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Credit card number generator in Java
import java.util.Random;
* A credit card number generator.
* @author Josef Galea
public class CreditCardNumberGenerator {
private Random random = new Random(System.currentTimeMillis());
* Generates a random valid credit card number. For more information about
* the credit card number generation algorithms and credit card numbers
* refer to <a
* href="">Everything
* you ever wanted to know about CC's</a>, <a
* href="">Graham King's blog</a>, and
* <a href=""
* >This is How Credit Card Numbers Are Generated</a>
* @param bin
* The bank identification number, a set digits at the start of the credit card
* number, used to identify the bank that is issuing the credit card.
* @param length
* The total length (i.e. including the BIN) of the credit card number.
* @return
* A randomly generated, valid, credit card number.
public String generate(String bin, int length) {
// The number of random digits that we need to generate is equal to the
// total length of the card number minus the start digits given by the
// user, minus the check digit at the end.
int randomNumberLength = length - (bin.length() + 1);
StringBuilder builder = new StringBuilder(bin);
for (int i = 0; i < randomNumberLength; i++) {
int digit = this.random.nextInt(10);
// Do the Luhn algorithm to generate the check digit.
int checkDigit = this.getCheckDigit(builder.toString());
return builder.toString();
* Generates the check digit required to make the given credit card number
* valid (i.e. pass the Luhn check)
* @param number
* The credit card number for which to generate the check digit.
* @return The check digit required to make the given credit card number
* valid.
private int getCheckDigit(String number) {
// Get the sum of all the digits, however we need to replace the value
// of the first digit, and every other digit, with the same digit
// multiplied by 2. If this multiplication yields a number greater
// than 9, then add the two digits together to get a single digit
// number.
// The digits we need to replace will be those in an even position for
// card numbers whose length is an even number, or those is an odd
// position for card numbers whose length is an odd number. This is
// because the Luhn algorithm reverses the card number, and doubles
// every other number starting from the second number from the last
// position.
int sum = 0;
for (int i = 0; i < number.length(); i++) {
// Get the digit at the current position.
int digit = Integer.parseInt(number.substring(i, (i + 1)));
if ((i % 2) == 0) {
digit = digit * 2;
if (digit > 9) {
digit = (digit / 10) + (digit % 10);
sum += digit;
// The check digit is the number required to make the sum a multiple of
// 10.
int mod = sum % 10;
return ((mod == 0) ? 0 : 10 - mod);

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@JLofgren JLofgren commented Feb 15, 2019

Thanks for this example. I just want to point out that it has a bug, though. It will only work correctly when given an odd-length string as input. The digit needs to be indexed odd-even from the right side of the number. So, one way to fix it is to loop from right to left rather than left to right. But, I'm realizing line 79 also needs a tweak...

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