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.

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.

#include<stdio.h>
#define bool int

/* 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(17)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(16)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(2)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(18)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(1)? printf("Yes\n"): printf("No\n");
  return 0;
}

Output:

No
No
Yes
Yes
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 http://geeksforgeeks.org/?p=1176 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 –> 01111

So, 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 Mohammad for adding this case).
Below is the implementation of this method.

#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(17)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(16)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(2)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(18)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(1)? printf("Yes\n"): printf("No\n");
  return 0;
}

Output:

No
No
Yes
Yes
No
Yes

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