Thread: Finding digits in a number

I supposed to write a recursive code that finds the position of a particular digit in a number, starting from the right. Eg position(12345,3) = 3, position(123513, 3) = 1 ...

My code goes like this

Code:

int position(int number, int digit) {
    if  ((number%10) == digit) return 1;
    return 1+ position(number/10, digit);
}

This works well and fine for all numbers except for cases when the digit is not found in the number. Eg position(12345, 8)
I am supposed to return a 0 value, not print nothing when I use printf to print out the value.
Eg position(1000, 4) = 0

It looks easy but I am stumped by it. I'm sure some of you experts out there can help me with this.
Thanks.

This is not the best way to do it, but it works.

Code:

int search(int num, int sf)
{
   char temp[100];
   sprintf(temp, "%d", num);
   char *pos = strrchr(temp, sf + '0');
   if (pos)
      return temp + strlen(temp) - pos;

   return 0;
}

To search from left to right, just change strrchr to strchr.

I was looking for a recursive solution along the lines of my solution.... I'm sure it can be done... it's just that I don't know how

Pier.

This is also not recursive, but works fine.

Code:

int position(int number, int digit)
{
   int pos = 0;
   do
   {
      pos++;
      if ((number%10) == digit)
         return pos;
   }
   while (number/=10);
   return 0;
}

Here a recursive function:

Code:

int position(int number, int digit) {
   int ret = 0;
   if(number == 0) return 0;
   if((number%10) == digit) return 1;
   ret = position(number/10, digit);
   return (ret==0 ? 0 : 1+ret);
}

Cool, Monster.... thanks a lot.

Pier.

Originally posted by Monster
Here a recursive function:

Code:

int position(int number, int digit) {
   int ret = 0;
   if(number == 0) return 0;
   if((number%10) == digit) return 1;
   ret = position(number/10, digit);
   return (ret==0 ? 0 : 1+ret);
}