Thread: sum of digits in a number (while loop) please explain

Hey all...also, i've started assembly on Intel 64 after dipping into C on FreeBSD for the past year. Would like to start using GDB.

Code:

// Program to give sum of digits of a number

#include <stdio.h>

int main(void)
{
    long num, temp, digit, sum = 0;

    printf("Enter your number.\n");
    scanf("%ld", &num);

    temp = num;

    // num = 103

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

    printf("Given number = %ld\n", temp);
    
    // output 103, 4

    printf("Sum of the digits %ld = %ld\n", temp, sum);
}

specifically, this while loop:

Code:

    while ( num > 0)
    {
        digit = num % 10;
        sum = sum + digit;
        num /= 10;
    {

Is this while loop looping num times until it reaches the value of greater than 0? Please expain.

103 % 10 = 3 (digit)
0 (sum) = 0 (sum) + 3
103 (num) = 103 (num) / 10 (compound operator, division)

Sincerely, lost

I was just going on about my progress with C and I mentioned GDB. Please excuse the confusion.


Was just looking or an explanation of the while loop line by line.

The explanation is right C Program to Compute the Sum of Digits in a given Integer - Sanfoundry