How to sum the digits of a number using Recursion [C]?

Question

For instance , the number is 123, the answer is 6. Only recursion!

Answer 1

While @SamGrondahl is on the right track, unfortunately, if you enter a negative number it gives unexpected results. This should work properly with negative numbers:

int sumdigits(int number) {
    int sign = number < 0 ? -1 : 1;
    number = sign * number; // abs the number

    if (number < 10) 
        return sign * number;
    else 
        return sign * (number % 10 + sumdigits(number / 10));
}

This will return the sum of the digits, and if the number is negative, negate that.

Answer 2

A correct answer is trickier to achieve than at first glance, since the negation of INT_MIN may not fit in an int. The posted solutions would work for the most part by changing their implementation to use unsigned int during recursion. Below is an alternative to that.

int sum_digits_recursively (char digits[], int i) {
    if (i == 0 && digits[i] == '-') return -sum_digits_recursively(digits, 1);
    if (digits[i] == '\0') return 0;
    return (digits[i]-'0') + sum_digits_recursively(digits, i+1);
}

int sum_digits (int x) {
    char digits[sizeof(x) * CHAR_BIT];
    snprintf(digits, sizeof(digits), "%d", x);
    return sum_digits_recursively(digits, 0);
}

Answer 3

This should do what you want:

int recurse( int number ) {
    if ( abs(number) < 10 ) return number;
    return number % 10 + recurse ( number / 10 );
}