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Re: Can free variable refers to a lexical environment?


From: IWAKI Hidekazu
Subject: Re: Can free variable refers to a lexical environment?
Date: Tue, 04 May 2010 15:42:03 -0000
User-agent: G2/1.0

On Mar 10, 3:21 am, Barry Margolin <bar...@alum.mit.edu> wrote:
> In article
> <180666d9-2ecb-4d99-a419-739650a8f...@c37g2000prb.googlegroups.com>,
>  IWAKI Hidekazu <i.hidek...@gmail.com> wrote:
>
>
>
>
>
> > Hello,
>
> > I need to make a table factory function in emacs with cl extension.
> > but I'm scheme user. I'm confusing emacs lisp behavior.
> > will you please tell me the emacs lisp's free variable issue.
>
> > my table factory function is following code:
> >  (defun mk-table-instance ()
> >   (let ((table nil)) ;; create a lexical value `table`
> >     (defun __temp__ (msg &rest value)
> >       ;; some operation changes the lexical variable `table`
> >       (case msg
> >    ((push) (push (car value) table))
> >    (otherwise table)))
> >     (function __temp__)))
> > ;; return the `__temp__`  procedure with the lexical variable `table`
> > ;; i.e. return a closure.
>
> > (fset 'table-object (mk-table-instance))
> > ;; 'table-object is an unique procedure object.
> > (table-object 'push 13)
> > ;; my plan => operate an unique `table` variable which was created by
> > involving `mk-table-instance`
> > ;; real        => "Debugger entered--Lisp error: (void-variable
> > table)"!!!!!!!
>
> > I researched this code. I guess the lexical binded variable `table` in
> > `mk-table-instance` procedure refer to the global environment.
> > In scheme, the variable refer to the lexical environment.
>
> > How to refer to a lexical environment?
>
> Emacs Lisp implements dynamic scoping, not lexical scoping.
>
> You can use lexical-let to emulate lexical binding.
>
> --
> Barry Margolin, bar...@alum.mit.edu
> Arlington, MA
> *** PLEASE post questions in newsgroups, not directly to me ***
> *** PLEASE don't copy me on replies, I'll read them in the group ***

Oh, it's just nice. I can use lexical binding in elisp programs.

I've tried some, this macro can not solve labeled procedures("defun"ed
procedures),
but only no labeled procedures (lambda expressions). I'll rewrite my
programs in the match style.

thank you for your advice.


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