The most efficient way to implement an integer based power function pow(int, int)

// 2^3
pow(2,3) == 8

// 5^5
pow(5,5) == 3125

int ipow(int base, int exp)
{
    int result = 1;
    while (exp)
    {
        if (exp & 1)
            result *= base;
        exp >>= 1;
        base *= base;
    }

    return result;
}

x^15 = (x^7)*(x^7)*x
x^7 = (x^3)*(x^3)*x
x^3 = x*x*x

n*n = n^2
n^2*n = n^3
n^3*n^3 = n^6
n^6*n^6 = n^12
n^12*n^3 = n^15

struct IeeeFloat
{

    unsigned int base : 23;
    unsigned int exponent : 8;
    unsigned int signBit : 1;
};


union IeeeFloatUnion
{
    IeeeFloat brokenOut;
    float f;
};

inline float twoToThe(char exponent)
{
    // notice how the range checking is already done on the exponent var
    static IeeeFloatUnion u;
    u.f = 2.0;
    // Change the exponent part of the float
    u.brokenOut.exponent += (exponent - 1);
    return (u.f);
}

2 ** 3 == 1 << 3 == 8
2 ** 30 == 1 << 30 == 1073741824 (A Gigabyte)

int pow( int base, int exponent)
{
    if (exponent == 0) return 1;  // base case;
    int temp = pow(base, exponent/2);
    if (exponent % 2 == 0)
        return temp * temp;
    else
        return (base * temp * temp);
}

int pow(int base, int pow) {
  int res = 1;
  for(int i=pow; i<pow; i++)
    res *= base;

  return res;
}