My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
#include <vector>
       
 
       
 
       
// note: my code only reversed odd numbers, they can't produce a leading zero

       
unsigned int reverse(unsigned int x)
       
{
       
  unsigned int result = 0;
       
  while (x > 9) // speed optimization: skip the last digit
       
  {
       
    auto digit = x % 10;
       
    result *= 10;
       
    result += digit;
       
    x /= 10;
       
  }
       
  // simple processing of the last digit, saves one division/modulo ;-)
       
  return result * 10 + x;
       
}
       
 
       
// return true if all digits are odd

       
bool onlyOdd(unsigned int x)
       
{
       
  while (x > 0)
       
  {
       
    // found an even digit ?
       
    if (x % 2 == 0)
       
      return false;
       
    x /= 10;
       
  }
       
  // yes, all odd
       
  return true;
       
}
       
 
       
int main()
       
{
       
  // if the first digit is odd, then the last has to be even
       
  // the same logic applies in reverse order, too
       
  // therefore analyze only odd numbers and multiply the result by 2
       
  const unsigned int factor = 2;
       
 
       
  unsigned int count = 0;
       
  // no solutions between 10^8 and 10^9
       
  for (unsigned int i = 11; i < 100*1000*1000; i += factor)
       
    if (onlyOdd(i + reverse(i)))
       
      count += factor;
       
 
       
  std::cout << count << std::endl;
       
  return 0;
       
}

This solution contains 7 empty lines, 9 comments and 2 preprocessor commands.