Finding if a number is the power of 2 in Scheme

Question

I'm fairly new to Scheme and am attempting to learn it on my own from scratch. I'm stuck on the syntax of this problem. I know that if I want to find out if a number is a power of 2, in C for instance, I would simply do:

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

which would return true or false. How would I be able to convert this into a couple simple lines in Scheme?

Answer 1

I'll give you a hint since you're trying to learn the language.

Scheme has a function called (bitwise-and ...) which is equivalent to the & operator in C (there is also (bitwise-xor ...), (bitwise-not ..), etc., which do the expected thing).

(Here is the documentation of the (bitwise-and ...) function)

Given that, would you be able to translate what you've written in your question into Scheme code?

N.B: For a problem like this, you really don't need to resort to bitwise operations when using Scheme. Realistically, you should be writing a (possibly probably tail) recursive function that will compute this for you.

Answer 2

Scheme also has biwise operators.

But if you really want to develop your scheme skills, you should write a function that decides if an integer is a power of 2 by recursively dividing it by 2 until you are left with either 2 or with an odd number. This would be very inefficient, but really cool nonetheless.

Answer 3

You can do this using a built in bitwise operator.

(define (pow2? x)
  (= (bitwise-and x (- x 1))
     0))