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Validate cc number field

neither lambic number doctor and welcome to my channel thank you for joining me for another mind workout I hope that you're staying indoors and that you keep safe in and keep learning today we will go over another cool little problem we will be using the C programming language and this one will be a bit more complex but we will break it down really nice and I hope it won't take us too long to solve of course whenever you feel that you can continue on your own please pause the video and do so so yeah let's dig right in what we need to do today is validate a credit card number how do we do that we run a checksum on it what does that mean we basically run some calculations on the digits of the credit card number and if a given condition checks out then we know that it's valid okay this also employs what's known as Lewin's algorithm you can look that up online and have a read it's pretty interesting so we would also have a few other basic conditions but we can leave these for the end we will have three examples American Express MasterCard and Visa American Express starts with 34 or 37 and it's 15 digits long MasterCard starts with anything between 51 and 55 and it's 16 digits long and visa starts with a four and it's it uses 13 or 16 digit numbers okay there are other cases of course that we will just focus on these ones for our problem and for example where would you need this to validate to evaluate a credit card number for example when you're having when you're making an online purchase so this would be when you have a form and you input your credit card number and if it's invalid it just says oh this number isn't good okay it's like a first line of checks to see if if if the number you provide this is valid okay so how does the checksum work basically there are two cases so first case and second case so for the first case we start with the last number and we add every other number okay so we would have four plus and then every other number so we have a bunch of zeros we're going to skip these so for next next next next next next and this is three plus so we were down here at three and next zero okay this equals to 7 okay so for the second case it's basically the same thing the only difference is that we start from the second to last number we had every other number and another small difference is that we multiplied each digit by two okay so we start with the 1 1 x 2 plus we're gonna skip all these zeros again so next next next next next we were at the 6 1 3 6 so 6 times 2 plus so we were down here at the six minutes is zero plus zero plus zero four okay okay so we have two plus 12 plus zero plus eight okay so one thing here whenever we have two digits numbers so if for example if our digit will be 5 or greater than 5 between 5 and 9 basically if we multiply it by 2 we would get a two digit number okay so whenever that that happens we add those two digits as well so here if we have 12 we will have 3 1 plus 2 okay so we actually have two plus three plus zero plus 8 equals 2 what do we get 13 okay good okay actually we can do this like 1 plus 2 so it'd be clear ok and then we add just add 7 and 13 so we get like a final son 7 plus 13 equals 20 wait and the condition says that this number should end with zero or in other words 20 modulo 10 equals to zero well what's modulo basically it's for example if you do modulo five modulo two actually the module as well I might know two equals one so what does that mean it's the remainder of the division okay so if we divide 5 by 2 2 goes into 5 twice so 2 times 2 that's 4 and 5 minus 4 the remainder is 1 so the quotient is 2 and the remainder is 1 okay so if our number ends with 0 modulo magical 10 would always be 0 ok so that's our main check here so how would we go about doing this so if we have the number here as an example how do we get the last number the last digit so again if we're not if you're not familiar with modulo you can look that up online it's it's pretty interesting as well how do we get the last digit we can use modulo so this number go to the next row right so we'd have it there 10 gives us 4 okay so if we divide this number by 10 we will have a remainder of 4 so that's how we get the last digit how do we move two digits over so we need to we need to add every other number okay how do we move two digits over basically we can use a division so this divided by 100 equals 2 remember then we just scrap the last two digits ok we're going to be using integers so we won't get like a dot decimal result we will just get this just this part here the integer number basically that's how we would scrap the last two digits so these are the two main methods were gonna be using so this is how we get the land digit we use modulo 10 and then to move to the jet over we divide the number by 100 okay so then for the first case we would need to one iterate to the CC number and then we would need to use modulo to get the last digit nice you would have like a some counter for example add last digit sum and then divide the number by 100 so this would allow us to move two digit over and then repeat okay just to test the first case the second case what do we do we can actually get this stuff here click the CC number actually first we will need to so if we want to start from the second last number we would actually divide the number by ten off the bat okay so divide the CC number by ten so we would start with the second last number okay and then it would be pretty much the same and just change the numbers right iterate through the CC number use module to get the last digit at last digit to sum divided the city number by ten and repeat the difference here is that we will we will multiply it by two just wanted to get the land digit and the last digit x - 2 and here we would have two smaller cases so here we will basically check if it's if it's made out of two digits how do we do that how do we get the last digit we use modulo okay so to get to here we use modulo 10 and we'll get two to get the first digit one we will divide it by ten okay because twelve marginal so for example Oh module 10 it's gonna be - okay so that's our last digit and 4/10 using integers we will get one point something so we will just get the one okay and then we can just add these up one plus two okay so use modulo four to use the patient for the first digit it will always be the second digit but I'm just calling it last and on four point one okay and four point three some leaves up that's it that's pretty much it that's the idea here then we divide the city number by 100 to move two digits over and then repeat okay so this is what we need to do to get started then we will do some other things as well of course but this is what we need so far so let's start implementing some of this so let's start here include standard and main class here so having these long numbers we would actually use guessed it the long data type okay wrong credit card number okay and now we will request it from the user do so here we have these input CC number so we will want to pump this up okay now we'll use scanf here we will have the long and we would send it to credit card number okay and the condition here would be while so what do we need to do we need to make sure that credit card number is greater than then zero or not equal to zero okay so while credit card number is less than or equal to zero so while credit card number is less than or equal to zero actually we will keep prompting for the credit card number okay so we have that down here so I propose we use like a like a working credit card number okay so we would always have this one here untouched okay because we will keep modifying it so it's it's it's it's better to have like a working one okay so the long as well working working CC equals with and we will initialize it with the credit card number okay so we have this down so let's start the first case while working CC so we need to make sure we don't go past zero so well this while working CC is greater than zero we will we will create int last digit equals to working CC modulo ten okay so this is what we said we're going to use to get the last digit we actually the sum here and some here we will add this up some plus last digit okie-doke and here we will basically just need to modify working number by dividing it with 100 by 100 okay so this should be pretty straightforward so if we have one two three four for example we will get the first digit the last digit that's four four so then we divided by 100 then we will get 12 and we get to 2 and then if we divide 12 by 100 we will get 0 so that's where we stop so we get 6 okay so I hope that makes sense so for the second case actually we we will need to replenish working CC so second case so working CC equals with credit card number and we will divide it by 10 okay so we will start with the second last number digit so here while working CC is again greater than 0 we will use this stuff again basically last digit then we actually have another one and let's say times and here we actually need to need to say something like sum equals with the Sun plus and then we want to add the last two digits right for example if it's so if we have for example 12 if we get if we do modulo 10 we get 2 if we divided by 10 we get 1 and we add those up if for example we have a single digit like 5 modulo 10 would simply return 5 because then you cannot do 10 doesn't go into doesn't go into 5 so whenever that's the case you just get 5 and dividing it by 10 would be zero and zero point something and then you just get 5 plus 0 so it's the original number okay so here we will have two cases so times 2 modulo 10 or the last digit we could do this in any order of course and then we can reuse this and say oh I don't have the first parenthesis and just divide it by 10 okay so that's how that's how we that's how we get the sum of the two digits if that's the case if it's one digit that's ok as well and then we divide working CC by 100 by 100 again to move two digits over yeah so I hope that makes sense and now basically we have we have the sum what do we need to do we need to so we will check for a condition as well but one one thing is we need to take care of these conditions as well so how do we get the first and second the first or the first two digits of the number so one way would be to for example if we have 1 2 3 4 1234 how do we get the first digit we can divide it by 1000 okay so that would be one point something so that's the first digit how do we get so divided by 1 and divided by 100 equals to 12 okay so if we divided by so I I propose we get the count the length of the number so for example if it's 4 we will just iterate through that and basically use 10 and multiply it by 10 by a set number of times okay and that's how we get we get the divisor here okay so if we need the first number for this case we would basically for example if we have an issue then we'll multiply it by 10 twice and we will get 1,000 okay how do we get 10 twice if the length is for some time equals 4 vizor people's 10 how do we multiply it by by by 10 2 times basically loop we use I that's less than length minus 2 okay so that will be 4 we will have to have 0 and 1 so 2 times that that's how we get one 1000 okay 10 times 10 times 10 okay this will the divisor being initialized with 10 so I hope that makes sense and for the 100 basically we can just use whatever we get in the final divisor basically just divided by 10 okay so we would we will instead of 1 tell you it would have one 100 okay so we can set that up here but how will we count how would we count the length of the number the number of digits okay so for that when I propose is we loop through it and we divided by 10 so we have 1 2 3 4 again we divided by 10 we get 1 2 3 and count counter it goes 1 goes up by 1 okay so we have one digit then we divide it by 10 again we have 12 and it's two digits then we divide it by 10 and it's one and it's three digits and then we divide it by 10 and then it's zero and it's four digits okay so the condition will be while the number is different than zero that's how we get that's how we get the length as well so let's implement that here so while we would actually need working Cece again replenished so to say so we will send credit card number to it again so while working CC is not equal to 0 we will let's create a variable here and count with 0 so here we will get sorry about that so working CC divided by 10 so working 60 equals to working CC divided by 10 and our count would be incremented okay so this is how I would get the count we can actually even test that out really quick so printf okeydoke so that's terminal so here I use the GCC compiler and we have credit [Applause] okay I'm gonna terminate this something is wrong here I'm gonna try to figure out what so while working she sees different than the zero so we divide this by 10 and then we increment count nothing count maybe I'm just gonna comment this out for for a second sure if that's the culprit something's not right here this should be good oh here we go okay so okay what I've been overlapping with something here commenters UCC yeah yeah we modified here so I neither replenished I hope we'll just not overlapping here or something well let's just go on I hope it's going to be fine so now we have the count so let's do the first digit equals CC actually we can use the credit card number this we won't be modifying it and times oh we need visor here for so we're gonna do the the loop that we talked about so for and equal zero in I sorry about that can't I put 0 is less than M minus 2 I plus plus okay so this might be a wrong number so we're gonna have another longer divisor equals to 0 it will equal 10 sorry we start with 10 just as we discussed so here advisor okay so for example this is for the first it is right and this is here and first second digit first Ashley first two digits that's that better okay equals two credit card number divided by the divisor divided by ten okay good so we have the first digit to the first two digits okay so what we need to do now is basically conduct the final checks okay what this up so here we get our number here we we create some of the variables that we need this is the iteration for the first case this is the second iteration for the second case this is for what do we get here the count the length of the number here we get the divisor that we that we will use here so we can get the first digit in the first two digits and here we can start on our final checks so if some marginal ten equals zero this is the main condition so modulo 10 well what's wrong here oh there's some issue must be a modifiable L value not sure what that is actually let's have a look oh okay we didn't initialize it here or that - in our checks so what's wrong here especially must be a modifiable overvalue not sure what this is I think we're missing something what we actually initialize it with zero change here which is here my using the increased amount of equals yeah that was it so this is what how we verify the condition I'm sorry about that okay so if this is the case and we need to go to the other cases so first digit equals to 22 equals I'm sorry about that okay so let's have the easiest one the visa so this is fourth digit equals to four and so here we have these I use just 13 or 16 digit numbers so count to 13 for count equals to 16 all right 13 and 16 yep okay but this will have to be one big condition so first so this will evaluate first I mean separately and the one that's value from this one need so this is also either 13 or 16 and the one from here we need to be valid alongside this one because here we have end okay so if the first digit is 4 and for example counted 13 working on 216 then we we need a string here for the result I think that's chart result set enough how long is MasterCard a are the MasterCard okay so here let's see p st RC r py change the string result it would pass the value V so either okay so here we would we would replace the result string with visa okay well if so the other ones actually need the first two digits so let's go with American Express 24 or 37 and if that's the first digit it was tubes 34 to 37 yes we need to be one big condition and cannot equal to so how many digits for American Express we have 15 okay here this will be a max okay okay for MasterCard if oh here we have first two digits I'm sorry about that okay first two digits greater than 50 anything and we start with the one year and less than 56 okay 600 so I hope you're following this along and that's 16 digits so the count needs to be 16 digits okay okie doke now we copy this here and we just need this line this will be good it needs to be so it's not or then it should be both be valid okay so first two digits must be greater than 50 or less than 56 and let it less than 56 okay yep okie doke so else we don't need the parentheses just in the curly brackets this will be invalid okey doke and for the big check here else and again we get invalid right if some modulo 10 isn't equal isn't equal to zero okay so I think this should be it so again here we have the the scan the variables the first case the second case then we get the count number and we get divisor okay then we get the first and the first the first digit in the first two digits here we have final checks okay so I hope this is right I'm probably going to see errors if you missed anything and I hope this makes sense so basically just that's what we did we got the last digit some added that then we got then we moved to digits over and so on for the second case we just got the we started from the second well digit moving two digits over multiplying by two checking that if the number that if the resulting number is it's two digits we sum those up as well and then we just got other things like the length of the number the count the digit count and then here and then we simply quit the divisor that we that helps us get the first or the first two digits of the number and that helps us validate some of the conditions for the first or the first the first digit or the first two digits then we use the count validate the length of the number and of course we have the big condition here and then we output the result that's pretty much it so let's have a go here and type printf and here we have a string and we use result okay so let's see how this goes I hope it's okay no we need GCC oh this there's an error here this up a bit writing Levin biased in for your size don't over for this oh okay so I think that was the number we need here see is about result when we send it MasterCard okay do this okay so we're missing something else okay so we all we need to string that each library as well okay so I'm gonna go ahead and add that up scream dot H so we can use the string copy okeydoke okay so it seems fine now credit so let's get our initial number so here okay it says Pisa we have a couple more here a few more here from PayPal so let's get American Express for example it makes nice let's get American Express for example let's get one that it's not in one of our keys it starts with six and you can see in salad okay that's good we're not using this case with two we're using for example this one five five MasterCard to get a visa let's get a 13 one this seems shorter so it might be the 13 version certain digits long version ease up okay invalid yeah so I think this is pretty much it so know what this up yep well this this is it so I hope this was fun for you as it was for me if you found this interesting in any way please give the video a like and subscribe and I'm looking forward to your comments in the comment section and yeah I'll see you world