java - 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(String args[])  
    {  

        Scanner in=new Scanner(System.in);
        System.out.println("Enter the number : ");
        int num = in.nextInt();

        int other = 1;  
        if(((~num) & 1) == 1)  
        {  
            System.out.println("The number is a power of two");  
        }  
        else  
        {
            System.out.println("The number is a  NOT A power of two");  
        }
    }  
} 

Let me know how can I find the power of two number.
For example 8 is a power of 2.
22 is not a power of 2, etc..

Answer 1

You can test if a positive integer n is a power of 2 with something like

(n & (n - 1)) == 0

If n can be non-positive (i.e. negative or zero) you should use

(n > 0) && ((n & (n - 1)) == 0)

---

If n is truly a power of 2, then in binary it will look like:

10000000...

so n - 1 looks like

01111111...

and when we bitwise-AND them:

  10000000...
& 01111111...
  -----------
  00000000...

Now, if n isn't a power of 2, then its binary representation will have some other 1s in addition to the leading 1, which means that both n and n - 1 will have the same leading 1 bit (since subtracting 1 cannot possibly turn off this bit if there is another 1 in the binary representation somewhere). Hence the & operation cannot produce 0 if n is not a power of 2, since &ing the two leading bits of n and n - 1 will produce 1 in and of itself. This of course assumes that n is positive.

This is also explained in "Fast algorithm to check if a positive number is a power of two" on Wikipedia.

---

Quick sanity check:

for (int i = 1; i <= 100; i++) {
    if ((i & (i - 1)) == 0)
        System.out.println(i);
}

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