From: Aldric Giacomoni <aldric@...>
Date: 2009-10-28T12:24:39+09:00
Subject: Closures / lambda question

This is something I don't understand, and did not understand when I
studied LISP. I just watched Dave Thomas' presentation, "Extending Ruby
for Fun and Profit", which I by the way highly recommend to everyone who
hasn't seen it...
And he has the following example:

def proc_with_enclosing_scope
  name = "Ruby"
  lambda { puts name }
end

the_proc = proc_with_enclosing_scope
the_proc.call

name = "Java"
the_proc.call

_____

I don't understand what Ruby is doing / what happens.
First question: the "name" variable is defined inside the method
proc_with_enclosing_scope, so why would changing the name outside the
method make a difference in the first place?

Second question: I tried to type this straight into irb and made a small
typo, so it came out as such -

>> def proc_with_enclosing_scope
>> name = "Ruby"
>> lamda { puts name }
>> end
=> nil
>> the_proc = proc_with_enclosing_scope
NoMethodError: undefined method `lamda' for main:Object
        from (irb):23:in `proc_with_enclosing_scope'
        from (irb):25

So.... When the_proc gets assigned the ... Value of the method
running... (?) What does it get assigned?

And lastly.. I know that "proc" exists, too. What is the difference /
what does it do?

I thank you very much in advance for the enlightenment you will provide
:)
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