Create a method to find if a number is a power of 2?

Question

I have this code to return true if num is a power of 2.

def is_power_of_two?(num)
  result = num.inject(0) {|n1, n2| n2 ** n1}
  if result == num
    true
  else
    false
  end
end

p is_power_of_two?(16)

I keep getting an error though. How could I fix and simplify this code?

Answer 1

Clearly, n is a non-negative integer.

Code

def po2?(n)
  n.to_s(2).count('1') == 1
end

Examples

po2?  0     #=> false
po2?  1     #=> true
po2? 32     #=> true
po2? 33     #=> false

Explanation

Fixnum#to_s provides the string representation of an integer (the receiver) for a given base. The method's argument, which defaults to 10, is the base. For example:

16.to_s     #=> "16" 
16.to_s(8)  #=> "20" 
16.to_s(16) #=> "10"
15.to_s(16) #=>  "f"

It's base 2 we're interested in. For powers of 2:

 1.to_s(2)  #=>      "1" 
 2.to_s(2)  #=>     "10" 
 4.to_s(2)  #=>    "100" 
 8.to_s(2)  #=>   "1000"
16.to_s(2)  #=>  "10000"

For a few natural numbers that are are not powers of 2:

 3.to_s(2)  #=>    "11" 
 5.to_s(2)  #=>   "101" 
11.to_s(2)  #=>  "1011" 

We therefore wish to match binary strings that contain one 1.

Another Way

R = /
    \A      # match beginning of string ("anchor")
    10*     # match 1 followed by zero or more zeroes
    \z      # match end of string ("anchor")
    /x      # free-spacing regex definition mode

def po2?(n)
  (n.to_s(2) =~ R) ? true : false
end

po2?(4)  #=> true
po2?(5)  #=> false

One more

This uses Fixnum#bit_length and Fixnum#[]:

def po2?(n)
  m = n.bit_length-1
  n[m] == 1 and m.times.all? { |i| n[i].zero? }
end

po2?  0     #=> false
po2?  1     #=> true
po2? 32     #=> true
po2? 33     #=> false

And a fourth, perhaps the best

def po2?(n)
  return false if n <= 0
  while n.even?
    n /= 2
  end
  n == 1
end

po2?  0     #=> false
po2?  1     #=> true
po2? 32     #=> true
po2? 33     #=> false