Bitwise Operators: Perform multiplication operation using Bitwise operators

The solution is very simple. We know the left shift operator is a 2 raised to the power of n.

Hence we do num << n, should give us the expected result.

Here we check the number of zero’s before the set bit. If the number of 0’s are even then it is a power of 4 else not.

16 in binary 10000

The number of 0’s is even before the set bit “1”. Hence it is a power of 4.

#include<iostream>
#include<vector>
#include<string>

using namespace std;

void multiply_2_n(int num, int n)
{
	cout<<"The number "<<num<< " raised to 2 power of "<<n<<" is = "<< (num <<n)<<endl;

}

void check_num_power_of_4(int num)
{
	int count = 0; 

	if ( num && !(num&(num-1)) ) 
	{ 
		while(num > 1) 
		{ 
			num  >>= 1; 
			count += 1; 
		}      
	} 

	if (count%2 == 0)
		cout<<"The number is a power of 4"<<endl;
	else
		cout<<"The number is not a power of 4"<<endl;

}

int main()
{
	int num = 16;
	int n = 2;
	
	multiply_2_n(num, n);
	check_num_power_of_4(num);

	return 0;

}

Output:

The number 16 raised to 2 power of 2 is = 64
The number is a power of 4