Find if a number is a power of two without math function or log function

Question

I want to find if a user entered number is a power of two or not.

My code doesn\'t work.

public class power_of_two
{  
    public static void main         


        

Answer 1

You can use the bitwise AND (&) operator:

return (num & -num) == num

Why this works?

Consider the number 8, what it is in binary (assuming 32-bits)?

0000 0000 0000 0000 0000 0000 0000 1000

Now let's see how -8 is represented? 1

1111 1111 1111 1111 1111 1111 1111 1000

Finally.. let's calculate 8 & -8:

0000 0000 0000 0000 0000 0000 0000 1000   8
↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓   &
1111 1111 1111 1111 1111 1111 1111 1000  -8
---------------------------------------
0000 0000 0000 0000 0000 0000 0000 1000   8    ¯\_(ツ)_/¯

Now let's take another example, let's say 7, which is not power of two.

0000 0000 0000 0000 0000 0000 0000 0111   7
↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓ ↓↓↓↓   &                       
1111 1111 1111 1111 1111 1111 1111 1001  -7
---------------------------------------
0000 0000 0000 0000 0000 0000 0000 0001  != 7  ¯\_(ة_ة)_/¯

As mentioned by @arshajii, think what will happen if num is zero.. I'll leave the solution for you :)

¹ A good way to remember how to calculate that: Begin from the rightmost bit, for each 0 you see, don't change it, when you see 1, leave it and proceed, but from now on, invert all bits. I tried to explain this more here.