Question

我创建了一个随机选择＆＃39;位的功能。从嵌套列表中删除，例如((1 0 0 1) (1 1 1 1) (0 1 0 1))，然后翻转它。如果是一个，则将其设为零，反之亦然。该函数运行良好，但我发现尽管我复制了它，它仍会改变原始参数。这是函数，最后两个写入演示了这个问题。如果我将((1 1 1 1))传递给此版本，我希望看到(write DNA-seq)打印的原始值，而是修改了原始版本，因此(write DNA-seq)和(write CDNA-seq)会打印相同的内容。< / p>

(defun rand-mutate (DNA-seq)
  (let ((CDNA-seq (copy-list DNA-seq)))
    (let ((gene (random-range 0 (length CDNA-seq))))
      (let ((base (random-range 0 (length (nth gene CDNA-seq)))))
        (cond ((= (nth base (nth gene CDNA-seq)) 0) (setf (nth base (nth gene CDNA-seq)) 1))
              (t (setf (nth base (nth gene CDNA-seq)) 0))) (write DNA-seq)(write CDNA-seq)))))

Answer 1

copy-list是浅副本，它只复制最外面的列表，而不是每个元素引用的列表。使用copy-tree执行深副本。

(defvar list '((1 0 0 1)))
(defvar list-copy (copy-list list))
(defvar list-copy-tree (copy-tree list))
(eq list list-copy) => NIL
(eq (car list) (car list-copy)) => T
(eq (car list) (car list-copy-tree)) => NIL

尽管创建了副本，但对lisp函数的参数已更改

1 个答案: