为什么这个mapcan会导致我的REPL冻结？

Question

在这个非常有用的答案中,有人建议我可以替换这段代码:

(defun describe-paths (location edges)
  (apply (function append) (mapcar #'describe-path
               (cdr (assoc location edges)))))

有了这个:

(defun describe-paths-mapcan (location edges)
  (mapcan #'describe-path
               (cdr (assoc location edges))))

我当然从概念上理解为什么这应该有效,但事实并非如此; 第二个变体冻结我的REPL,CL提示永远不会返回.我必须重启SLIME.所以我查了一下,我想知道mapcan不使用的事实list,而是nconc原因？因此,这些实际上不是功能相同的代码块？

对于好奇,我传递了这个:

(describe-paths-mapcan 'living-room *edges*)

在哪里*edges*:

(defparameter *edges* '((living-room (garden west door)
             (attic upstairs ladder))
            (garden (living-room east door))
            (attic (living-room downstairs ladder))))

和:

(defun describe-path (edge)
  `(there is a ,(caddr edge) going ,(cadr edge) from here.))

Answer 1

我认为这与此有关describe-edges.它被定义为:

(defun describe-path (edge)
  `(there is a ,(caddr edge) going ,(cadr edge) from here.))

那里的quasiquote我们可以macroexpand..你得到:

(macroexpand '`(there is a ,(caddr edge) going ,(cadr edge) from here.)) ; ==>
(CONS 'THERE
 (CONS 'IS
  (CONS 'A
   (CONS (CADDR EDGE) (CONS 'GOING (CONS (CADR EDGE) '(FROM HERE.)))))))

根据mapcan的文档,破坏性地完成了连接.查看返回的最后一个元素describe-path将与它返回的下一个元素共享结构,因此nconc将进行无限循环.

如果您要更改describe-edges为以下内容,它将起作用:

(defun describe-path (edge)
  (list 'there 'is 'a (caddr edge) 'going (cadr edge) 'from 'here.))