I'm trying to implement Luhn's algorithm. For every other digit you have to take it, multiply it by 2, and add it to a running total. If the ith digit * 2 is greater than 10, you split it up, and just add it to the running total. I create a void function that does this and pass in 123456789 as a test value.

#include <stdio.h>
// #include <cs50.h> // 

void LuhnsAlgorithm(long); // needs to be changed to str return type when submitted
int main()
{


    int num = 123456789;
    LuhnsAlgorithm(num);
}

void LuhnsAlgorithm(long cc) // return type will be str when submitting
{
    int sum = 0;
    // case 1

    // case 2 (every other odd digit, multiplied by 2, and then added)
    long case2 = cc;
    while (case2 > 0)
    {
        if ((case2 / 10 % 10) * 2 >= 10) // checks if the every-other-digit's product when multiplied by 2 is bigger than 10 (aka, has 2 digits)
        {
            // creating this stupid temp variable 
            int digitBreakUp = case2 / 10 % 10 * 2;
            sum  = (digitBreakUp / 10)   (digitBreakUp % 10);
        }
        else // if the product is just 1 digit then add it to the sum
        {
            sum  = (case2 / 10 % 10) * 2;
        }
        case2 = case2 / 10;
    }

    printf("The sum of every last other digit in 123456789 is %i\n", sum);
}

The expected sum should be 22 (8 --> 1 6, 6 --> 1 2, 4 --> 8, 2 --> 4). But, I get 36. What's wrong? How do I get to 0 / get it to 'stop iterating' until it reaches the beginning of the number?

Thanks

CodePudding user response：

Since your test is for every other digit, the line of code:

case2 = case2 / 10;

Should be:

case2 = case2 / 100;

If you do that, your summation of every other digit does come out to 22.

Give that a try.