Emirps in C++

Design a program called emirps.cpp which will identify all of the emirps between two positive integers N1 and N2 exclusively. We will assume N1 < N2.
An emirp is a prime which is also a prime when all of its digits are reversed, excluding palindromic primes. For example, 13 and 17 are emirps because 31 and 71 are also prime. The prime number 131 is not an emirp because it is palindromic.
So, the idea is, find the first prime after N1, reverse it, then determine if the reversed number is also prime. Continue this until you reach N2.
You might try to do this top down.

1. Imagine a loop variable ii from N1+1 up to N2-1
2. Imagine you have a boolean function called prime with an integer parameter (the prime candidate) which returns true if the integer parameter is a prime number
3. Imagine you have an integer function called reverse with an integer parameter (the number you want to reverse) which returns the input number reversed.
4. With these three imaginary building blocks, the program is just:

   for ii = (N1 + 1) up to (N2 - 1)
     if prime(ii) and prime( reverse(ii) ) and ( ii != reverse(ii) ) print out ii
   end for
   

5. Now build the loop and stubs for prime and reverse.
6. Get prime working and test it out.

   The prototype for the function is:
   bool prime(int number);
   
   The pseudocode for this function is:
   set isprime equal to true
   set divisor equal to two
   while divisor is less than number and isprime is true
     if divisor goes into number evenly, set isprime equal to false
     increment divisor by one
   end while
   return isprime
   

7. Get reverse working and test it out.

   The prototype for the function is:
   int reverse(int number);
   
   The pseudocode for this function is:
   set total equal to zero
   set digits equal to one
   figure out how many digits there are in the number:
     while (int(number / pow(10, digits)) != 0 ) ++digits
   for ii equal to 1 up to ii equal to digits
     add (number % 10)*pow(10,digits - ii) to total
     divide number by 10 and store the result back in number
   end for
   return total
   

8. Put it all together