How does shovel (<<) operator work in Ruby Hashes?

Question

I was going through Ruby Koans tutorial series, when I came upon this in about_hashes.rb:

def test_default_value_is_the_same_object
  hash = Has         


        

Answer 1

When you're doing hash = Hash.new([]) you are creating a Hash whose default value is the exact same Array instance for all keys. So whenever you are accessing a key that doesn't exist, you get back the very same Array.

h = Hash.new([])
h[:foo].object_id # => 12215540
h[:bar].object_id # => 12215540

If you want one array per key, you have to use the block syntax of Hash.new:

h = Hash.new { |h, k| h[k] = [] }
h[:foo].object_id # => 7791280
h[:bar].object_id # => 7790760

Edit: Also see what Gazler has to say with regard to the #<< method and on what object you are actually calling it.

Answer 2

You have mixed up the way this works a bit. First off, a Hash doesn't have a << method, that method in your example exists on the array.

The reason your code is not erroring is because you are passing a default value to your hash via the constructor. http://ruby-doc.org/core-1.9.3/Hash.html#method-c-new

hash = Hash.new([])

This means that if a key does not exist, then it will return an array. If you run the following code:

hash = {}
hash[:one] << "uno"

Then you will get an undefined method error.

So in your example, what is actually happening is:

hash = Hash.new([])

hash[:one] << "uno"   #hash[:one] does not exist so return an array and push "uno"
hash[:two] << "dos"   #hash[:two] does not exist, so return the array ["uno"] and push "dos"

The reason it does not return an array with one element each time as you may expect, is because it stores a reference to the value that you pass through to the constructor. Meaning that each time an element is pushed, it modifies the initial array.