/* ---------- Example 
    Iterative function to raise a double  x  to a positive power  n.
     Uses bitwise operations to check parity (even/odd) and to divide by 2,
     thus explicitly avoiding extra divisions.  
     Note that n is odd if and only if the 1's bit in its binary 
     representation is 1 and not zero. Thus, if the bitwise AND operator &
     is applied to n and the number 1, the result will be zero if n is 
     even and 1 if n is odd. Let y be a binary digit, so that y is zero or 1.
     Then:
                                         xxxxxxxxxxy
                                       & 00000000001
                                         -----------
                                       = 0000000000y

    (The values of the 'x' digits are irrelevant.)
   ---------- 
*/
#include<iostream>
#include<cassert>
using namespace std;

double intPower( double x, unsigned int n )
{
    assert ( n >= 0 );
    if ( n == 1 ) return x;  else if (n == 0) return 1.0;  // base cases.

    double prod = 1, power2 = x;    // Initialize.
    while ( n > 1 ) {
       if ( n & 1 )       // True if last bit of n is 1 (i.e., if n is odd).
          prod *= power2;

       power2 *= power2;
       n >>= 1;           // Shift n right 1 bit ( divide n by 2; n /=2 . ).
    }
    return prod*=power2;  // The final value of n is always 1.
}


int main()
{
  double x;
  unsigned int n;
  cout << "\nEnter the base x: ";
  cin  >> x;
  cout << "\nEnter the positive integer exponent n: ";
  cin  >> n;

  cout << "\n\n" << x << " raised to the power " << n << " is " 
       << intPower( x, n ) << "\n\n";
  return 0;
}