# 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 2^{2}= 4 Input : n = 7 Output : No Input : n = 32 Output : Yes 2^{5}= 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 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

**Time Complexity: **O(log_{2}n)

**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;` `}` |

**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 <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;` `}` |

**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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