How to check if a number is a power of 2

Today I needed a simple algorithm for checking if a number is a power of 2.

The algorithm needs to be:

Simple
Correct for any ulong value.

I came up with this simple algorithm:

private bool IsPowerOfTwo(ulong number)
{
    if (number == 0)
        return false;

    for (ulong power = 1; power > 0; power = power << 1)
    {
        // This for loop used shifting for powers of 2, meaning
        // that the value will become 0 after the last shift
        // (from binary 1000...0000 to 0000...0000) then, the 'for'
        // loop will break out.

        if (power == number)
            return true;
        if (power > number)
            return false;
    }
    return false;
}

But then I thought: How about checking if log₂ x is an exactly a round number? When I checked for 2^63+1, Math.Log() returned exactly 63 because of rounding. So I checked if 2 to the power 63 is equal to the original number and it is, because the calculation is done in doubles and not in exact numbers.

private bool IsPowerOfTwo_2(ulong number)
{
    double log = Math.Log(number, 2);
    double pow = Math.Pow(2, Math.Round(log));
    return pow == number;
}

This returned true for the given wrong value: 9223372036854775809.

Is there a better algorithm?

Solution 1:

There's a simple trick for this problem:

bool IsPowerOfTwo(ulong x)
{
    return (x & (x - 1)) == 0;
}

Note, this function will report true for 0, which is not a power of 2. If you want to exclude that, here's how:

bool IsPowerOfTwo(ulong x)
{
    return (x != 0) && ((x & (x - 1)) == 0);
}

Explanation

First and foremost the bitwise binary & operator from MSDN definition:

Binary & operators are predefined for the integral types and bool. For integral types, & computes the logical bitwise AND of its operands. For bool operands, & computes the logical AND of its operands; that is, the result is true if and only if both its operands are true.

Now let's take a look at how this all plays out:

The function returns boolean (true / false) and accepts one incoming parameter of type unsigned long (x, in this case). Let us for the sake of simplicity assume that someone has passed the value 4 and called the function like so:

bool b = IsPowerOfTwo(4)

Now we replace each occurrence of x with 4:

return (4 != 0) && ((4 & (4-1)) == 0);

Well we already know that 4 != 0 evals to true, so far so good. But what about:

((4 & (4-1)) == 0)

This translates to this of course:

((4 & 3) == 0)

But what exactly is 4&3?

The binary representation of 4 is 100 and the binary representation of 3 is 011 (remember the & takes the binary representation of these numbers). So we have:

100 = 4
011 = 3

Imagine these values being stacked up much like elementary addition. The & operator says that if both values are equal to 1 then the result is 1, otherwise it is 0. So 1 & 1 = 1, 1 & 0 = 0, 0 & 0 = 0, and 0 & 1 = 0. So we do the math:

100
011
----
000

The result is simply 0. So we go back and look at what our return statement now translates to:

return (4 != 0) && ((4 & 3) == 0);

Which translates now to:

return true && (0 == 0);

return true && true;

We all know that true && true is simply true, and this shows that for our example, 4 is a power of 2.

Solution 2:

Some sites that document and explain this and other bit twiddling hacks are:

http://graphics.stanford.edu/~seander/bithacks.html
(http://graphics.stanford.edu/~seander/bithacks.html#DetermineIfPowerOf2)
http://bits.stephan-brumme.com/
(http://bits.stephan-brumme.com/isPowerOfTwo.html)

And the grandaddy of them, the book "Hacker's Delight" by Henry Warren, Jr.:

http://www.hackersdelight.org/

As Sean Anderson's page explains, the expression ((x & (x - 1)) == 0) incorrectly indicates that 0 is a power of 2. He suggests to use:

(!(x & (x - 1)) && x)

to correct that problem.

Solution 3:

return (i & -i) == i

Solution 4:

bool IsPowerOfTwo(ulong x)
{
    return x > 0 && (x & (x - 1)) == 0;
}