C Program to find whether a no is power of two - GeeksforGeeks

19 Jul, 2021
Given a positive integer, write a function to find if it is a power of two or not.Examples :  
Input : n = 4
Output : Yes
22 = 4

Input : n = 7
Output : No

Input : n = 32
Output : Yes
25 = 32
1. 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.  
C
// C Program to find whether a// no is power of two#include <math.h>#include <stdbool.h>#include <stdio.h> /* Function to check if x is power of 2*/bool isPowerOfTwo(int n){    return (ceil(log2(n)) == floor(log2(n)));} // Driver programint main(){    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");    return 0;} // This code is contributed by bibhudhendra
No
Yes
 
Time Complexity: O(log2n)
Auxiliary Space: O(1)
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2. 
C
#include <stdbool.h>#include <stdio.h> /* Function to check if x is power of 2*/bool isPowerOfTwo(int n){    if (n == 0)        return 0;    while (n != 1) {        if (n % 2 != 0)            return 0;        n = n / 2;    }    return 1;} /*Driver program to test above function*/int main(){    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");    return 0;}
No
Yes
 
3. 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. Please see Count set bits in an integer for counting set bits.4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.For example for 4 ( 100) and 16(10000), we get following after subtracting 1 3 –> 011 15 –> 01111So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case). Below is the implementation of this method.  
C
#include <stdio.h>#define bool int /* Function to check if x is power of 2*/bool isPowerOfTwo(int x){    /* First x in the below expression is for the case when x is 0 */    return x && (!(x & (x - 1)));} /*Driver program to test above function*/int main(){    isPowerOfTwo(31) ? printf("Yes\n") : printf("No\n");    isPowerOfTwo(64) ? printf("Yes\n") : printf("No\n");    return 0;}
No
Yes
 
Please refer complete article on Program to find whether a no is power of two for more details! 
souravmahato348
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