## Introduction :

In this C++ program, we will learn how to check if a number is power of *2* or not using its binary representation. The first method will use bitwise *AND* operation and the second one will count the number of *1* in its binary form.

### Method 1: Using bitwise operation :

Letâ€™s take a look at the binary representation of *0* to *16.*

```
0 0
1 1 *
2 10 *
3 11
4 100 *
5 101
6 110
7 111
8 1000 *
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
16 10000 *
```

The rows with star marked are the rows for the power of *2*. As you can see here, if a number is *n* power of *2*, its binary representation will be *1* followed by *n* times *0*. For example, *16* is *2* to the power *4*. So, its binary representation is *10000* or *1* followed by *4* zeros.

One more thing to notice here is that the number prior to the power of *2* has its binary representation with only *1*. Check *1, 7* and *15* in the above list. *0* is an exception here.

Using the above two concepts, we can find out if a number is the power of two or not using a single logical *AND* operation. If a number is the power of *2*, we can do one logical *AND* with the number and the previous number and the result will be *0* always. For example :

```
15 : 0 1 1 1 1
16 : 1 0 0 0 0
--------------
& : 0 0 0 0 0
```

Here, the logical *AND* of *15* and *16* is *0*.

### C++ program :

In C++, for *AND* operation, *&* is used. We can write the above logic in code like below :

```
#include <iostream>
using namespace std;
int main()
{
int number;
cout << "Enter a positive number : "; cin >> number;
if (number <= 0)
{
cout << "Please enter a valid number.";
}
else
{
if ((number & (number - 1)) == 0)
{
cout << number << " is a power of two number" << endl;
}
else
{
cout << number << " is not a power of two number" << endl;
}
}
return 0;
}
```

### Sample Output :

```
Enter a positive number : 16
16 is a power of two number
Enter a positive number : 22
16 is not a power of two number
```

### Method 2: By counting the number of 1 in its binary form :

We have seen before that if a number is the power of two, it has only one *1* in its binary form. So, we can calculate the total number of *1* in its binary to find out if it is the power of two or not.

The easiest way to calculate the total number of *1* is to keep calculating the *AND* of the number and its previous number until the result is *0*. The number of steps required to get *0* is the total number of *1*.

### C++ program :

```
#include <iostream>
using namespace std;
int main()
{
int number;
cout << "Enter a positive number : "; cin >> number;
if (number <= 0)
{
cout << "Please enter a valid number.";
}
else
{
int count = 0;
while(number != 0){
number = number & (number - 1);
count ++;
}
cout << count<<endl;
if (count == 1)
{
cout << "It is a power of two" << endl;
}
else
{
cout << "It is not a power of two" << endl;
}
}
return 0;
}
```

Actually, this is the same as the first method. If the value of count is greater than *1* in the while loop, we can just skip and say that this is not a power of two. In the previous solution, we were checking if the *AND* of a number and its previous number is *0* or not i.e. if the value of *count* is *1* or not as per the second solution.

It will give the same result for a number as the first solution.

### Conclusion :

We have learned two different ways to verify if a number is the power of two or not. Try to run these examples on your machine and drop one comment below if you have any queries.