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 = 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

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.

Answer 2

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.