How do you do *integer* exponentiation in C#?

Question

The built-in Math.Pow() function in .NET raises a double base to a double exponent and returns a double result.

W

Answer 1

I cast the result into int, like this:

double exp = 3.0;
int result = (int)Math.Pow(2.0, exp);

In this case, there are no rounding errors because base and exponent are integer. The result will be integer too.

Answer 2

Here's a blog post that explains the fastest way to raise integers to integer powers. As one of the comments points out, some of these tricks are built into chips.

Answer 3

A pretty fast one might be something like this:

int IntPow(int x, uint pow)
{
    int ret = 1;
    while ( pow != 0 )
    {
        if ( (pow & 1) == 1 )
            ret *= x;
        x *= x;
        pow >>= 1;
    }
    return ret;
}

Note that this does not allow negative powers. I'll leave that as an exercise to you. :)

Added: Oh yes, almost forgot - also add overflow/underflow checking, or you might be in for a few nasty surprises down the road.

Answer 4

LINQ anyone?

public static int Pow(this int bas, int exp)
{
    return Enumerable
          .Repeat(bas, exp)
          .Aggregate(1, (a, b) => a * b);
}

usage as extension:

var threeToThePowerOfNine = 3.Pow(9);

Answer 5

Two more...

    public static int FastPower(int x, int pow)
    {
        switch (pow)
        {
            case 0: return 1;
            case 1: return x;
            case 2: return x * x;
            case 3: return x * x * x;
            case 4: return x * x * x * x;
            case 5: return x * x * x * x * x;
            case 6: return x * x * x * x * x * x;
            case 7: return x * x * x * x * x * x * x;
            case 8: return x * x * x * x * x * x * x * x;
            case 9: return x * x * x * x * x * x * x * x * x;
            case 10: return x * x * x * x * x * x * x * x * x * x; 
            case 11: return x * x * x * x * x * x * x * x * x * x * x; 
            // up to 32 can be added 
            default: // Vilx's solution is used for default
                int ret = 1;
                while (pow != 0)
                {
                    if ((pow & 1) == 1)
                        ret *= x;
                    x *= x;
                    pow >>= 1;
                }
                return ret;
        }
    }

    public static int SimplePower(int x, int pow)
    {
        return (int)Math.Pow(x, pow);
    }

I did some quick performance testing

---

• mini-me : 32 ms

• Sunsetquest(Fast) : 37 ms

• Vilx : 46 ms

• Charles Bretana(aka Cook's): 166 ms

• Sunsetquest(simple) : 469 ms

• 3dGrabber (Linq version) : 868 ms

(testing notes: intel i7 2nd gen, .net 4, release build, release run, 1M different bases, exp from 0-10 only)

Conclusion: mini-me's is the best in both performance and simplicity

very minimal accuracy testing was done

Answer 6

How about:

public static long IntPow(long a, long b)
{
  long result = 1;
  for (long i = 0; i < b; i++)
    result *= a;
  return result;
}