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 = Hash.new([])

  hash[:one] << "uno"
  hash[:two] << "dos"

  assert_equal ["uno", "dos"], hash[:one]
  assert_equal ["uno", "dos"], hash[:two]
  assert_equal ["uno", "dos"], hash[:three]

  assert_equal true, hash[:one].object_id == hash[:two].object_id
end

The values in assert_equals, is actually what the tutorial expected. But I couldn’t understand how there is a difference between using << operator and = operator?

My expectation was that:

• hash[:one] would be ["uno"]
• hash[:two] would be ["dos"]
• hash[:three] would be []

Can someone please explain why my expectation was wrong?

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.