64 Bit digitsum (C) [Feel Free to edit my English]

Question

This is part of my assignment, but I don't know why the output is not correct. Help?

/**
 * Create a function called digitsum that takes an long integer (64                     bits) and
 * sums the value of each of its digits. i.e. the number 316 should sum to
 * 3+1+6 or 10. Return the sum as the value of the function.
 * Hints:
 * - If the number is negative, make it positive before anything else.
 * - n % 10 will give you the value of the first digit.
 * - n / 10 will shift the integer to the right one digit.
 * - You're done summing when n is zero.
 */

Example input/output:

 * ./p2 316
 * num = 316, sum = 10
 * ./p2 -98374534984535
 * num = -98374534984535, sum = 77
 */

#include <stdio.h>
#include <stdlib.h>

int digitsum(int n){ //first try using function
    if (n % 10)
        return digitsum == n % 10;
    else if (n / 10)
        return digitsum == n / 10; 
    else     
        return 0;  //done summing when n is zero.
}

 // Read down in the comments and do **NOT** modify main below.


int main(int argc, char **argv)
{
    int64_t n;

    if (argc < 2) {
        printf("usage: %s <int>\n", argv[0]);
        exit(1);
    }
    n = atoll(argv[1]);

    printf("num = %ld, sum = %d\n", n, digitsum(n));
    return 0;
}

When I use gcc to compile, and it is only show the output "sum is 310" instead of "sum is 10"? I am new to C programming and I'm still studying..

Answer 1

The function of int digitsum(int n) is wrong.

You should add each digits in loop, like the follow code :

int digitsum(int64_t n){ //first try using function
    int ret = 0;
    if (n < 0)
        n = -n;
    while (n != 0) {
        ret += n % 10;
        n /= 10;
    }
    return ret;  //done summing when n is zero.
}