Check whether given number is Power of 2 using bitwise operators in C

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Problem : Check whether given number is power of 2 using bitwise operators in C

Algorithm :

Let us see the binary representation of numbers which are power of 2:

2     - 10 
4     - 100
8     - 1000
16    - 10000
32    - 100000

As we can see in above example,any number which is power of 2 has only 1 bit set. Now let us subtract 1 from above power of 2 numbers. 

1    - 01 
3    - 011
7    - 0111
15   - 01111
31   - 011111

Now we see that when we subtract by 1, we get all 1’s expect for which was originally set for power of 2.

So when we do bitwise-AND between a number which is power of 2 and number which is power of 2 minus 1, we will get 0. 

// C program to check whether the given number is power of 2 using bitwise operators
#include <stdio.h>
#include <stdbool.h> //for bool

/*Function which returns true if given number is power of 2*/
bool power_of_2(int number)
{

        /*If the number is 1, directly return false*/
        if(number == 1)
                return false;

        /*Check whether the number is power of 2*/
        if((number & (number - 1)) != 0)
                return false;

        /*return true when number is power of 2*/
        return true;
}

int main() {
        int number = 0;
        bool result = false;

        printf("\r\nEnter a number:");
        scanf("%d", &number);

        result = power_of_2(number);

        if(result == true)
                printf("\r\n%d is Power of 2\n", number);
        else
                printf("\r\n%d is NOT Power of 2\n", number);

        return 0;
}

You can check more about power of 2 here : https://en.wikipedia.org/wiki/Power_of_two

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