Summing Even Digits in an Integer with C++ Loops

Welcome to another exciting session! In today's lesson, we face a unique coding challenge. We will traverse the digits of a number using a while loop under a specific condition. You will hone your skills in working with C++ loops and conditional statements, both of which are fundamental building blocks of programming. Shall we begin?

We begin by setting the basic structure for our function. In this step, we define a variable, digit_sum, that will accumulate the sum of the even digits.

Here's the initial framework of our function:

int solution(int n) {
    int digit_sum = 0;
    // The function expands from here.
}

The most effective tool for iterating through the digits of n is a while loop. The loop will run as long as n is greater than zero. Integrating this into our function produces:

int solution(int n) {
    int digit_sum = 0;
    while (n > 0) {
        // Further development of the function will proceed here.
    }
}

Within our loop, we'll extract the last digit of n using the modulo operation (n % 10). If this digit is even, we add it to the digit_sum.

After we process a digit, we'll truncate the last digit of n using integer division (n / 10). This step readies the while loop for the next digit.

This is what the code looks like now:

int solution(int n) {
    int digit_sum = 0;
    while (n > 0) {
        int digit = n % 10;
        if (digit % 2 == 0) {  // Check if the digit is even.
            digit_sum += digit;
        }
        n = n / 10;  // Remove the last digit.
    }
}