在this very useful answer中,建议我替换此代码:
(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了为什么应该这样做,但它没有;第二个变体冻结了我的REPL,CL提示符永远不会返回我得重新启动粘液所以我想知道,mapcan不使用
list
,而是使用nconc
,这是否是原因因此,这些代码块实际上不是功能相同的代码块?对于好奇的人,我传递这个:
(describe-paths-mapcan 'living-room *edges*)
其中
*edges*
是:(defparameter *edges* '((living-room (garden west door)
(attic upstairs ladder))
(garden (living-room east door))
(attic (living-room downstairs ladder))))
以及:
(defun describe-path (edge)
`(there is a ,(caddr edge) going ,(cadr edge) from here.))
最佳答案
我认为这与describe-edges
有关定义如下:
(defun describe-path (edge)
`(there is a ,(caddr edge) going ,(cadr edge) from here.))
我们可以在那里找到准液你会得到:
(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documentation for mapcan的规定,浓缩是破坏性的查看从
macroexpand
返回的内容的最后一个元素将与它返回的下一个元素共享结构,因此describe-path
将形成一个无限循环。如果您将
nconc
更改为以下值,它将起作用:(defun describe-path (edge)
(list 'there 'is 'a (caddr edge) 'going (cadr edge) 'from 'here.))