Difference between append and cons racket

问题

I'm trying to understand what the difference is between cons and append in terms of how the lists are being stored underneath. consider the case:

(define x '(hi, hello))
(define y '(randomvalue1, randomvalue2))

if I'm assuming that calling (cons x y) on these two lists will generate a dotted pair where the first pointer points to x and the second points to y.

all in all neither x or y has been changed. Only thing added memory wise is the dotted pair.

However when I call append with the two lists:

(append x y)

I would assume that in order for racket to avoid changing x and y it will have to create a new list with for entries with copies of the values of x and y in order. This list does not have any pointers pointing to neither x or y.

Are my assumptions correct?

回答1:

I prefer to write lists, using (list ...) where (list <a> <b> <c> <d>) is a shorthand for (cons <a> (cons <b> (cons <c> (cons <d> empty))))

Before getting to your question, there is an issue I would like to mention:

(define x '(hi, hello))
(define y '(randomvalue1, randomvalue2))

will result in:

(list 'hi ',hello)
(list 'randomvalue1 ',randomvalue2)

Do you mean to write:

(define x '(hi hello))
(define y '(randomvalue1 randomvalue2))

which results in:

(list 'hi 'hello)
(list 'randomvalue1 'randomvalue2)

?

---

Now, for your question, here's an implementation of append:

(define (append x y)
  (cond
    [(empty? x) y]
    [(cons? x) (cons (first x) (append (rest x) y))]))

Intuitively, it destructs x to elements and rearrange elements to form a new list. However, y stays the same throughout the function. Thus, the memory address of y in the result of (append x y) still points to the same y as before. For x, it does not.

来源：https://stackoverflow.com/questions/35115348/difference-between-append-and-cons-racket