Ruby local_variable keeps referenced to @instance_variable

Question

class Foo
  def bar
    @instance_variable = [['first']]

    # make a duplicate object with the :dup method
    local_variable=@instance_variable.dup

    # They have different object_id
    p @instance_variable.object_id
    p local_variable.object_id


    local_variable.each{|n|n.push('second')}
    @instance_variable
  end
end

f=Foo.new
p f.bar

=> 2000
=> 2002
=> [["first", "second"]]

It seems that the local_variable still references to the @instance_variable, although it is a different object. This behaviour is both with the push and unshift in the each block. With a normal assignment like local_variable='second', the result is as expected => [['first']]

I don't understand why local_variable.each{|n|n.push('second')} has an effect on the @instance_variable

Using Ruby-1.9.2p318

trptcolin · Accepted Answer · 2013-02-10 23:26:14Z

2

Both local_variable and @instance_variable have references to the same object, the inner array ['first']. And because it's a mutable Array, you can effect changes to one array through the other.

Object#dup in Ruby provides a shallow copy. In order to make a deep copy of an Array, you'd need to write some code (or find a library) that recursively walks the data structure, deep-cloning its pieces of mutable state.

answered Feb 10, 2013 at 23:26

trptcolin

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1 Comment

Chuck Over a year ago

The easy way to do a deep copy is to marshal and unmarshal the array. That will completely reconstruct the object graph, assuming everything in there is amenable to marshaling (not every object necessarily is).

matt · Accepted Answer · 2013-02-11 00:21:05Z

1

The problem is you're not testing the right object. You say:

p @instance_variable.object_id
p local_variable.object_id

But that's not the object you're going to push onto. Try this instead:

p @instance_variable[0].object_id
p local_variable[0].object_id

They are the same object.

In other words, it is not the case that changing local_variable changes @instance_variable, but it just so happens that they both contain a reference to the same object, so obviously changing that object as pointed to by one changes that object as pointed to by the other.

edited Feb 11, 2013 at 0:21

answered Feb 10, 2013 at 23:30

matt

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