Reversing an integer using recursive method in C++

Question

I have this recursive function to reverse a positive integer. Anybody having an efficient algorithm to do the task using fewer recursive calls (but still using recursion), please post here!

int countDigit (const int & _X) {
    if (_X < 10)
        return 1;
    return (countDigit(_X / 10) + 1);
}

int getPower (const int & _X, const int & _Y) {
    if (_Y == 0)
       return 1;
    int ans = getPower(_X, _Y / 2);
    if (_Y % 2)
       return _X * ans * ans;
    return ans * ans;
 }

 int reverseDigit (const int & digit)  {
    if (digit < 10) {
      return digit;
    }
    int num = (digit % 10) * getPower(10, countDigit(digit / 10));
    return num + reverseDigit(digit / 10);
 }

Answer 1

Sure ... effectively read in digits from the number and add them to the end of a temporary number:

EDIT: Based on the questioner's bizarre modification of the question (to require recursion), I have edited my clean iterative solution to include an arbitrary amount of recursion, denoted by the positive integer [recurse].

int reverseDigit(int digit, int recurse=1) {
    if (recurse>0) reverseDigit(digit,recurse-1);

    int num=0;
    while (digit!=0) {
        num*=10;
        num+=digit%10;
        digit/=10;
    }
    return num;
}

Answer 2

This is recursive

void reverse(int number, int& result)
{
if (number == 0) return;
else
   {
    result *= 10;
    result += number % 10; 
    reverse(number / 10, result ); 
   }
}

This is how you call it with (545146548) example

   void main()
   {
       int result = 0;
       reverse(545146548, result);
       cout << result << endl;
   };