Sum of digits in C

Hello,

I have written in C a program that calculate the sum of digits (for a number).

For Example for the number 199 the checksum is 19.

But now i want that my program also calculate the checksum of 19.

How can i do that in C? Can i make a nested function in C?

Here is my code

Code:

---

int dsum(int z)
{
    int q=0;
    while(z>0)
    {
        q+= z%10;
        z /=10;
    }
    return q;
}



int main(){
    int z =199;
    int i;
    int qs= dsum(z);
    printf("%d", qs);

return 0;}

---

What's your question?
Your code gives the right answer.

For my example the checksum is 19. But i want now that my function dsum calculate the checksum again, until the sum of the digits contains only one number

For example:
199 => 19
19 => 10
10 => 1

In that case my solution is = 1

For me is not very clear how i can extend my function to do that.

Simples!

Code:

---

qs = z;
do {
  qs = dsum(qs);
} while ( qs >= 10 );

---

In this particular case there happens to be a mathematical shortcut:

Code:

---

int dsum(int z)
{
    if (z == 0)
    {
        return 0;
    }
    int q = z % 9;
    return (q == 0) ? 9 : q;
}

---

(though you need to decide what should happen if the input is a negative number)

Quote:

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Originally Posted by laserlight

In this particular case there happens to be a mathematical shortcut

---

Witchcraft!
How does that work?

Informally, consider that (10^i)d = d (mod 9) for a non-negative integer i and a decimal digit d (but of course 9 = 0 (mod 9)), where ^ denotes exponentiation and = denotes congruence.

Now, suppose we're trying to (non-recursively) sum the digits of an N digit non-negative integer x = (10^0)d_0 + (10^1)d_1 + ... + (10^i)d_i + ... + (10^(N-1))d_(N-1). This means we want to compute (d_0 + d_1 + ... + d_(N-1)). Since (10^i)d = d (mod 9), it follows that (10^0)d_0 + (10^1)d_1 + ... + (10^i)d_i + ... + (10^(N-1))d = (d_0 + d_1 + ... + d_i + ... + d_(N-1)) (mod 9).

But (d_0 + d_1 + ... + d_i + ... + d_(N-1)) is itself an integer y = (10^0)d_0 + (10^1)d_1 + ... + (10^i)d_i + ... + (10^(N-1))d_(N-1) for another value of N and another sequence of digits d_i. So it follows that y = (d_0 + d_1 + ... + d_i + ... + d_(N-1)) (mod 9) for this new sequence of digits, so on and so forth. Hence computing the remainder of x divided by 9 gives us the recursive sum of digits of x, with the caveat that when x is not 0, a remainder of 0 means that the recursive sum of digits is 9, as per the earlier observation that 9 = 0 (mod 9).

That's very interesting. After reading it through carefully I guess I understand it, but it still seems like magic.

Quote:

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Originally Posted by Salem

Simples!

Code:

---

qs = z;
do {
  qs = dsum(qs);
} while ( qs >= 10 );

---

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Thank you. It is clear now for me. To do while loop is very helpful.