# How to check if a number is a power of 2

## 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 `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?