function that splits digits then sums them - c

Question

I am trying to solve a problem that requires me to take a number. Use the % and / to take the rightmost number and sum the digits that are separated. Then tell see if the number is divisible by 9.
I have created a function that will separate the rightmost number and then I have tried to take that number and run it through a while loop. The problem that exist is, when I run the while loop. It creates an infinite loop or the output will not be printed.
#include <stdio.h>
int loopnum(int n);
int main(void)
{
int num;
int sum = 0;
int d = loopnum(num);
printf("Enter a number:\n");
scanf("%d", &num);
while (loopnum(num) > 0) {
printf("d = %d", d);
printf(",sum = %d\n", sum);
}
if (num % 9 == 0) {
printf("n = %d is divisible by 9\n", num);
}
else {
printf("n = %d is not divisible by 9\n", num);
}
return 0;
}
int loopnum(int n)
{
n = n % 10;
n = n / 10;
return n;
}
Enter a number:
9
n = 9 is divisible by 9
The result of this code is suppose to output d = "the digit" ,sum =
"digit + sum" and n = "digit" is divisible by 9. for example if I
input 9. The output would be d = 9, sum = 9.

Answer 1

The function does not make sense.
int loopnum(int n)
{
n = n % 10;
n = n / 10;
return n;
}
For example for n equal to 27 you will get
n = n % 10;
now n is equal to 7 and then
n = n / 10;
now n is equal to 0.
So for the number 27 the function returns 0.
And moreover inside the loop
while (loopnum(num) > 0) {
printf("d = %d", d);
printf(",sum = %d\n", sum);
}
neither num, nor d, nor sum is changed.
There is no great sense to define such a function except that it could itself to output intemediate sums and digits.
Without the function the program can look the following way.
#include <stdio.h>
int main(void)
{
while ( 1 )
{
const unsigned int Base = 10;
const unsigned int DIVISOR = 9;
printf( "Enter a non-negative number (0 - exit): " );
unsigned int n;
if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
unsigned int sum = 0;
unsigned int tmp = n;
do
{
unsigned int digit = tmp % Base;
sum += digit;
printf( "d = %u, sum = %u\n", digit, sum );
} while ( tmp /= Base );
printf( "\nn = %u is %sdivisble by %u.\n\n",
n,
sum % DIVISOR == 0 ? "" : "not ", DIVISOR );
}
return 0;
}
Its ouutput might look like
Enter a non-negative number (0 - exit): 9
d = 9, sum = 9
n = 9 is divisble by 9.
Enter a non-negative number (0 - exit): 123456
d = 6, sum = 6
d = 5, sum = 11
d = 4, sum = 15
d = 3, sum = 18
d = 2, sum = 20
d = 1, sum = 21
n = 123456 is not divisble by 9.
Enter a non-negative number (0 - exit): 0

Answer 2

The loopnum code is wrong. It will always return zero and therefor the while will not loop.
int loopnum(int n)
{
n = n % 10; // After this n is a number between 0 and 9
n = n / 10; // Consequently, when you divide by 10 you'll end up with zero
return n;
}
Your design requires two things:
1) Return the remainder
2) Change n
To do that you need to pass a pointer to n.
Something like:
int loopnum(int *n)
{
int res = *n % 10;
*n = *n / 10;
return res;
}
int main(void)
{
int x = 123;
int sum = 0;
while(x) sum += loopnum(&x);
printf("%d\n", sum);
return 0;
}