How do i make a for loop for the 11 remaining digits after that (

Question

The final digit of a Universal Product Code is a check digit computed as follows:

Add the digits in the odd-numbered positions from the right (first, third, fifth, etc. - not including the check digit) together and multiply by three.

Add the digits (up to but not including the check digit) in the even-numbered positions (second, fourth, sixth, etc.) to the result.

Take the remainder of the result divided by 10 (ie. the modulo 10 operation). If the remainder is equal to 0 then use 0 as the check digit, and if not 0 subtract the remainder from 10 to derive the check digit.

Algorithm Hints

1. You can retrieve the last digit using modulo 10.
2. You can remove the last digit by dividing it by 10.
3. The check digit is the last digit, you’ll want to save that value first.
4. There are exactly 11 remaining digits after that (which calls for a for loop).
5. Even number indexes represent odd-numbered positions.

Answer 1

Hi Diana, I'm offering some help by associating some of the code you'll need with comments based on the instructions. I hope this helps. Let me know if you have any further questions.

uint64_t shift_digits(uint64_t digits) {

// You can remove the last digit by dividing it by 10

return digits / 10;

}

uint64_t last_digit(uint64_t digits) {

// You can retrieve the last digit using modulo 10

return digits % 10;

}

void process_upc(uint64_t upc) {

int i;

// The check digit is the last digit, you’ll want to save that value first

uint64_t check = last_digit(upc);

uint64_t digits = shift_digits(upc);

uint64_t odd_sum = 0;

uint64_t even_sum = 0;

uint64_t result; // TODO: compute result

uint64_t computed_check;

// There are exactly 11 remaining digits after that (which calls for a for loop)

for (i = 0; i < 11; i++) {

uint64_t last = last_digit(digits);

// Even number indexes represent odd-numbered positions

if (i % 2 == 0) {

// TODO: Add the digits in the odd-numbered positions

} else {

// TODO: Add the digits in the even-numbered positions

}

digits = shift_digits(digits);

}

// TODO: multiply sum of odd-numbered positions by three

// TODO: Add the digits in the even-numbered positions to the result.

// TODO: Take the remainder of the result divided by 10 (ie. the modulo 10 operation).

//TODO: If the remainder is equal to 0 then use 0 as the check digit, and if not 0 subtract the remainder from 10 to derive the check digit.

if (result % 10 == 0) {

computed_check = ??;

} else {

computed_check = ??;

}

}