How to check if a number is a power of 2

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

The algorithm needs to be:

1. Simple
2. 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 round number? But 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 `double`s 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?

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.