Test I ran to see if there were any faster ways of finding the sum of digits in an integer than looping and dividing by 10.

test.c

#include <stdio.h>
#include <time.h>
#include <string.h>

/**
 *  Test to see how fast it takes to fund the sum of 
 *  the digits of an integer.
 */

/**
 * Traditional method
 */
unsigned int test1(unsigned int n){
    unsigned int sum = 0;
    while (n){
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

/**
 *  Using sprintf() for int to char.
 */
unsigned int test2(unsigned int n){
    char str[] = "0000000000"; // 2^32 base 10 has 10 digits
    sprintf(str, "%u", n);
    unsigned int sum = 0, max = strlen(str);
    int i;
    for (i = 0; i < max; i++)
        sum += str[i]-'0';
    return sum;
}

/**
 *  Using sprintf() for int to char.
 *  This time with known length.
 */
unsigned int test3(unsigned int n){
    char str[] = "000000"; // 2^32 base 10 has 10 digits
    sprintf(str, "%u", n);
    unsigned int sum = 0, max = 6;
    int i;
    for (i = 0; i < max; i++)
        sum += str[i]-'0';
    return sum;
}

/**
 * Using the trick divide by 10
 */
unsigned int test4(unsigned long n){
    unsigned int sum = 0;
    while (n){
        sum += n % 10;
        // n /= 10;
        n = (n * 0x1999999A) >> 32;
    }
    return sum;
}

/**
 * Using the trick divide by 10
 */
unsigned int test5(unsigned long n){
    unsigned int sum = 0, q, r;
    while (n){
        sum += n % 10;
        // n /= 10;
        // n = (n * 0x1999999A) >> 32;
        q = (n >> 1) + (n >> 2);
        q = q + (q >> 4);
        q = q + (q >> 8);
        q = q + (q >> 16);
        q = q >> 3;
        r = n - q*10;
        n = q + ((r + 6) >> 4);
    }
    return sum;
}

/**
 * Using the trick divide by 10
 */
unsigned int test6(unsigned long n){
    unsigned int sum = 0;
    unsigned long t;
    while (n){
        // sum += n % 10;
        // n /= 10;
        t = (n * 0x1999999A) >> 32;
        sum += n-t;
        n = t;
    }
    return sum;
}

int main(int argc, char *argv[]){
    clock_t t; // microsecond resolution
    unsigned int n = 123456, i, num = 10000;
    unsigned int sum;
    unsigned long timesum, j, times = 1000;

    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test1(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test1: \t%luus\n", timesum/times);
    else
        printf("Test1 returned incorrect answer\n");
    
    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test2(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test2: \t%luus\n", timesum/times);
    else
        printf("Test2 returned incorrect answer\n");

    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test3(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test3: \t%luus\n", timesum/times);
    else
        printf("Test3 returned incorrect answer\n");

    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test4(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test4: \t%luus\n", timesum/times);
    else
        printf("Test4 returned incorrect answer\n");

    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test5(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test5: \t%luus\n", timesum/times);
    else
        printf("Test5 returned incorrect answer\n");

    timesum = 0;
    for (j = 0; j < times; j++){
        t = clock();
        for (i = 0; i < num; i ++)
            sum = test6(n);
        timesum += clock()-t;
    }
    if (sum == 21)
        printf("Test6: \t%luus\n", timesum/times);
    else
        printf("Test6 returned incorrect answer\n");

    return 0;
}