http://graphics.stanford.edu/~seander/bithacks.html#DetermineIfPowerOf2
You have many useful bit hacks here.
Visar Zejnullahu
On Sat, Sep 1, 2012 at 1:17 AM, Visar Zejnullahu wrote:
> 2^n in binary is 10...0 (with n 0s), and 2^n - 1 is 11...1 (with n 1s). So
> if you do bitwise and (&am
2^n in binary is 10...0 (with n 0s), and 2^n - 1 is 11...1 (with n 1s). So
if you do bitwise and (&) to 2^n and 2^n-1 you get all 0s. That's why you
check if (num & (num - 1)) == 0.
Visar Zejnullahu
On Sat, Sep 1, 2012 at 12:54 AM, Lazar wrote:
> Hello,
>
> I'm fai
