6.001 Recitation #7                                        Fall 1997
Prof. Albert R. Meyer

                      Solution to 5-minute-problem #14

FOR PROBLEM STATEMENT, SEE RECITATION NOTES, WED, 11/12.

From: xela@MIT.EDU
To: section7-forum@theory.lcs.mit.edu
Subject: Today's 5 minute problem
Date: Fri, 14 Nov 1997 12:08:49 EST

Here's the code for today's problem.  My apologies for not having time to
get it all up on the board.

; 5 minute problem for 6.001 Recitation 7, 14 Nov 1997, Prof. Meyer

(define (make-uninit-me-and-installer)
  (let ((name 'me)                       ; fills NAME slot
	(me   'to-be-me)
	(you  'to-be-you))
    (let ((uninit-me
	   (lambda (msg)
	     (cond
	      ((eq? 'name msg) (lambda () name))
	      ((eq? 'me msg) (lambda () me))
	      ((eq? 'you msg) (lambda () you))
	      (else (fail))))))
      (begin
	  (set! me uninit-me)            ; fills ME slot
	  (list
	   uninit-me                     ; Returns a list of uninit-me with
	   (lambda (new-you)             ; the name and me slots filled,
	     (set! you new-you)))))))    ; and a procedure that will later
                                         ; fill the you slot, below:

;Value: make-uninit-me-and-installer

(define me-and-installer
  (make-uninit-me-and-installer))        ; make the list
;Value: me-and-installer

(define me (car me-and-installer))       ; me = uninit-me --- 3 slots, 2 filled
;Value: me

(define me-you-installer (cadr me-and-installer))  ; (lambda (new-you)...
;Value: me-you-installer

; Now, do the same with you:

(define (make-uninit-you-and-installer)           
  (let ((name 'you)                               
	(me   'to-be-me)                         
	(you  'to-be-you))                       
    (let ((uninit-you                             
	   (lambda (msg)                         
	     (cond                               
	      ((eq? 'name msg) (lambda () name)) 
	      ((eq? 'me msg) (lambda () me))  
	      ((eq? 'you msg) (lambda () you))
	      (else (fail))))))
      (begin
	(set! you uninit-you)
	  (list                
	   uninit-you
	   (lambda (new-me)
	     (set! me new-me)))))))
;Value: make-uninit-you-and-installer

(define you-and-installer
  (make-uninit-you-and-installer))
;Value: you-and-installer

(define you (car you-and-installer))               
;Value: you

(define you-me-installer (cadr you-and-installer))
;Value: you-me-installer

; Now we have both, each inited except for the slot pointing to the other.

(me-you-installer you)   ; initializes me's you slot to you
;Value: to-be-you        ; For some reason I don't understand, the value
                         ; returned is the 'before' value....  But it
(you-me-installer me)    ; does the right thing:
;Value: to-be-me

   ;;;THE VALUE RETURNED BY SET! IS UNSPECIFIED IN R4RS SCHEME.  MIT-SCHEME
   ;;;CHOOSES TO RETURN THE "BEFORE" VALUE OF THE ASSIGNED VARIABLE. --ARM

(ask me 'name)
;Value: me

(ask me 'me)
;Value: #[compound-procedure 2]

(ask (ask me 'me) 'name)
;Value: me

(ask me 'you)
;Value: #[compound-procedure 3]

(ask (ask me 'you) 'name)
;Value: you

(ask you 'name)
;Value: you

(ask you 'me)
;Value: #[compound-procedure 2]

(ask (ask you 'me) 'name)
;Value: me

(ask you 'you)
;Value: #[compound-procedure 3]

(ask (ask you 'you) 'name)
;Value: you


--Alex

-----------
Carl Alexander      Course VI (special student)   SIPB (prospective)   MITSFS
xela@mit.edu        Mitgaard ("honorary mold")    LSC (sysadmin; night worker)
http://www.mit.edu/~xela
-----------
(define (hosed alex)
  (+ (real-job alex) (6.001 alex) (trying-to-have-a-life alex)))