From: Hal Fulton <hal9000@...>
Date: 2006-08-02T08:40:14+09:00
Subject: Re: I thought this was the one that worked?

Chad Perrin wrote:
> On Tue, Aug 01, 2006 at 11:39:29AM +0900, Hal Fulton wrote:
> 
>>Chad Perrin wrote:
>>
>>>Am I to assume that some_local_variable is going out of scope outside of
>>>the lambda declaration at some point, to create that closure?
>>>
>>>
>>>
>>>
>>>>  All things that are called "blocks" in Ruby are closures.
>>>>  Do you still have any questions on the matter?
>>>
>>>
>>>Yes.  Are you just failing to explain how these things are closures, or
>>>do you not know what a closure is?  You keep telling me blocks are
>>>closures somehow, but you're not explaining how.  Saying it doesn't make
>>>it so.  Please explain what I'm missing.
>>>
>>
>>I'm unfamiliar with your assertion that the context has to go
>>out of scope before a closure exists.
>>
>>Not trying to argue, just stating my perception.
>>
>>Maybe you can point us to a formal definition somewhere.
> 
> 
> How's this?:
>   
>   Closure
>       A "closure" is an expression (typically a function) that can have
>       free variables together with an environment that binds those
>       variables (that "closes" the expression).
> 
> I found it on this page:
> 
>   http://jibbering.com/faq/faq_notes/closures.html
> 
> I also found this:
> 
>   http://www.perl.com/pub/a/2002/05/29/closure.html
> 
> 
> 
>>I would have said that a closure *will* retain its variables
>>*if/when* they go out of scope.
>>
>>Bad analogy: We might loosely define a car as "a machine that
>>goes places," but it's still a car even when it sits still.
> 
> 
> My understanding is that the vehicle is a car, and the use of it is a
> drive (to the country, or whatever).  Similarly, a block of code that
> can be passed around like a first-class member is a block, or lambda, or
> whatever you want to call it today: if you pass it to a function as an
> argument, it's a callback, and if you lexically close it, it's a
> closure.

I think "lexically" might be the key word there. Isn't it true
that Perl has only lexical closures, not "real" closures?

> 
>>My impression is that you're applying the definition too
>>rigidly.
> 
> 
> I suppose that's possible, but I really don't think so.  Do you have a
> formal definition of a closure that specifically contradicts my
> understanding?

Well, I would say the definition you offered above. It says that
a closure *can* have free variables, not that it must. All of the
internal housekeeping (as I understand it) is done the same for a
Ruby block with variables as for one without. (Besides, as someone
pointed out, there is always a "self" at the very least.)


Hal