append 和 cons 球拍的区别 [英] Difference between append and cons racket

问题描述

我试图了解 cons 和 append 在列表如何存储在下面方面的区别.考虑这种情况:

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))

如果我假设调用(cons x y) 在这两个列表上会生成一个点对其中第一个指针指向 x，第二个指针指向 y.

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.

总而言之，x 或 y 都没有改变.唯一增加记忆的是点对.

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

但是，当我使用两个列表调用 append 时:

However when I call append with the two lists:

(append x y)

我认为，为了让球拍避免更改 x 和 y，它必须为条目创建一个新列表，其中包含 x 和 y 的顺序.这个列表没有任何指向 x 或 y 的指针.

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.

我的假设是否正确?

推荐答案

我更喜欢写列表，使用 (list ...) where (list ; <c> <d>) 是 (cons <a> (cons <b> (cons <c> (cons <d> empty)))) 的简写

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))))

在回答你的问题之前，我想提一个问题:

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

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

将导致:

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

你的意思是写:

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

导致:

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

?

现在，对于您的问题，这里是 append 的实现:

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))]))

直观地，它将 x 解构为元素并重新排列元素以形成一个新列表.但是，y 在整个函数中保持不变.因此，(append x y) 的结果中y 的内存地址仍然指向与之前相同的y.对于 x，它不会.

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.

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