross sum of a given number n. The cross sum of a number is the sum of its individual digits.

The crossSum method uses a while loop to keep adding the last digit of n to total until n becomes 0. This is done by using the modulus operator % to get the last digit of n, and dividing n by 10 in each iteration of the loop to remove the last digit.

In the main method, crossSum is called with the argument 174, and the result is printed. The expected output is 12, since 174 has digits 1, 7, and 4, and the sum of these is 12.

public class CrossSumIterative {
    static int crossSum(int n) {
        int total = 0;
        while (0 != n) {
            total = total + (n % 10);
            n = n / 10;
        }
        return total;
    }

    public static void main(String[] args) {
        System.out.print(crossSum(174));
    }
}

Output

12

Recursive version

This is a Java program that calculates the cross-sum of an integer using recursion. The cross-sum of a number is the sum of its digits.

The program defines a static method crossSum that takes an integer n as its argument. The method uses recursion to sum the digits of the number.

The method checks if n is less than or equal to 9, in which case it returns n. This is the base case of the recursion, where the calculation stops.

If n is not less than or equal to 9, the method returns the last digit of n (obtained by n % 10) plus the cross-sum of the rest of n (obtained by n / 10).

The main method calls the crossSum method with an argument of 915, and prints the result.

public class CrossSumRecursive {
    static int crossSum(int n) {
        return n <= 9 ? n : (n % 10) + crossSum(n / 10);
    }

    public static void main(String[] args) {
        System.out.print(crossSum(915));
    }
}

Output

15