C++ Program to find whether a no is power of two

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++

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// C++ Program to find whether a
// no is power of two
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if x is power of 2
bool isPowerOfTwo(int n)
{
    return (ceil(log2(n)) == floor(log2(n)));
}
  
// Driver program
int main()
{
    isPowerOfTwo(31) ? cout << "Yes" << endl : cout << "No" << endl;
    isPowerOfTwo(64) ? cout << "Yes" << endl : cout << "No" << endl;
  
    return 0;
}
  
// This code is contributed by Surendra_Gangwar

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Output:

No
Yes

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.

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 –> 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 https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).

Please refer complete article on Program to find whether a no is power of two for more details!



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