Sum of digits in C#

Question

What\'s the fastest and easiest to read implementation of calculating the sum of digits?

I.e. Given the number: 17463 = 1 + 7 + 4 + 6 + 3 = 21

Answer 1

I would suggest that the easiest to read implementation would be something like:

public int sum(int number)
{
    int ret = 0;
    foreach (char c in Math.Abs(number).ToString())
        ret += c - '0';
    return ret;
}

This works, and is quite easy to read. BTW: Convert.ToInt32('3') gives 51, not 3. Convert.ToInt32('3' - '0') gives 3.

I would assume that the fastest implementation is Greg Hewgill's arithmetric solution.

Answer 2

int n = 17463; int sum = 0;
for (int i = n; i > 0; i = i / 10)
{
sum = sum + i % 10;
}
Console.WriteLine(sum);
Console.ReadLine();

Answer 3

I thought I'd just post this for completion's sake:

If you need a recursive sum of digits, e.g: 17463 -> 1 + 7 + 4 + 6 + 3 = 21 -> 2 + 1 = 3
then the best solution would be

int result = input % 9;
return (result == 0 && input > 0) ? 9 : result;

Answer 4

If one wants to perform specific operations like add odd numbers/even numbers only, add numbers with odd index/even index only, then following code suits best. In this example, I have added odd numbers from the input number.

using System;
                    
public class Program
{
    public static void Main()
    {
        Console.WriteLine("Please Input number");
        Console.WriteLine(GetSum(Console.ReadLine()));
    }
    
    public static int GetSum(string num){
        int summ = 0;
        for(int i=0; i < num.Length; i++){
            int currentNum;
            if(int.TryParse(num[i].ToString(),out currentNum)){
                 if(currentNum % 2 == 1){
                    summ += currentNum;
                }
            }
       } 
       return summ;
    }
}

Answer 5

#include <stdio.h>

int main (void) {

    int sum = 0;
    int n;
    printf("Enter ir num ");
    scanf("%i", &n);

    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }

    printf("Sum of digits is %i\n", sum);

    return 0;
}

Answer 6

A while back, I had to find the digit sum of something. I used Muhammad Hasan Khan's code, however it kept returning the right number as a recurring decimal, i.e. when the digit sum was 4, i'd get 4.44444444444444 etc. Hence I edited it, getting the digit sum correct each time with this code:

 double a, n, sumD;
 for (n = a; n > 0; sumD += n % 10, n /= 10);
 int sumI = (int)Math.Floor(sumD);

where a is the number whose digit sum you want, n is a double used for this process, sumD is the digit sum in double and sumI is the digit sum in integer, so the correct digit sum.