SPOJ ADDREV Wrong Answer - c

I'm trying to solve the Adding Reversed Numbers problem (ADDREV) at the Sphere Online Judge but my submission keeps coming up wrong answer.
I've tried int, unsigned int, long, and unsigned long for my variables and they all work equally well on my computer with some test data (also below) but they all fail the SPOJ.
I'm hoping someone might be able to shed some insight into why my program would be failing on their system. I've also left a message on their forum but there doesn't seem to be a lot of traffic.
Here's my code:
#include <stdio.h>
#define FIRST 1
#define SECOND 2
int main()
{
int c, k, x, y, state, place, total, reverse = 0;
do
{
c = getchar();
if (c < 48 || c > 57)
{
continue;
}
else
{
k = k * 10;
k = k + (c - 48);
}
} while (c != '\n');
state = FIRST;
place = 1;
do
{
c = getchar();
if (c == ' ')
{
state = SECOND;
place = 1;
continue;
}
else if (c == '\n')
{
total = x + y;
place = 1;
while ((total / place) >= 10)
{
place = place * 10;
}
while (place > 0)
{
reverse = reverse + ((total % 10) * place);
total = total / 10;
place = place / 10;
}
printf("%d\n", reverse);
state = FIRST;
place = 1;
reverse = 0;
x = 0;
y = 0;
k--;
continue;
}
if (state == FIRST && (c >= 48 && c <= 57))
{
x = x + ( (c - 48) * place );
place = place * 10;
}
else if (state == SECOND && (c >= 48 && c <= 57))
{
y = y + ((c - 48) * place );
place = place * 10;
}
} while (k > 0);
return 0;
}
And... here's the test data I'm using:
12
24 1
4358 754
305 794
2762 2563
435 4320
0 0
123 456
20 20
10000 10000
999999 999999
321 583
9999999 999999
And here's the results my program gives on my computer:
34
1998
1
4236
867
0
579
4
2
8999991
805
89999901
Any help would be appreciated :)

At this point:
int c, k, x, y, state, place, total, reverse = 0;
you create a variable k but give it no value. Shortly after:
k = k * 10;
you use this variable. At that point, your program exhibits undefined behaviour - anything might happen.

Related

CS50 Credit Assignment - Floating Point Exception

any ideas why this is generating a floating point exception? Please ignore the bad coding. This is very rough and I am just trying to experiment different things on this assignment. Also, I trying to do this assignment with only the operations learned in class (I am awere there are others that would make this easier)
Thanks!
#include <stdio.h>
#include <cs50.h>
int main(void)
{ long c = get_long("What is your credit card number?\n");
long i = 10;
long j = 100;
long n = 0;
long m = 0;
long x = 2;
do
{ long a = (c % i - (c % i)/10)/(i/10);
n = a + n;
i = i*100;
}
while (c/i >= 0.1 || c/i == 0);
do
{ long b = (c%j - (c%j)/10)/(j/10);
m = b + m;
j = j*100;
}
while (c/j >= 0.1 || c/j == 0);
long sum = n + 2*m;
if (i > j)
{
x = i;
}
else
{
x = j;
}
if (sum % 10 == 0)
{ if((x == 10000000000000000) && ((c % (x/100) - c % (x/1000))/(x/100000) == (34)))
{
printf("Amercan Express\n");
}
else if((x == 100000000000000000) && ((c % (x/100) - c % (x/1000))/(x/100000) == (51)))
{
printf("MasterCard\n");
}
else if((x == 10000000000000) && ((c % (x/100) - c % (x/1000))/(x/100000) == 4))
{
printf("Visa\n");
}
else if((x == 10000000000000000) && ((c % (x/100) - c % (x/1000))/(x/100000) == 4))
{
printf("Visa\n");
}
else
{
printf("Invlid\n");
}
}
else
{
printf("Invlid\n");
}
}'''
This is the part that is the problem:
while (c/i >= 0.1 || c/i == 0);
in your first do while loop.
Your data types are long and so you are saying that as long as c/i is greater than 0.1 or == 0 then keep going. Unfortunately this will go on forever as a long will always be >= 0.1 or <= 0 because it cant hold decimal places.If you play around with debug50 by using debug50 ./filename you can place a red dot and step through. What you will see is the first do while loop will keep running increasing the value of i every cycle. Eventually the value of i is bigger than the memory limit for a long and so the code creates unexpected results.If you step through you find that it eventually assigns 0 to i and so you suffer a divide by 0 error.
You can fix this by changing data types or altering that while loop.
I hope that helps. Id suggest stepping through that do while loop using debug50 in the cs50 ide to see whats going on.

Wanted to check if a credit card number is valid or not

I wanted to check if a credit card number is valid or not but when i run the code, every number I give as input, the output comes as invalid.
The example given below is what I should i do.
Example with David’s Visa: 4003600000000014.
For the sake of discussion, let’s first underline every other digit, starting with the number’s second-to-last digit:
4003600000000014
Okay, let’s multiply each of the underlined digits by 2:
1•2 + 0•2 + 0•2 + 0•2 + 0•2 + 6•2 + 0•2 + 4•2
That gives us:
2 + 0 + 0 + 0 + 0 + 12 + 0 + 8
Now let’s add those products’ digits (i.e., not the products themselves) together:
2 + 0 + 0 + 0 + 0 + 1 + 2 + 0 + 8 = 13
Now let’s add that sum (13) to the sum of the digits that weren’t multiplied by 2 (starting from the end):
13 + 4 + 0 + 0 + 0 + 0 + 0 + 3 + 0 = 20
, the last digit in that sum (20) is a 0, so David’s card is legit!
#include <stdio.h>
int main()
{
int no;
printf("Visa number: ");`
scanf("%d", &no);
int d_1, d_2, d_3, d_4, d_5, d_6, d_7, d_8, d_9, d_10, d_11, d_12, d_13, d_14, d_15;
d_15 = no%10;
d_14 = ((no%100)/10)*2;
d_13 = (no%1000)/100;
d_12 = ((no%10000)/1000)*2;
d_11 = (no%100000)/10000;
d_10 = ((no%1000000)/100000)*2;
d_9 = (no%10000000)/1000000;
d_8 = ((no%100000000)/10000000)*2;
d_7 = (no%1000000000)/100000000;
d_6 = ((no%10000000000)/1000000000)*2;
d_5 = (no%100000000000)/10000000000;
d_4 = ((no%1000000000000)/100000000000)*2;
d_3 = (no%10000000000000)/1000000000000;
d_2 = ((no%100000000000000)/10000000000000)*2;
d_1 = (no%1000000000000000)/100000000000000;
int d[7] = {d_2, d_4, d_6, d_8, d_10, d_12, d_14};
int n,add;
for (n=1; n<=7; n++)
if(d[n]>10)
{
d[n] = (d[n]%10);
d[(15-n)+1] = ((d[n]%100)/10);
int sum=0;
for (int i=0; i<7; i++)
sum += d[i];
}
else
{
add = d_14 + d_12 + d_10 + d_8 + d_6 + d_4 + d_2;
}
int sum = add + d_15 + d_13 + d_11 + d_9 + d_7 + d_5 + d_3 + d_1;
if ((sum % 10) == 0)
{
printf("%s\n", "The card is valid");
}
else
{
printf("%s\n", "The card is invalid");
}
}
every number I give as input, the output comes as invalid.
Too big
OP's int is likely 32-bit.
Reading text input that would attempt to form an int outside the int range is undefined behavior. Rest of code is irrelevant.
int no;
scanf("%d", &no); // attempt to read "4003600000000014" leads to UB.
Consider reading user input into a string first and then process the characters. #Weather Vane
char buf[100];
if (fgets(buf, sizeof buf, stdin)) {
int i;
sum[2] = { 0, 0 }; // sums of even indexed digits and odd indexed digits.
// Note: only 1 sum really needed, but using 2 sums to mimic OP's approach
for (i = 0; isdigit((unsigned char) buf[i]); i++) {
digit = buf[i] - '0';
if (i%2 == 0) {
digit *= 2;
if (digit >= 10) {
digit = (digit/10 + digit%10);
}
}
sum[i%2] += digit;
}
// reject bad input: too long or missing expected end
if (i > 16 || (buf[i] != '\n' && buf[i] != '\0')) {
puts("Bad input");
} else {
// pseudo code to not give everything away.
// do math on sum[0], sum[1]
// if as expected --> success
}
}
#include <stdio.h>
#include <cs50.h>
long credit;
int getting_the_final_total_number (void);
void checking_which_kind (void);
int main(void)
{
credit = get_long("Number: ");
int i = 0;
long number_count = credit;
//finding how many numbers are there.
while(number_count > 0)
{
number_count /= 10;
i++;
}
//we use and because (using or make once true, the code block will work and always telling INVALID)
if(i != 13 && i != 15 && i != 16)
{
printf("INVALID\n");
return 0;
}
int total = getting_the_final_total_number(); //adding sum_1 and sum_2
if(total % 10 != 0)
{
printf("INVALID\n");
return 0;
}
checking_which_kind();
}
//assigning the credit to another variable for the loop
int getting_the_final_total_number (void)
{
long credit_one = credit;
int mod_1;
int mod_2;
int sum_1 = 0;
int m;
int d;
int sum_2 = 0;
do
{
//cutting the number into two pieces with all the last numbers and all the second-last-numbers
//cutting the last numbers.
mod_1 = credit_one % 10;
credit_one = credit_one / 10;
sum_1 += mod_1;
//cutting the second-last-numbers.
mod_2 = credit_one % 10;
credit_one = credit_one / 10;
//doubling the mod_2 (the second-last-numbers)
mod_2 = mod_2 * 2;
//making them into one number (if there is 16 or 18 in the product then make them 1 and 6 or 1 and 8. After that add them all together ).
m = mod_2 % 10; //This is for only one standing numer like 1 or 2 or 9 etc (but no 12 or 14 or 16)
d = mod_2 / 10; //This is for ten's digit to make sure to become ONE standing digit
sum_2 = sum_2 + m + d;
}
while(credit_one > 0);
return sum_1 + sum_2;
}
//checking the first two number of credit
void checking_which_kind (void)
{
long cc = credit;
do
{
cc = cc / 10;
}
while(cc > 100);
if(cc / 10 == 5 && (cc % 10 > 0 || cc % 10 < 6))
{
printf("MASTERCARD\n");
}
else if(cc / 10 == 3 && (cc % 10 == 4 || cc % 10 == 7))
{
printf("AMERICAN EXPRESS\n");
}
else if(cc / 10 == 4 && cc % 10 >= 0)
{
printf("VISA\n");
}
else
{
printf("ERROR");
}
}

c - find palindrome

I'm a newcomer to programming, and I chose C as my first language(been learning it for a month or so).
I've been trying to solve this palindrome question for hours and still couldn't come up with a satisfying solution.
The question is here (from SPOJ), and here's my code:
#include <stdio.h>
#include <string.h>
void plus_one(char *number);
int main(void)
{
char number[1000001];
int i, j, m, k, indicator;
int a;
scanf("%d", &j);
for (i = 0; i < j; i++) {
scanf("%s", number);
k = 1;
while (k != 0) {
plus_one(number);
a = strlen(number);
indicator = 1;
for (m = 0; m < a / 2; m++) {
if (number[m] != number[a - m - 1]) {
indicator = 0;
break;
}
}
if (indicator != 0) {
printf("%s\n", number);
k = 0;
}
}
}
return 0;
}
void plus_one(char *number)
{
int a = strlen(number);
int i;
number[a - 1]++;
for (i = a; i >= 0; i--){
if (number[i - 1] == ':') {
number[i - 1] = '0';
number[i - 2]++;
}
else
break;
}
if (number[0] == '0') {
number[0] = '1';
strcat(number, "0");
}
return;
}
My idea was to examine every number greater than the input until a palindrome is found, and it worked well on my computer. But SPOJ responded "time limit exceeded", so I guess I need to find the next palindrome possible myself instead of using brute force. Can someone please give me a hint about how I can make this go faster? Thanks!
Since you're asking for a hint and not for C code (which I'm notoriously bad at), here's what I would do:
Determine if the number k has an even or odd number of digits, store that in a boolean called odd.
Take the first half of the number k, including the middle digit if odd is true, and store it in a variable called half.
808 -> 80
2133 -> 21
Mirror the half variable, taking care to not duplicate the middle digit if odd is true, and store it in a variable called mirror.
80 -> 808
21 -> 2112
Check if mirror > k
If true: you found your result
If false: increment half and start over from step 3.
(After maximum one increment you're guaranteed to have found your result.)
80 -> 81 -> 818
21 -> 22 -> 2222
Here's a JavaScript implementation for your reference:
const palin = (k) => {
const mirror = (half, odd) => half + Array.from(half).reverse().join('').substring(odd);
const s = k.toString();
const odd = s.length % 2;
const half = s.substring(0, Math.floor(s.length / 2) + odd);
let mirrored = mirror(half, odd);
if (+mirrored <= k) {
mirrored = mirror((+half + 1).toString(), odd);
}
return mirrored;
}
console.log(palin(5));
console.log(palin(808));
console.log(palin(2133));
Welcome to the site. What you have posted is commendable for someone who has only been using C for a month! Anyway ... I think your suspicion is correct. Using 'brute force' to find the next palindrome is probably the not to way go.
This question is as much about algorithm design as about C. Nonetheless, how you handle char[] representations of integers in C is interesting and relevant. FWIW, my attempt is pasted below.
It accepts a char[] representation of the number (n) and the number of digits (k) as arguments, and returns 1 on success or 0 on failure (another pass needed).
static int next_palindrome(char *n, size_t k) {
unsigned i = 0, carry = 0;
char tmp = 0;
int finished = 1;
for (i = 0; i < k; i++) {
if (carry) {
finished = 0;
*(n + k - i - 1) = *(n + k - i - 1) + 1;
if (*(n + k - i - 1) == 10) {
*(n + k - i - 1) = 0;
carry = 1;
} else
carry = 0;
continue;
}
if (i >= k / 2) continue;
if (*(n + k - i - 1) == *(n + i)) continue;
tmp = *(n + k - i - 1);
*(n + k - i - 1) = *(n + i);
if (tmp > *(n + i)) {
carry = 1;
}
}
return finished;
}
I have only tested it on numbers with < 64 digits so far, but have no reason to believe it will fail for larger numbers of digits.
Sample usage: http://codepad.org/3yyI9wEl

Write a program in C to apply Luhn's algorithm for credit card validation

I need my program to prompt a user for an input and re-prompt in case the input doesn't follow a credit card format (ex: negative numbers or letters, etc.) and then apply the algorithm to see if the number is a valid credit card number and if yes, whether it's Visa, MasterCard or AmEx.
I know that this question has been answered with different codes on this website, I swear I read everything that I could possibly find (on this site and elsewhere on the net) but I'm having a really hard time understanding the C syntax and I wanted to try to come up with something myself instead of copying bits of codes I don't understand from other answers. If someone can help me out and look at what I've done so far and tell me what I'm doing wrong I would be really grateful. Also, any tips that could help me make sense of the C syntax logic better would be extremely appreciated.
My program is compiling but when I run it it's acting up in a very weird way: when I enter an input sometimes it will say that it is invalid (even if it's a valid number) and sometimes it will just not return anything after I press enter and won't stop running no matter how many times I press the return key.
Here is my code so far:
#include <stdio.h>
#include <cs50.h>
#include <math.h>
int main(void)
{
printf("Please give me your credit card number:\n") ;
long long card_num ;
do
{
card_num = GetLongLong() ;
}
while (card_num < 1 || card_num > 9999999999999999) ;
// Make a copy of the card number to be used and modified throughout the process.
long long temp_num = card_num ;
int digit = 0 ;
int count = 0 ;
int sum_a = 0 ;
int sum_b = 0 ;
// Isolate every digit from the credit card number using a loop and the variable 'digit'.
// Keep track of the amount and position of each digit using variable 'count'.
while (card_num >= 0)
{
digit = card_num % 10 ;
count++ ;
temp_num = (card_num - digit) / 10 ;
break ;
// Apply Luhn's algorithm using two different 'for' loops depending on the position of each digit.
for (count = 0 ; count % 2 == 0 ; count++)
{
sum_a = sum_a + ((card_num % 10) * 2) ;
while ((card_num % 10) * 2 >= 10)
{
sum_a = (sum_a % 10) + 1 ;
}
}
for (count = 0 ; count % 2 != 0 ; count++)
{
sum_b = sum_b + digit ;
}
return sum_a ;
return sum_b ;
return count ;
}
// Checking the validity of the number according to Luhn's algorithm
int total_sum = sum_a + sum_b ;
if (total_sum % 10 != 0)
{
printf("This is an invalid number.\n") ;
}
// If the number entered doesn't have the right amount of digits according
// to variable 'count', declare the number as invalid.
if (count != 13 || count != 15 || count != 16)
{
printf("This is an invalid number.\n") ;
}
// Reset value of variable 'temp_num' and apply calculations that will isolate the first two digits.
// Store the results in a variable 'company_id'.
temp_num = card_num ;
int company_id ;
while (temp_num > 100)
{
temp_num = card_num - (card_num % 10) ;
company_id = temp_num / 10 ;
}
return company_id ;
// Print the type of credit card depending on the company ID and amount of digits.
if (company_id > 50 && company_id < 56 && count == 16)
{
printf("MASTERCARD\n") ;
}
else if ((company_id == 4) && (count == 13 || count == 16))
{
printf("VISA\n") ;
}
else if ((company_id == 34 || company_id == 37) && (count == 15))
{
printf("AMEX\n") ;
}
else
{
printf("This is an invalid number.\n") ;
}
return 0 ;
}
Your answer is a out of order pastiche with sections that don't follow logically from the previous.
Specific issues:
This logic:
if (count != 13 || count != 15 || count != 16)
invalidates every card, the ors (||) should be ands (&&) for this to work.
This loop makes no sense:
while (card_num >= 0)
{
digit = card_num % 10 ;
count++ ;
temp_num = (card_num - digit) / 10 ;
break ;
...
}
The break is unconditional so it exits the loop and ignores the next twenty lines.
You appear to have spliced in subroutines from elsewhere as you call return five times, only the last of which is valid:
return sum_a ;
return sum_b ;
return count ;
return company_id ;
return 0 ;
In several places you use card_num when you should be using temp_num.
You fail to exit the program once you know the card is invalid -- instead you just keep on testing. You fail to acknowledge when a card is valid.
You count the number of digits in the card number but wait until after you run other checks before testing if that digit count was valid or not.
What follows is my rework of your code to address the above and some style issues:
#include <stdio.h>
#include <cs50.h>
#include <math.h>
int main(void)
{
printf("Please give me your credit card number: ") ;
long long card_num = 0LL;
while (card_num < 1LL || card_num > 9999999999999999LL)
{
card_num = GetLongLong();
}
// Make a copy of the card number to be used and modified throughout the process.
long long temp_num = card_num;
// Isolate every digit from the credit card number using a loop and the variable 'digit'.
// Keep track of the amount and position of each digit using variable 'count'.
int count = 0;
while (temp_num > 0LL)
{
temp_num = temp_num / 10LL;
count++;
}
// If the number entered doesn't have the right amount of digits according
// to variable 'count', declare the number as invalid.
if (count != 13 && count != 15 && count != 16)
{
printf("This is an invalid number (# of digits).\n");
return 1;
}
// Reset value of variable 'temp_num' and apply calculations that will isolate the first two digits.
// Store the results in a variable 'company_id'.
temp_num = card_num;
while (temp_num > 100LL)
{
temp_num = temp_num / 10LL;
}
int company_id = temp_num;
// Print the type of credit card depending on the company ID and amount of digits.
if (company_id > 50 && company_id < 56 && count == 16)
{
printf("MASTERCARD\n") ;
}
else if ((company_id == 34 || company_id == 37) && (count == 15))
{
printf("AMEX\n") ;
}
else if ((company_id / 10 == 4) && (count == 13 || count == 16 || count == 19))
{
printf("VISA\n") ;
}
else
{
printf("This card was issued by an unknown company.\n");
}
// Apply Luhn's algorithm.
int sum = 0;
temp_num = card_num;
for (int i = 1; i <= count; i++)
{
int digit = temp_num % 10LL;
if (i % 2 == 0)
{
digit *= 2;
if (digit > 9)
{
digit -= 9;
}
}
sum += digit;
temp_num /= 10LL;
}
// Checking the validity of the number according to Luhn's algorithm
if (sum % 10 != 0)
{
printf("This is an invalid number (Luhn's algorithm).\n");
return 1;
}
printf("This is a valid number.\n");
return 0;
}
This is not a finished program -- there's error checking and other details needed. Rather than summing the digits when a doubled card number is greater than 9, I used the simpler approach of subtracting 9.
Here my solution, this method received the var long credit card number, I hope to be helpful.
void check_card(long n)
{
long temp_n = n;
int count = 2;
while(temp_n > 100)
{
temp_n = temp_n / 10;
count ++;
}
long temp_n2 = n;
int sum = 0;
for (int i = 1; i <= count; i++)
{
int digit = temp_n2 % 10;
if (i%2 == 0)
{
if (digit * 2 > 9)
{
sum += (digit * 2) - 9;
}
else
{
sum += digit * 2;
}
}
else
{
sum += digit;
}
temp_n2 /= 10;
}
bool flag = (sum % 10 == 0) ? true : false;
if (count == 15 && (temp_n == 34 || temp_n == 37) && flag)
{
printf("AMEX\n");
}
else if(count == 16 && (temp_n > 50 && temp_n < 56) && flag)
{
printf("MASTERCARD\n");
}
else if((count == 13 || count == 16) && (temp_n / 10 ==4) && flag)
{
printf("VISA\n");
}
else
{
printf("INVALID\n");
}

'if' statement in C not executing even though conditions are met

I'm a first time programmer trying to complete a simple command line program as part of the first assignment for an online course I am taking, but I seem to have hit a roadblock that I can't figure out with GDB or my own research.
After hours of rewrites, and hours of debugging, I finally got the code below to compile. The program is supposed to take a credit card number as an input, and then check whether it's valid per the specifications of the assignment. I used a test number from here: PayPal Test Credit Cards
The odd thing is, when I enter an AMEX card number, it correctly produces the text "AMEX", but when I try a Visa or a Master Card, it prints "INVALID".
In GDB I broke at the Verify function and it seems to incorrectly skip these two if/else if statements without proceeding to the Checksum function even though conditions appear to be met.
if (firstDigit == 4 && totalDigits == (13 | 16) && Checksum(cardNumber, totalDigits) == 0) // checks for a valid Visa.
...
else if (firstDigit == 5 && secondDigit == (1 | 2 | 3 | 4 | 5) && totalDigits == 16 && Checksum(cardNumber, totalDigits) == 0) // checks for a valid Mastercard.
...
The AMEX line of code that correctly executes is:
else if (firstDigit == 3 && secondDigit == (4 | 7) && totalDigits == 15 && Checksum(cardNumber, totalDigits) == 0) // checks for a valid American Express.
The arguments for all three lines seem to be formatted exactly the same. That is far as I could get in GDB though. I would print totalDigits, firstDigit, and secondDigit in GDB right before stepping through the above two non-executing lines and everything looked correct. So I'm stumped, why is the AMEX line executing, but not the others?
Thanks in advance everyone. This is the first program after hello.c that I've tried to write, so I am open to absolutely any criticism or suggestions if it looks like I'm doing something weird/wrong.
Full code:
checker.c
#include <stdio.h>
#include <stdlib.h>
int MAX = 16;
int* DigitSort(unsigned long long x, int* array);
int Verify(int* array);
int main (void)
{
int* output = malloc (sizeof(int) * (MAX + 2)); // creates a blank array for the individual digits of the card number.
unsigned long long userInput = 0;
do
{
printf("Please enter a credit card number:\n");
scanf("%lld", &userInput);
}
while (userInput <= 0); // checks to make sure the user entered a number.
switch(Verify(DigitSort(userInput, output))) // sorts the user's input into individual digits and verifies the card type and validity.
{
case 1 :
printf("VISA\n");
break;
case 2 :
printf("MASTERCARD\n");
break;
case 3 :
printf("AMEX\n");
break;
case 0 :
printf("INVALID\n");
break;
default :
printf("INVALID\n");
}
free(output);
return 0;
}
int Verify(int* array) // verifies whether or not a card number is valid. Must pass the function a sorted array of individual digits.
{
int* cardNumber = array;
int firstDigit = cardNumber[0];
int secondDigit = cardNumber[1];
int totalDigits = 0;
int Checksum(int* cardNumber, int totalDigits);
int i = 0;
while (firstDigit >= 1 && cardNumber[i] >= 0) // this step counts the number of digits in the array.
{
totalDigits = totalDigits + 1;
i++;
}
if (firstDigit == 4 && totalDigits == (13 | 16) && Checksum(cardNumber, totalDigits) == 0) // checks for a valid Visa.
{
return 1;
}
else if (firstDigit == 5 && secondDigit == (1 | 2 | 3 | 4 | 5) && totalDigits == 16 && Checksum(cardNumber, totalDigits) == 0) // checks for a valid Mastercard.
{
return 2;
}
else if (firstDigit == 3 && secondDigit == (4 | 7) && totalDigits == 15 && Checksum(cardNumber, totalDigits) == 0) // checks for a valid American Express.
{
return 3;
}
else // if the card number doesn't match any of the above conditions or fails the checksum, an 'I' for Invalid is returned.
{
return 0;
}
}
int* DigitSort(unsigned long long x, int* array) // takes a long long as input and sorts it into individual digits
{
int* arrayReversed = malloc (sizeof(int) * (MAX + 2)); // creates a new array to hold the reversed order of digits.
int i = 0;
arrayReversed[0] = 0;
if (i < (MAX - 1) && x >= 10)
{
do
{
arrayReversed[i] = x % 10;
x = x / 10;
i++;
}
while (i < (MAX -1) && x >= 10);
}
if (i < MAX && x >= 1 && x <= 9)
{
arrayReversed[i] = (int) x;
x = (x - x);
}
if (x == 0)
{
int j = 0;
do
{
array[j] = arrayReversed[i]; // sorts the digits from the reversed array and places them into the sorted array.
j++;
i--;
}
while (j < MAX && i >= 0);
array[j] = -1;
}
free(arrayReversed);
return array;
}
int Checksum(int* cardNumber, int totalDigits)
{
int sum1 = 0;
int sum2 = 0;
int i = (totalDigits - 2);
int j = (totalDigits - 1);
while (i >= 0)
{
sum1 = ((cardNumber[i] * 2)%10) + ((cardNumber[i] * 2)/10) + sum1;
i -= 2;
}
while (j >= 0)
{
sum2 = (cardNumber[j] + sum2);
j -= 2;
}
if (((sum1 + sum2) % 10) == 0)
{
return 0;
}
else
{
return 1;
}
}
Your first problem is here:
if (firstDigit == 4 && totalDigits == (13 | 16) && ...
You need to write:
if (firstDigit == 4 && (totalDigits == 13 || totalDigits == 16) && ...
Your first check is looking for 0x1D == 29 as the number of digits (because, as paisanco points out in a comment, the | operator is the bitwise OR operator), and no credit card needs 29 digits (yet, and not for a long time to come). Note the extra parentheses for clarity and accuracy. Don't mess around risking removing them — the code won't work properly again. And in general, be explicit if your condition has both && and || operators and use parentheses to group terms explicitly.
You have similar problems elsewhere. As it happens, (4 | 7) is the same value as 7, so the condition works when the second digit is 7 (but not when it is 4). But it doesn't mean what you intended it to mean.
Computer languages don't work the same as human languages. Get used to writing out the condition somewhat more verbosely. Some other languages provide shorthands for these conditions; C is not such a language.

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