Simple mathematical approach for summing the digits in a number

digitsum.c

#include <stdio.h>
#include <math.h>

/*
The algorithm is:
d_i = (c mod 10^(i+1)) / 10^i
S = sum_{i=0}^{floor(log_10(c))+1} d_i = sum_{i=0}^{floor(log_10(c))+1} (c mod 10^(i+1)) / 10^i

Where c is the number in question, d_i is the digit at position i counted from the least significant digit
and S is the sum of the digits in the number.
*/

int main(void)
{
	int c = 1000;

	printf("The number is %d.\n", c);

	int s = 0;

/*
	// Special case for 3 digits:
	s += (c % 10) / 1;
	s += (c % 100) / 10;
	s += (c % 1000) / 100;
*/

	printf("%f\n", log(c));

	// General case for n digits
	int i;
	for (i = 0; i < (int)(log(c) / log(10)) + 1; i++) {
		s += (c % (int)pow(10, i + 1)) / (int)pow(10, i);
	}

	printf("The sum of the digits is %d.\n", s);

	return 0;
}