# Hi, I'm Richard Macias.

Software Developer, Teacher, Gamer, Musician, technology enthusiast.

# [Bit Manipulation] Bitwise Operations in Ruby

So today I decided to learn about bit manipulation. I only learned about them today so i’ll be keeping it super simple K.I.S.S.. Lets start with setting up our playground for learning.

``````1.to_s(2)  #=> "1"
2.to_s(2)  #=> "01"
20.to_s(2) #=> "10100"
``````

As shown above, you can convert an integer to a string of binary using the Fixnum#to_s method.

For the sake of convenience, we’re gonna make a new method for the Integer class

``````class Integer
def bits
self.to_s(2)
end
end

a = 20
a.to_s(2) #=> "10100"
a.bits    #=> "10100"
``````

In case this confused you, we’re just adding a method to the Integer class in Ruby. Cool? Cool, moving on!

## AND & OR operators

Lets quickly go over AND & OR operators. In binary, `1` represents `true` and `0` represents `false`

``````true && true    #=> true
true && false   #=> false
false && true   #=> false
false && false  #=> false

true || true    #=> true
true || false   #=> true
false || true   #=> true
false || false  #=> false
``````

This will be important as we move forward. You do not need to be a Ruby guru to learn bit manipulation, but if these didn’t make sense, I recommend going on Code Academy and taking the ruby course or simply opening `irb` and coming back since this will be pretty important moving forward.

## AND operator &

The AND operater enumerates through the binary and returns values where both integers are equal to `1` at both indexes. If they don’t match, they will be set to `0`

``````(a = 20).bits   #=> "10100"
(b = 17).bits   #=> "10001"
(a & b).bits    #=> "10000"

# index 0 was the only index where the value was 1 for both arguments
# thus everything else was set to 0, returning 10000
``````

## OR operator |

The OR operater enumerates through the binary and returns values where either integers are equal to `1` at that index. They will set to `0` only if both are 0

``````(a = 20).bits   #=> "10100"
(b = 17).bits   #=> "10001"
(a | b).bits    #=> "10101"

# index 0, 2 and 4 all had values of 1 in either comparisons
# they returned true and resulted in "10101"
``````

## The XOR operator ^

The XOR operator will specifically take the value where values do not match, return `true` and set that value to 1. It will also omit any leading 0’s in the process

``````(a = 20).bits   #=> "10100"
(b = 17).bits   #=> "10001"
(a ^ b).bits    #=> "101"

#index 0: matched, it's set to false
#index 1: are both false, moving on
#index 2: one of the oporators are true, other is false. The XOR operator registers this as a true value and returns 1
#index 3: are both false, moving on
#index 4: same as index(2), registers true, sets it to one
#
#we now have 00101, it will omit the leading 0's and return "101"
``````

## The NOT operator ~

The NOT operator flips

``````(a=17).bits #=> "10001"
~a          #=> -18
(~a).bits   #=> "-10001"
``````

Alright, so that’s not working like we expected… It’s literally just prepending a `-` into the string

``````-17.bits        #=> "-10001"
"-" + 17.bits   #=> "-10001"
``````

Here are some examples of positive and negative integers and their binary:

binary - - decimal
`0000` `0`
`0001` `1`
`0010` `2`
`0111` `7`
`1111` `-1`
`1110` `-2`
`1101` `-3`
`1000` `-8`

So it looks like our `#to_s(2)` method isn’t going to work here. We’ll have to figure something else out.

While it’s true that represent negative numbers for human readable integers, the is impossible in computer land. There is an alternative however.

Using the two’s complement method, we have a clever way of getting negative integers following these simple rules.

• for zero, use all 0’s
• for positive integers, start counting up, with a max of 2(number of bits - 1 ) -1.
• for negative integers, do exactly the same but switch the role of 0’s and 1’s (so instead of starting with `0000`, you’ll start with `1111`, that’s the “complement” part).

With this in mind, let’s sample this with a nibble (1/2 byte)

• `0000` - `0`
• `0001` - `1`
• `0010` - `2`
• `0011` - `3`
• `0111` - `7`

That’s as far as you can go with positives 2^(3)-1 = 7 As for negatives

• `1111` - `-1`
• `1110` - `-2`
• `1101` - `-3`
• `1000` - `-8`

It’s a bit confusing to look at for a moment, but if you stop to think about it, it actually makes a lot of sense.

## Left and right shift operators

Left and right shift operators shift integer bits to the left or right by the given number of positions.

``````a = 18

(a >> 2).to_s(2)       #=>     "100"
(a >> 1).to_s(2)       #=>    "1001"
(a).to_s(2)            #=>   "10010"
(a << 1).to_s(2)       #=>  "100100"
(a << 2).to_s(2)       #=> "1001000"
``````

Notice what’s happening here? By shifting to the right, we shorten the bits. By shifting left, we extend the bits, filling the new bits with `0`’s.

## That’s it!

I learned this pretty quickly and decided to write a post about it same day. If you have any questions or feedback, please email me. 