Re: Elisp lexical-let

[Daniel Kraft <d@domob.eu>]

Hi Andy,

thanks for the clarifications!

> It's actually fairly simple, imo. Alpha-equivalence says that
> (lexical-let ((x a)) x) is the same as (lexical-let ((y a)) y). (Note
> that this lexical-let corresponds to Scheme's let.) So your program is
> the same as:
> 
>> (lexical-let ((y 2))
>>   y ; -> 2
>>   (foo) ; -> 1
>>   (setq y 3)
>>   y ; -> 3
>>   (foo) ; -> 1
>>   (let ((x 4))
>>     x ; -> 4?
>>     (foo) ; -> 4
>>     (setq x 5)
>>     x ; -> 5
>>     (foo) ; -> 5
>>     ) ; end the let
>>   y ; -> 3?
>>   (foo) ; -> 4
>>   )
>> x ; -> 4
>> (foo) ; -> 4

Yes of course, my main question was how the inner let is handled, and 
what dynamic values x will attain (the results of (foo)) -- and here's 
this funny stuff that the inner let will be like lexical-let for those 
variables already lexically bound.

> I haven't reviewed your compiler yet, and for that I apologize. But. You
> need to make sure that when you compile, you build up a compile-time
> environment, mapping symbols to locations -- if a variable is free,
> lexically bound, or dynamically bound. When your compiler sees a lexical
> binding, replace that identifier with a gensym, and add an entry to your
> compile-time environment to map that identifier to that lexical gensym.
> That way you will effectively be processing my example, which is
> equivalent to yours, but perhaps clearer because the names have been
> uniquified. Then your problems go away.

This is how I was thinking about implementing the lexical-let semantics. 
  To give more details:

I want to keep track of identifiers that are lexically bound in the 
current scope while compiling as well as to what gensym they are, and 
then instead of the fluid-ref's just reference/set this 
gensym-identifier when I come across an access.  It seems this is 
exactly what you describe above :)

So, that's it with regards to messages from me :D  Thanks for all your 
comments again,

Daniel

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