Find the sum of the digits of a given number

Question

I need to find the sum of the digits of the given number and repeat the process until the value lies between 1 to 9.

e.g if the input is 72457 then,

7+2+4+5+7 = 25

2+5 = 7

so , the function should return 7.

Also, if the input is a negative number like -72457 the function should return -7.

Here's what i tried,

int digitSum(int input1){
int flag=0;
if(input1<0)
 flag=1;
 int rem,sum=0;
 int x=abs(input1);
 while(x>0){
 rem=x%10;
 sum+=rem;
 x=x/10;
 }
 int value = sum%9;
 if(value==0){
 if(flag==1)
 return -9;
 else
 return 9; 
 }
 else{
 if(flag==1)
 return -value;
 else
 return value;
 }
}

• Peter Taylor: What if the input is 0?
• DJHenjin: You can shuffle the logic around a bit to simplify this by first having an int which represents the sign of the input number by dividing input / abs(input), which will give you either -1 or 1. Then, during your summation loop, do a comparison of != 0. so that you don't have to worry about the sign. And finish off by returning sum * sign
• Toby Speight: n % 9 gets you most of the way there; you only need to adjust zero results to ±9, and you're done.

Answer 1

A simplified pure numerical function may be:

int digitSum(int input) {
 int n,m,s = input<0 ? -1:1;
 for(n=input*s; n>9; ) {
 for(m=n, n=0; m>0; m/=10)
 n+=m%10;
 }
 return n*s;
}

• Bo R: Thanks, @holger, I missed the recursivness of the problem.