Traversing Scheme function as a list

Question

Isn't it possible to treat functions in Scheme as any other list?

Basically, what I want do to is something like this:

(define (foo) "hello")

(cdr foo)  ; or similar, should return the list ((foo) "hello")

I've found a similar discussion about this, and I feel a bit disappointed if this is not possible with Scheme. If so, why is this impossible? Is it possible in other lisps?

EDIT: Changed (cdr 'foo) to (cdr foo) -- it was misleading. I'm asking, why can't I access a function as a list?

Answer 1

I have often wanted to be able to do the same thing csl. Below is a quick example of how you could go about doing this in mzscheme.

DrScheme 4.2

(module qdefine mzscheme
  (provide ;(all-from-except mzscheme let)
   (rename define olddefine)
   (rename quote-define define)
   (rename quote-cdr cdr)
   (rename quote-car car))

  (define define-list '())
  (define define-list-add 
    (lambda (key value)
      (set! define-list (cons `(,key ,value) define-list))))

  (define-syntax quote-define
    (syntax-rules ()
      ((_ (pro-name args ...) body ...) 
       (begin
         (define (pro-name args ...) body ...)
         (define-list-add pro-name  '((pro-name args ...) body ...))))
      ((_ pro-name pro) (begin
                          (define pro-name pro)
                          (define-list-add pro-name 'pro)))

      ))

  (define quote-cdr (lambda (lst)
                      (if (procedure? lst)
                          (cdr (cadr (assoc lst define-list)))
                          (cdr lst))))

  (define quote-car (lambda (lst)
                      (if (procedure? lst)
                          (car (cadr (assoc lst define-list)))
                          (car lst))))
  )
(require 'qdefine)

(define testfunc (lambda (args) args))
(cdr testfunc)
(car testfunc)

(define (testfunc2 test) 'blah)
(cdr testfunc2)
(car testfunc2)

(define testlist '(1 2 3 4 5 6 'a))
(cdr testlist)
(car testlist)

Outputs:

((args) args)
lambda
('blah)
(testfunc2 test)
(2 3 4 5 6 'a)
1
>