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

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..

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);
}

1
2
4
8
16
32
64

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.

double input = 22;

while(((input != 2) && input % 2 == 0) || input == 1) {
 input = input /2;
}

return input == 2;

Keep dividing it by 2 until you reach 1 or an odd number. If it reaches 1 it's a power of 2, otherwise it isn't.

The straightforward solution:

bool isPowerOfTwo(int n) {
    // All values < 1 cannot be (positive, at least) powers of two.
    if (n < 1) return false;

    // Keep shifting bits.
    while (n > 1) {
        // Since n > 1, the low bit set means at least two bits must
        // be set, making n no longer a power of two.
        if (n & 0x1) return false;
        // Throw away the bottom (zero) bit.
        n >>= 1;
    }
    // Only one bit was set, therefore, n is a power of two.
    return true;
}

Of course, this is not as optimal as some of the other bit-trickery solutions (which are indeed quite clever), but it's very easy to see how it works, and verify it works in your head.

For the input 4, we get:

n = 4 (0x100)
run loop
n = 2 (0x10)
run loop
n = 1 (0x1)
return true

For an invalid input, like 5, we get:

n = 5 (0x101)
return false (0x101 & 0x1 => 0x1, which is truthy)