Math.Pow taking an integer value

Question

From http://msdn.microsoft.com/en-us/library/system.math.pow.aspx

int value = 2;
for (int power = 0; power <= 32; power++)
    Console.WriteLine("{0}^{1} = {2:N0}",
                      value, power, (long) Math.Pow(value, power));

Math.Pow takes doubles as arguments, yet here we are passing in ints.

Question: Is there any danger of floating point rounding errors if there is an implicit conversion to double happening?

If yes, it is better to use something like:

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

Related question: http://stackoverflow.com/questions/383587/how-do-you-do-integer-exponentiation-in-c Some good reading in that question and related links — dash, Jun 25 '12 at 20:22
See also: http://stackoverflow.com/questions/936541/math-pow- — dash, Jun 25 '12 at 20:28
And [this one](http://stackoverflow.com/questions/4297454/c-sharp-math-pow-is-not-calculating-correctly) on rounding errors. — Anders Gustafsson, Jun 25 '12 at 20:32
Your first link dash is a good one - thats where I initially got my IntPow code from. — Dave Mateer, Jun 25 '12 at 20:48

score 8 · Answer 1 · answered Jul 02 '15 at 06:09

8

In your special case, when you are calculating 2 to the power x, you can use a simple left shift. This would simplify your code to:

public static int TwoPowX(int power)
{
    return (1<<power);
}

answered Jul 02 '15 at 06:09

user5072377

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And in floating-point, powers of two can be trivially calculated with `ldexp`. Which .NET doesn't have, but is pretty easy to write using `BitConverter.Int64BitsToDouble`. – Ben Voigt Jul 15 '17 at 16:29

score 5 · Answer 2 · answered Jun 25 '12 at 21:02

5

No, there's no possibility of rounding error caused by the conversion to double. double can exactly represent all integers which fall in the domain of the power function.

answered Jun 25 '12 at 21:02

Ben Voigt

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score 4 · Accepted Answer · answered Jun 25 '12 at 20:20

4

Yes, there is an implicit conversion to double happening, and yes there is a possibility of floating point rounding errors as a result.

As to whether it's worth using the alternate method you propose, that's specific to the application. Is a floating point rounding error entirely unacceptable? Will you be using numbers that fit within int32 (it doesn't take a whole lot for powers to overflow)?

answered Jun 25 '12 at 20:20

Servy

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floating point rounding error when converting int to double? please write more on that. – Mare Infinitus Jun 25 '12 at 20:24
1

It's not just converting an int to a double, it's converting an int to a double and then doing stuff with it. As soon as you start performing any operations (add/multiply/whatever) there is at least a possibility of rounding errors. Since raising to a power is likely to be broken up into a number of add/multiplies, it increases the probability and magnitude of floating point errors. – Servy Jun 25 '12 at 20:27
you mean as they exist with any IEEE 754 number? – Mare Infinitus Jun 25 '12 at 20:40

score -1 · Answer 4 · answered Jun 25 '12 at 21:00

-1

public static int IntPow(int number, uint power)
        {
            int result = 1;
            for (int i = 0; i < power; i++)
            {
                result *= number;
            }
            return result;
        }

for readability!

answered Jun 25 '12 at 21:00

Dave Mateer

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There's a reason for the extra code -- this answer is `O(power)`, while the code in the question if `O(log(power))`. – Ben Voigt Jun 25 '12 at 21:03

Math.Pow taking an integer value

4 Answers4