# How to check if a number is power of 2 or not in C++ (different methods)?

In this article, we are going t learn how to check, whether a given number is power of 2 or not using C++ program? Here, we are using different 4 methods to check it.
Submitted by Shubham Singh Rajawat, on February 27, 2018

Suppose if a number N is given and you have to find out if N is power of 2 or not.

There are many solutions to this problem

1) By simply repeatedly diving N by 2 if N is even number. If it end up at 1 then N is power of 2

```#include <iostream>

using namespace std;
int main()
{
int n;
cout<<"Enter the number :";
cin>>n;

if(n>0)
{
while(n%2 == 0)
{
n/=2;
}
if(n == 1)
{
cout<<"Number is power of 2"<<endl;
}
}
if(n == 0 || n != 1)
{
cout<<"Number is not power of 2"<<endl;
}
return 0;
}
```

Output

```	First run:
Enter the number :10
Number is not power of 2

Second run:
Enter the number :16
Number is power of 2
```

2) By taking log2 of N and then pass it to floor and ceil if both gives same result then N is power of 2

```#include <iostream>
#include <cmath>
using namespace std;

int main()
{
int n;
cout<<"Enter the number :";
cin>>n;
if(ceil(log2(n))== floor(log2(n)))
{
cout<<"Number is power of 2"<<endl;
}
else
{
cout<<"Number is not power of 2"<<endl;
}

}
```

Output

```	First run:
Enter the number :10
Number is not power of 2

Second run:
Enter the number :16
Number is power of 2
```

3) By using bit manipulation

```    Suppose N = 8 = (1000)2
Then N-1 = 7 = (0111)2
N & (N-1)= (1000)2 & (0111)2 = (0000)2

N = 5 = (0101)2
N-1 = 4 = (0100)2
N & (N-1) = (0101)2 & (0100)2 = (0001)2
```

If a number is power of 2 then in binary representation the count of 1 will be one.

```#include <iostream>
using namespace std;

int main()
{
int n;
cout<<"Enter the number :";
cin>>n;
if(n != 0 && (n & (n-1)) == 0)
{
cout<<"Number is power of 2"<<endl;
}
else
{
cout<<"Number is not power of 2"<<endl;
}
}
```

Output

```	First run:
Enter the number :10
Number is not power of 2

Second run:
Enter the number :16
Number is power of 2
```

4) By counting the number of 1’s in the binary form of N. If count is 1 then N is power of 2.

```#include <iostream>
using namespace std;

int main()
{
int n,count1=0;
cout<<"Enter the number :";
cin>>n;
while(n)
{
n = n & (n-1);
count1++;
}
if(count1 == 1)
{
cout<<"Number is power of 2"<<endl;
}
else
{
cout<<"Number is not power of 2"<<endl;
}

}
```

Output

```	First run:
Enter the number :10
Number is not power of 2

Second run:
Enter the number :16
Number is power of 2
```

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