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 `log`

is an exactly round number? But when I checked for 2^63+1, _{2} x`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);
}
```

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.

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.:

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.

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