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

After posting the question I thought of the following solution:

We need to check if exactly one of the binary digits is one. So we simply shift the number right one digit at a time, and return true if it equals 1. If at any point we come by an odd number ((number & 1) == 1), we know the result is false. This proved (using a benchmark) slightly faster than the original method for (large) true values and much faster for false or small values.

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;
}

---

Of course, Greg's solution is much better.

Answer 2

Find if the given number is a power of 2.

#include <math.h>

int main(void)
{
    int n,logval,powval;
    printf("Enter a number to find whether it is s power of 2\n");
    scanf("%d",&n);
    logval=log(n)/log(2);
    powval=pow(2,logval);

    if(powval==n)
        printf("The number is a power of 2");
    else
        printf("The number is not a power of 2");

    getch();
    return 0;
}

Answer 3

This is another method to do it as well

package javacore;

import java.util.Scanner;

public class Main_exercise5 {
    public static void main(String[] args) {
        // Local Declaration
        boolean ispoweroftwo = false;
        int n;
        Scanner input = new Scanner (System.in);
        System.out.println("Enter a number");
        n = input.nextInt();
        ispoweroftwo = checkNumber(n);
        System.out.println(ispoweroftwo);
    }
    
    public static boolean checkNumber(int n) {
        // Function declaration
        boolean ispoweroftwo= false;
        // if not divisible by 2, means isnotpoweroftwo
        if(n%2!=0){
            ispoweroftwo=false;
            return ispoweroftwo;
        }
        else {
            for(int power=1; power>0; power=power<<1) {
                if (power==n) {
                    return true;
                }
                else if (power>n) {
                    return false;
                }
            }
        }
        return ispoweroftwo;
    }
}

Answer 4

A number is a power of 2 if it contains only 1 set bit. We can use this property and the generic function countSetBits to find if a number is power of 2 or not.

This is a C++ program:

int countSetBits(int n)
{
        int c = 0;
        while(n)
        {
                c += 1;
                n  = n & (n-1);
        }
        return c;
}

bool isPowerOfTwo(int n)
{        
        return (countSetBits(n)==1);
}
int main()
{
    int i, val[] = {0,1,2,3,4,5,15,16,22,32,38,64,70};
    for(i=0; i<sizeof(val)/sizeof(val[0]); i++)
        printf("Num:%d\tSet Bits:%d\t is power of two: %d\n",val[i], countSetBits(val[i]), isPowerOfTwo(val[i]));
    return 0;
}

We dont need to check explicitly for 0 being a Power of 2, as it returns False for 0 as well.

OUTPUT

Num:0   Set Bits:0   is power of two: 0
Num:1   Set Bits:1   is power of two: 1
Num:2   Set Bits:1   is power of two: 1
Num:3   Set Bits:2   is power of two: 0
Num:4   Set Bits:1   is power of two: 1
Num:5   Set Bits:2   is power of two: 0
Num:15  Set Bits:4   is power of two: 0
Num:16  Set Bits:1   is power of two: 1
Num:22  Set Bits:3   is power of two: 0
Num:32  Set Bits:1   is power of two: 1
Num:38  Set Bits:3   is power of two: 0
Num:64  Set Bits:1   is power of two: 1
Num:70  Set Bits:3   is power of two: 0

Answer 5

return ((x != 0) && !(x & (x - 1)));

If x is a power of two, its lone 1 bit is in position n. This means x – 1 has a 0 in position n. To see why, recall how a binary subtraction works. When subtracting 1 from x, the borrow propagates all the way to position n; bit n becomes 0 and all lower bits become 1. Now, since x has no 1 bits in common with x – 1, x & (x – 1) is 0, and !(x & (x – 1)) is true.

Answer 6

Mark gravell suggested this if you have .NET Core 3, System.Runtime.Intrinsics.X86.Popcnt.PopCount

public bool IsPowerOfTwo(uint i)
{
    return Popcnt.PopCount(i) == 1
}

Single instruction, faster than (x != 0) && ((x & (x - 1)) == 0) but less portable.