From: Josh Cheek <josh.cheek@...>
Date: 2011-12-30T17:59:38+09:00
Subject: Re: Hash access

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On Fri, Dec 30, 2011 at 1:41 AM, Yong Li <gilbertly@gmail.com> wrote:

> Hi Josh,
>
> Here is how Hash in Ruby works when it tries to determine if two keys are
> equal:
> * the #hash method on both objects are called to calculate their hash codes
> * if their hash codes are not equal, they are not equal
> * if their hash codes are equal, then #== is called to determine if
> two objects are equal
>
> In your example, all three objects actually return the same hash
> codes, so #== (instead of eql?) is used to check their equality.
>
> The "first Josh" and the "second Josh" are equal because their #==
> (inherited from Object#==) simply calls #eql? which you have
> overridden to make them equal.
>
> The "first Josh" is not equal to "Josh" because they are of different
> classes, and User#== (inherited from Object#==) does not allow objects
> of different classes to be equal.
>
> As a side note: you should always define #hash and #== together and
> make sure whenever #== returns true #hash mush return the same number,
> otherwise, using these objects as hash keys will break the hash
> semantics. Also, avoid using mutable objects as hash keys unless their
> #hash number is immutable.
>
> I hope this helps
>
>
> On Fri, Dec 30, 2011 at 1:37 PM, Josh Cheek <josh.cheek@gmail.com> wrote:
> > Hi, was playing around with an idea after reading the thread about
> defining
> > #hash. My understanding was that #hash gives a unique identifier, and
> that
> > #eql? allows the hash to determine whether the two objects are equal in
> > terms of being the same hash key. So I wrote some code that should take
> an
> > equivalent instance, or a string for quick access. But it behaves in a
> way
> > that I completely don't understand. Hoping someone can help:
> >
> >
> > User = Struct.new :name, :age, :identifier do
> >  def hash
> >    name.hash
> >  end
> >
> >  def eql?(other)
> >    puts "#{name} was asked if they were equal to #{other.inspect}"
> >    (other == name) || (other.name == name && other.age == age)
> >  end
> > end
> >
> > josh = User.new 'Josh', 28, 'first Josh'
> > hash = {josh => josh}
> >
> > hash[josh]                                # => #<struct User name="Josh",
> > age=28, identifier="first Josh">
> > hash[User.new 'Josh', 28, 'second Josh']  # => #<struct User name="Josh",
> > age=28, identifier="first Josh">
> > hash['Josh']                              # => nil
> >
> > # >> Josh was asked if they were equal to #<struct User name="Josh",
> > age=28, identifier="first Josh">
> >
> >
> >
> > So I would have expected all three to go through eql? Instead, we see
> that
> > only the case where the key was the same object goes through. However, it
> > identifies that the second Josh is the same key, without invoking
> User#eql?
> > How does it do this?
> >
> > And why does the string "Josh" not find the instance?
> >
> > This is all probably in my copy of the Pickaxe, but it's in Chicago and
> I'm
> > out of town :/
>
>
I see. The confusion for me was that the comparison goes in the other
direction. (ie hash["Josh"] turns into "Josh".eql?(#<struct User ...>) but
I was thinking it would be #<struct User ...>.eql?("Josh")). This becomes
apparent if I change the log line to `puts "#{inspect} was asked if they
were #eql? to #{other.inspect}"` I just didn't do that in the name of
brevity, and it masked the discrepancy.

So it's probably implemented something like this (ignoring nuances like
collisions and default values)

# expectation
class Hash
  def [](key)
    potential_key, potential_value = at_hash key.hash
    return potential_value if potential_key.eql? key
  end
end


# actual
class Hash
  def [](key)
    potential_key, potential_value = at_hash key.hash
    return potential_value if key.equal? potential_key
    return potential_value if key.eql? potential_key
  end
end

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