方案：比较列表大小

发布于 2024-11-01 23:38:21 字数 442 浏览 1 评论 0原文

我的计划程序有问题。我试图获取 2 个列表并比较它们的大小，如果大小相等则返回 true，如果不相等则返回 false。每个原子的值并不重要。

示例：

(structeq '(a (b(c))) '(1(2(3)))) => #t
(structeq '(x) '(()) => #f

这是我的代码：

(define (structeq list1 list2)
    (cond ((null? list1) list2)
    (eq? (length list1) (length list2))))

(structeq '(a b c d) '(a b c))

但是，这会返回最后一个列表的大小。我哪里出错了？

编辑：取消这个问题。我想通了，我只需要删除 cond 语句即可。

原文

I am having an issue with my Scheme program. I am trying to take in 2 lists and compare their sizes and return true is the sizes are equal, and false if they are not. The value of each atom doesn't matter.

Example:

(structeq '(a (b(c))) '(1(2(3)))) => #t
(structeq '(x) '(()) => #f

Here is my code:

(define (structeq list1 list2)
    (cond ((null? list1) list2)
    (eq? (length list1) (length list2))))

(structeq '(a b c d) '(a b c))

However, this returns the size of the last list. Where am I going wrong?

EDIT: Cancel this question. I figured it out, i just needed to remove the cond statement.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

梦太阳 2024-11-08 23:38:21

请注意：

(define (same-length a b)
  (if (and (null? a) (null? b)) #t
      (if (or (null? a) (null? b)) #f
          (same-length (cdr a) (cdr b)))))

一旦找到较短列表的末尾就会停止。

note that:

(define (same-length a b)
  (if (and (null? a) (null? b)) #t
      (if (or (null? a) (null? b)) #f
          (same-length (cdr a) (cdr b)))))

would stop as soon as it found the end of the shorter list.

回复收藏 0 原文

深海夜未眠 2024-11-08 23:38:21

(eq? (length list1) (length list2))))

代码中的这一行有一个谓词，但没有结果。如果它们相等，您想返回#t。
当列表的长度不相等时，添加一个 else 情况来捕获也是很好的。例如：(else #f)

在此处了解有关条件的更多信息。

(eq? (length list1) (length list2))))

This line in your code has a predicate but no consequent. If they are equal, you want to return #t.
Also would be good to add an else case to catch when the lists' length are not equal. eg: (else #f)