How to check if a number is a power of 2

Question

Today I needed a simple algorithm for checking if a number is a power of 2.

The algorithm needs to be:

1. Simple
2. Correct for any ulong

Answer 1

This program in java returns "true" if number is a power of 2 and returns "false" if its not a power of 2

// To check if the given number is power of 2

import java.util.Scanner;

public class PowerOfTwo {
    int n;
    void solve() {
        while(true) {
//          To eleminate the odd numbers
            if((n%2)!= 0){
                System.out.println("false");
                break;
            }
//  Tracing the number back till 2
            n = n/2;
//  2/2 gives one so condition should be 1
            if(n == 1) {
                System.out.println("true");
                break;
            }
        }
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner in = new Scanner(System.in);
        PowerOfTwo obj = new PowerOfTwo();
        obj.n = in.nextInt();
        obj.solve();
    }

}

OUTPUT : 
34
false

16
true

Answer 2

Here's a simple C++ solution:

bool IsPowerOfTwo( unsigned int i )
{
    return std::bitset<32>(i).count() == 1;
}

Answer 3

int isPowerOfTwo(unsigned int x)
{
    return ((x != 0) && ((x & (~x + 1)) == x));
}

This is really fast. It takes about 6 minutes and 43 seconds to check all 2^32 integers.

Answer 4

Here is another method I devised, in this case using | instead of & :

bool is_power_of_2(ulong x) {
    if(x ==  (1 << (sizeof(ulong)*8 -1) ) return true;
    return (x > 0) && (x<<1 == (x|(x-1)) +1));
}

Answer 5

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

Answer 6

bool isPowerOfTwo(int x_)
{
  register int bitpos, bitpos2;
  asm ("bsrl %1,%0": "+r" (bitpos):"rm" (x_));
  asm ("bsfl %1,%0": "+r" (bitpos2):"rm" (x_));
  return bitpos > 0 && bitpos == bitpos2;
}