EQ #38

Here’s a solution that iterates over the number of strings, rather than the number of bits in the word.

int nstrings (unsigned long int x)
{
   int result = 0;
   /* convert x into a word that has a '1' for every
    * transition from 0 to 1 or 1 to 0 in the original
    * word.
    */
   x ^= (x << 1);
   /* every pair of ones in the new word represents
   * a string of ones in the original word. Remove
    * them two at a time and keep count.
    */
   while (x) {
     /* remove the lowest set bit from x; this
      * represents the start of a string of ones.
      */
     x &= ~(x & -x);
     ++result;
     /* remove the next set bit from x; this
      * represents the end of that string of ones.
      */
     x &= ~(x & -x);
   }
   return result;
}