Re: The Paul Graham's "Revenge of the nerds" cummulator function and the solution in Emacs Lisp

[Barry Margolin <barmar@alum.mit.edu>]

In article <mailman.243.1244499530.2239.help-gnu-emacs@gnu.org>,
 Tiago Charters de Azevedo <tca@cii.fc.ul.pt> wrote:

> I've been reading  Paul Graham's "Revenge of the nerds"
> (http://www.paulgraham.com/icad.html) particularly  the appendix. It states 
> the
> problem of writing "a function that generates accumulators-- a function that 
> takes
> a number n, and returns a function that takes another number i and returns n
> incremented by i. (That's incremented by, not plus. An accumulator has to
> accumulate.)"
> 
> Should not the function foo in Common Lisp  work with Emacs Lisp?

Emacs Lisp has dynamic scoping, not lexical scoping.  If you want to 
emulate lexical scoping, (require 'cl-macs) and use lexical-let:

(defun foo (n)
  (lexical-let ((lexn n))
    (lambda (i)
      (incf lexn n))))

> 
> (defun foo (n)
>   (lambda (i)     
>     (incf n i))) 
> 
> by setting
> 
> (setq a (foo 3))
> 
> followed  by 
> 
> (funcall a 1)
> 
> When I try to evaluate (funcall a 1) in Emacs (in emacs-lisp mode) I get:
> 
> Debugger entered--Lisp error: (void-variable n)
>   (+ n i)
>   (setq n (+ n i))
>   (incf n i)
>   (lambda (i) (incf n i))(1)
>   funcall((lambda (i) (incf n i)) 1)
>   eval((funcall a 1))
>   eval-last-sexp-1(nil)
>   eval-last-sexp(nil)
>   call-interactively(eval-last-sexp
> 
> 
> It seems that emacs considers that n is a local variable because it is inside 
> a
> lambda expression?
> 
> Is this a bug? Or I'm missing something?
> 
> Thanks
> 
> tca

-- 
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
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