Code to check if an integer is a power of 2 or not in a single line

Problem

Check if integer is power of 2 or not

FOLLOW UP - Preferable solution to get it in 1 line.

Solution

Method1
All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. We have discussed how to count number of bits set here.

Method 2
Note that the bit pattern of a power of two is of the form 10…0 and that of a number just one less is 011…1. Hence if we have & operator between the 2, we will get 0.We have discussed the solution here. Also, as 0 is not in power of 2, we have to explicitly check if n is greater than 0.

C code :

bool isPowerOfTwo(int n)  
{  
    return (n > 0) && ((n & (n - 1)) == 0);  
}

Method 3 - Complement and compare

C code

bool isPow2 = ((x & ~(x-1))==x)? x : 0;

Method 4

C code

bool isPowerOfTwo(int i){  
   return i>0 && ((i & -i) == i);  
}

Method 5 - Shift right

private static bool IsPowerOfTwo(ulong number)  
{  
    while (number != 0)  
    {  
        if (number == 1)  
            return true;  
  
        if ((number & 1) == 1)  
            // number is an odd number and not 1 - so it's not a power of two.  
            return false;  
  
        number = number >> 1;  
    }  
    return false;  
}

Method 6 - Divide the number by 2 until it becomes 0

bool isPowerOfTwo(int n)  
{  
  if (n == 0)  
    return 0;  
  while (n != 1)  
  {  
    if (n%2 != 0)  
      return 0;  
    n = n/2;  
  }  
  return 1;  
}

Method 7 -Take log of number and check if it is integer
A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.

FOLLOW UP - Clearly Method 2,3 and 4 one-liners and are using bit hacks.

Please share if you have some method not covered above.

See also