Algorithms Quiz | Bit Algorithms | Question 4

Consider the following code snippet for checking whether a number is power of 2 or not. 

C

/* Incorrect function to check if x is power of 2*/
bool isPowerOfTwo (unsigned int x)
{
  return (!(x&(x-1)));
}

What is wrong with above function?

(A)

It does reverse of what is required

(B)

It works perfectly fine for all values of x.

(C)

It does not work for x = 0

(D)

It does not work for x = 1

Answer: (C)

Explanation:

The logic behind this approach is that if a number is a power of 2, it will have only one bit set to 1 in its binary representation, and subtracting 1 from it will result in a binary number with all bits flipped to the right of that one bit. Performing a bitwise AND between these two numbers will result in 0, indicating that the number is a power of 2. 

However, there is a problem with this approach when the input is 0. In that case, the function will return true, even though 0 is not a power of 2. This is because 0 is the only number whose binary representation has no bits set to 1, so subtracting 1 from it will result in a binary number with all bits set to 1, and performing a bitwise AND between 0 and (0-1) will also result in 0. 

Therefore, the given function is incorrect as it does not handle the special case of 0. To fix this, we need to add a separate check for 0 at the beginning of the function and return false if the input is 0.

Please see https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/

Hence Option (C) is the correct answer.

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