I am confused about the difference between '(()) and (cons null null) in scheme.
The code below show that b and c are completely the same thing.
(define (dup2 x)
(let ((d '(())))
(set-car! d (car x))
(set-cdr! d (cdr x))
d))
(define a '(1 2))
(define b (dup2 a))
(define c (dup2 a))
(set-car! b 2)
> c ;; --> (2 2)
However, when I used dup instead of dup2:
(define (dup x)
(let ((d (cons null null)))
(set-car! d (car x))
(set-cdr! d (cdr x))
d))
(define a '(1 2))
(define b (dup a))
(define c (dup a))
(set-car! b 2)
> c ;; --> (1 2)
Variable b and c are different. I have done some experiments, but I haven't understand yet.
The value of d in the first implementation is literal data, and is modified with undefined consequences. To highlight what's happening, consider the following code:
(define (incorrect-list-null-and-x x)
(let ((l '(()))) ; a list of the form (() . ())
(set-cdr! l (cons x (cdr l))) ; (cdr l) is (), so (cons x (cdr l)) should be (x . ()) == (x), right?
; and now l should be (() . (x . ())) == (() x), right?
l))
The expected result is that (incorrect-list-null-and-x n) should return a list of the form (() n), and it does the first time, but successive calls are still accessing the same data:
(incorrect-list-null-and-x 1) ;=> (() 1)
(incorrect-list-null-and-x 2) ;=> (() 2 1)
(incorrect-list-null-and-x 3) ;=> (() 3 2 1)
(incorrect-list-null-and-x 4) ;=> (() 4 3 2 1)
The same problem manifests itself a bit differently in your dup2. Every value returned from dup2 is actually the same pair:
(let* ((x (dup2 (cons 1 2)))
(y (dup2 (cons 3 4))))
(display x)
(display y))
outputs:
(3 . 4)(3 . 4)
because the call (dup2 (cons 3 4)) modifies the same structure that was previously returned by (dup2 (cons 1 2)).
Data literals, like '(()), are meant to be read-only, and modifying it using set-car! or set-cdr! has undefined behaviour. For predictable behaviour, use the (cons '() '()) version if you want to use set-car! or set-cdr! on it.
In particular, cons creates a new cons cell, whereas a data literal usually won't.
Still, for the purposes of implementing dup, why are you using set-car! and set-cdr! anyway? Just use cons directly:
(define (dup x)
(cons (car x) (cdr x)))
In your first code snippet you use (d '(())) which ends up binding a literal to d. You then modify the literal which is generally undefined. In your second code snippet you use (d (cons null null)) which binds d to a newly created 'cons cell' which you then modify. There is no problem modifying that.
Note: you've not defined null. Perhaps you meant '()?
