CODEWITHSUNDEEP
cpseudocode
Haldir87
Haldir87

Reputation: 375

Credit Card validation using Luhn's algorithm in C

#include <stdio.h>


int main(int argc, string argv[])
{
   Get input from the user;
   Store it in a long long int;

   // iterate through the number
   for each odd number in the input multiply by 2
   if the number is bigger than 9 then
   do module of it and add it to the first number;
   store it into metasum;

   for each even number in the input
   add it to the others;
   store the result into metasum2;

   // is it a valid card?
   if (metasum + metasum2)%10 == 0
   {
   card is valid;
   // check which kind of cc is
   if the input begins with a 3 then it's an AMEX;
   else if the input begins with 4 it is a VISA;
   else if the input begins with 5 it is a MASTERCARD;
   }
   else
   {
   card is not valid;
   }

}

It is actually a sort of pseudocode. I'm attending CS50x and I would like to know if my pseudocode implementation is good enough.

I have another problem though, when I try to implement the code in C I do not know how to iterate through a long long int, so I can't implement Luhn's algorithm.

How can I do this without using a string and then converting each char into an int?

Thanks for your response

Upvotes: 1

Views: 4440

Answers (2)

Kos
Kos

Reputation: 72241

Use modulo and integral division.

The first digit from the right (youngest digit) of a number is number % 10.
The second is (number/10) % 10. The third would be (number/100) % 10. Easy pattern - you can write it as a function. (Just avoid mixing in floats here.)

If you want to iterate over a number's digits, you might want to check how many digits a number has (see: base-10 logarithm) - the easiest way is to do repeatedly divide that number by 10 until it's 0.

Anyway, remember that credit card numbers are quite long! An unsigned 64-bit number can contain like 19 significant digits (max is 2**64-1), so it's enough for credit cards, but that won't work for bank accounts for instance. An approach using strings (some form of) will be more general.

Upvotes: 1

Yann Ramin
Yann Ramin

Reputation: 33177

To "iterate" a non-array, you need to separately keep track of your position (to count through all possible digits), using division and modulus to advance and extract the digit, for example:

long long number = 9999999999;
for (int digitPos = 0; digitPos < 16; digitPos++) {
  int thisDigit = number % 10;
  number = number / 10; 
}

As for applicability to your course, I can't vouch for that :)

Upvotes: 2

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