# 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

 `// C Program to find whether a ` `// no is power of two ` `#include ` `#include ` `#include ` ` `  `/* 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) ? ``printf``(``"Yes\n"``) : ``printf``(``"No\n"``); ` `    ``isPowerOfTwo(64) ? ``printf``(``"Yes\n"``) : ``printf``(``"No\n"``); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by bibhudhendra `

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.

## C

 `#include ` `#include ` ` `  `/* 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; ` `} `

Output:

```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 –> 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).
Below is the implementation of this method.

## C

 `#include ` `#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; ` `} `

Output:

```No
Yes
```

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

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