可以使用 cons 来做 ((a. b). (c. d)) 吗？如果不是，任何其他方式或点对不能有这样的第二个元素？

Question

;;; <- can one use cons to do  ((a . b) . (c . d))?


(define x (cons a b)); nil -- should it be error
(define x (cons 'a 'b)); (a . b)

(define y (cons 'c 'd)); (c . d)

(define z00 (cons x y)) ; (((a . b) c . d) <- cannot use cons to do  ((a . b) . (c . d))?
(define z01 (cons x 'y)) ; ((a . b) . y) 
(define z10 (cons 'x y)) ; (x c . d) 
(define z11 (cons 'x 'y)); (x . y))

(define z (list x y z00 z01 z10 z11)) 
        ; ((a . b) (c . d) ((a . b) c . d) ((a . b) . y) (x c . d) (x . y))

;;; and if not any other means or dot pair cannot have 2nd element like this?

Answer 1

Yes, you can. And the language has a wonderful predicate called equal? which will allow you to test this:

> (equal? (cons (cons 'a 'b) (cons 'c 'd))
          '((a . b) . (c . d)))
#t
> (equal? '((a . b) . (c . d))
          '((a . b) c . d))
#t

And you can even write a little display function which will confirm this:

(define (display-thing thing)
  (if (cons? thing)
      (begin
        (display "(")
        (display-thing (car thing))
        (display " . ")
        (display-thing (cdr thing))
        (display ")"))
      (display thing)))

And now

> (display-thing (cons (cons 'a 'b) (cons 'c 'd)))
((a . b) . (c . d))
> (display-thing '((a . b) . (c . d)))
((a . b) . (c . d))
> (display-thing '((a . b) c . d))
((a . b) . (c . d))

What this should all be telling you is that ((a. b). (c. d)) and ((a. b) c. d) are merely different ways of writing a structurally identical object .

Answer 2

Pairs visualize differently based on their content. If the cdr of a pair contains the empty list it is a proper list an dthe dot and extra empty list is not shown:

(cons 'a '())
'(a . ())
; ==> (a)

A pair that has pair as it's cdr can be visualized as a list element without the . and extra parenthesis:

(cons 'b '(a))
'(b . (a))
; ==> (b a)
(cons 'b '(a . c))
'(b . (a . c))
; ==> (b a . c)

These are just made so that we can have (1 2 3) displayed instead of (1. (2. (3. ()))) which is how it really is made.

If you were to not have a pair or a empty list in the cdr then it falls back to showing the dotted pair:

(cons 'a 'b)
'(a . b)
; ==> (a . b)

In your example '((a. b). (c. d)) because there is a pair after a dot (eg. the cdr if the pair the visualization will remove the dot and one pair of parentheses and show it like ((a. b) c. d) . This is the only acceptable correct way for a REPL to display this even though both your example and the display will be read in as the same structure.

There is a similar issue with numbers. In code you can use 10 , #xa and #o12 to get the number 10 and the value will have no idea what format is was read in as and only show the base 10 in the REPL.

Answer 3

;;; ```
;;; <- can one use cons to do  ((a . b) . (c . d))?


(define x (cons a b)); nil -- should it be error
(define x (cons 'a 'b)); (a . b)

(define y (cons 'c 'd)); (c . d)

(define z00 (cons x y)) ; (((a . b) c . d) <- cannot use cons to do  ((a . b) . (c . d))?
(define z01 (cons x 'y)) ; ((a . b) . y) 
(define z10 (cons 'x y)) ; (x c . d) 
(define z11 (cons 'x 'y)); (x . y))
(define z22 (cons '(f g) '(h i)))
(define z2c (cons (cons 'f 'g) (cons 'h 'i)))

(define fgc (cons ('f 'g))); should it be error (nil) 
; actually not as 'f is an exoression (quote f) and f for some reason is nil it becomes nil from quote of nil abd so is the second one.  now (nil . nil) is (nil)

(define z (list x y z00 z01 z10 z11 z22 z2c fgc)) 
        ; ((a . b) (c . d) ((a . b) c . d) ((a . b) . y) (x c . d) (x . y))



;;; ```

;;; and if not any other means or dot pair cannot have 2nd element like this?

;;; possibly not

;;; as the cons join 2 pairs of dotted pairs and can generate one dotted pair
;;; ((a . b) . (c . d)) but the printing rule is reflected the list bias 

;;; this new dotted pair will have the first element as (( a . b) ... print as
;;;   ((a . b) ...
;;; the 2nd element it will consider whether it is an atom or another dotted pair 
;;;      (other possibilities like loop back or something else ... not sure)
;;; as (c . d) is a dotted pair the "printing" continues as a list would
;;; 
;;;      ((a . b) c ... 

;;; however the second element of (c . d) is not a dotted pair but an atom and print as 

;;;               . d) will it becomes

;;; hence even though you form the binary tree head the dot pair would display as a partial list

;;; you can have a list of dotted pairs like z 
;;; but not dotted pair of dotted pairs