Sum of square of each digit of an integer

My first try that worked:


int S(int x){
   int sum=0, y=0;
   while (x!=0){
      if (x < 10) {
       y = x;
      } else {
       y = x % 10;
      }
      x = (x - y) / 10;
      sum += y * y;
   }
   return sum;
}

Returns:
DigitsSum(55)=50 [5*5 + 5*5]
DigitsSum(230)=13 [2*2 + 3*3 + 0*0]
DigitsSum(37)=58 [3*3 + 7*7]

U P D A T E

A better solution:
int ur_num;
int sum_of_sq=0;
while(ur_num!=0){
remainder=ur_num%10;
sum_of_sq += remainder*remainder;
ur_num = ur_num/10;}