**I**n this post will be discussing how to calculate if a number is a power of two or not. As an example, 8 is a power of two but the number 10 is not.

There are many ways we can solve this.

**First**, we will take an approach which is simple and iterative. In this case, we will calculate the power of two one by one and check with the supplied number. The below code illustrates it.

bool isPowerofTwo(unsigned num)

{

auto y = 1;

while (0 != y)

{

if (num == y) return true;

if (num < y) return false;

y <<= 1;

}

return false;

}

**Second**, assuming, the number is a 32-bit number, this is also an iterative solution. In this scenario, iterating all bits and counting the set bits. Any number which is a power of 2 will have only one bit set and the rest will be zeros. As an example, 8 in binary representation is 1000. Using this observation, we can implement an iterative solution.

bool isPowerofTwo(unsigned num)

{

auto one_count = 0;

for (auto index = 0; index < 32; ++index) {

if ( num & 1) ++one_count;

num >>= 1;

}

return one_count == 1;

}

So, from the client code, the above code returns true for the number 8388608.

std::cout << "Is 8,388,608 power of two? Ans:- " << std::boolalpha << isPowerofTwo(8388608) << "\n";

Though the above solution works, however, both solutions are not efficient as it's looping through.

So, we can use a trick, as we know if a number is a power of two, in that case, only one bit will be set. Hence, here is the

**third**solution and it's the most optimized one:bool isPowerOfTwo(unsigned num)

{

return (num & (num - 1)) == 0;

}

Let the number be 8 and its bit representation is 1000 and (num - 1) = 7. 7's bit representation is 0111. The bitwise '&' result is 0000 = 0. Hence, 8 is a power of two.

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