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.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
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 );
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
In this particular case there happens to be a mathematical shortcut:
(though you need to decide what should happen if the input is a negative number)Code:int dsum(int z) { if (z == 0) { return 0; } int q = z % 9; return (q == 0) ? 9 : q; }
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Witchcraft!
How does that work?
A little inaccuracy saves tons of explanation. - H.H. Munro
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).
Look up a C++ Reference and learn How To Ask Questions The Smart Way
That's very interesting. After reading it through carefully I guess I understand it, but it still seems like magic.
A little inaccuracy saves tons of explanation. - H.H. Munro
Thank you. It is clear now for me. To do while loop is very helpful.
