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) 

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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.

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Quick sanity check:

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

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