Implementing a written algorithm is something you do frequently as a C language programmer. You’re given a set of directions and your job is to translate that English into computer code — and make that code work.
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\nFor this month’s exercise, the challenge was to apply the MOD 10 algorithm to a series of sample credit card numbers and report whether each is valid. As a review, here is the algorithm:<\/p>\n
<\/p>\n <\/strong><\/p>\n For my solution, I used two for<\/em> loops. The first processes elements of the First comes the for<\/em> loop that tallies the digits:<\/p>\n int<\/em> variable The for<\/em> loop works through every other (odd) digit; loop variable Variable The After the loop is complete, digit tally is held in variable The \n
card[]<\/code> array. A second for<\/em> loop tallies the digits to get the sum (Step 3). That’s followed by some math to calculate the result (Step 4) and compare it with the check digit (Step 5). A switch-case<\/em> structure then examines the card number’s first digit to report the card type.<\/p>\n\r\n \/* tally the digits *\/<\/span>\r\n sum = 0;\r\n for(d=0; d<15; d+=2)\r\n {\r\n \/* double every odd digit *\/<\/span>\r\n odd = (card[c][d] - '0') * 2;\r\n \/* Any values greater than 9, sum the digits *\/<\/span>\r\n if( odd > 9)\r\n odd = odd % 10 + 1;\r\n \/* fetch the even digit *\/<\/span>\r\n even = (card[c][d+1] - '0');\r\n \/* Add all the digits together except check digit*\/<\/span>\r\n if( d < 14 )\r\n sum += odd + even;\r\n else\r\n sum += odd;\r\n }<\/pre>\nsum<\/code> holds the digit tally. It's initialized to zero.<\/p>\nd<\/code> is incremented by 2 with d+=2<\/code> as the loop step value.<\/p>\nodd<\/code> holds the result of a doubled digit in the card number. Character value '0'<\/code> is subtracted from the character digit to obtain the integer value, which is then doubled. If the value of odd<\/code> is greater than 9, a quick digit-sum is performed to make the results single-digit. The calculation is odd % 10 + 1<\/code>, which is sneaky, but it works.<\/p>\nif( d < 14 )<\/code> statement tallies the digits in the card number, both the unchanged even digits and the doubled odd<\/code> digits. This statement ensures that for the last iteration of the loop, only the final odd<\/code> digit is added to the sum<\/code>, not the check digit.<\/p>\nsum<\/code>. The remainder of the algorithm is implemented as follows:<\/p>\n\r\n \/* multiply the tally by 9 *\/<\/span>\r\n sum *= 9;\r\n \/* compare the last digit of the result to the check digit *\/<\/span>\r\n last_digit = sum % 10;\r\n check_digit = card[c][15] - '0';\r\n if( check_digit == last_digit)\r\n printf(\"Valid \");\r\n else\r\n printf(\"Invalid \");<\/pre>\nsum<\/code> is multiplied by 9, then the last digit is peeled off and stored in the last_digit<\/code> variable. That value is compared with the final digit of the card number, stored in variable check_digit<\/code>. The if<\/em> statement displays the results.<\/p>\n