MoonOwl22 MoonOwl22 - 3 months ago 16
C Question

How is there the arithmetic error of a division by zero in this function?

I have written an implementation of Luhn's algorithm. When I test the program with numbers that are invalid by the number of digits I get no program error. Instead, the program executes the

printf()
call in the main function. However, whenever I test the program using valid card numbers as per the algorithm I get a division by 0 error. Debugging with DBG gives me an arithmetic error while MSVC gives me a floating point exception, which narrow down to division by 0. This is the code:

#include <stdio.h>
#include <math.h>
#include "cs50.h" // For MSVC, I used scanf_s() from stdio.h instead since cs50.h is not supported

long long Prompt(void); // Returns credit card number
int Validate(long long); // Returns checksum
void Report(long long); // Prints success

int main()
{
long long CardNumber = Prompt();
int CheckSum = Validate(CardNumber);
if (!(CheckSum % 10)) {
Report(CardNumber);
} else {
printf("INVALID\n");
}
return 0;
}

long long Prompt(void)
{
printf("CARD NUMBER: ");
long long CardNumber;
CardNumber = GetLongLong();
return CardNumber;
}

int Validate(long long CardNumber)
{
int CheckSum = 0;
int Digit = 0;
int DigitCount = (int)(floor(log10l(CardNumber) + 1));
for (int i = 2; i < DigitCount; i += 2) {
Digit = 2 * ((CardNumber % (10 ^ i)) / (10 ^ (i - 1))); // BUG
if (Digit > 9) {
int BiDigit = Digit;
Digit = 0;
int SubDigit = 0;
for (int j = 0; j < 2; i++) {
SubDigit = (BiDigit % (10 ^ j)) / (10 ^ (j - 1)); // BUG
Digit += SubDigit;
}
}
CheckSum += Digit;
}
for (int i = 1; i < DigitCount; i += 2) {
Digit = (CardNumber % (10 ^ i)) / (10 ^ (i - 1));
CheckSum += Digit;
}
return CheckSum;
}

void Report(long long CardNumber)
{
int DigitCount = (int)(floor(log10l(CardNumber) + 1));
int Digit1 = (CardNumber % (10 ^ (DigitCount - 1))) / (10 ^ (DigitCount - 2)); // BUG
int Digit2 = (CardNumber % (10 ^ (DigitCount - 2))) / (10 ^ (DigitCount - 3)); // BUG
if (Digit1 == 4 && (DigitCount == 13 || DigitCount == 16))
printf("VISA\n");
else if (Digit1 == 3 && (Digit2 == 4 || Digit2 == 7) && DigitCount == 15)
printf("AMERICAN EXPRESS\n");
else if (Digit1 == 4 && (Digit2 >= 1 || Digit2 <= 5) && DigitCount == 16)
printf("MASTERCARD\n");
else
printf("INVALID\n");
return;
}

Answer
Digit = 2 * ((CardNumber % (10 ^ i)) / (10 ^ (i - 1)));

^ is not a power operator, but bitwise xor in C and C++. Once i reaches 10, the expression 10 ^ i becomes zero.


BTW, in order to get digits from an integral number, it is better to avoid floating point operations (floor, log10l). A possible algorithm may look like this:

int DigitIndex = 0;
while (CardNumber > 0)
{
    int Digit = CardNumber % 10;
    CardNumber /= 10;

    // calculate check sum depending on parity of DigitIndex

    DigitIndex++;       
}