尽管创建了副本，但 lisp 函数的参数发生了变化

Question

我创建了一个函数来从嵌套列表中随机选择一个“位”，例如((1 0 0 1) (1 1 1 1) (0 1 0 1))，然后翻转它。如果是 1，则将其设为 0，反之亦然。该函数运行良好，但我发现尽管我制作了副本，但它会改变原始参数。这里是函数，最后写的两篇演示了这个问题。如果我将((1 1 1 1)) 传递给它，我希望看到(write DNA-seq) 打印的原始值，但是原始值被修改了，所以(write DNA-seq) 和(write CDNA-seq) 打印相同的东西。

(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 是一个浅 副本，它只复制最外层的列表，而不是每个元素所引用的列表。使用copy-tree 执行深复制。

(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