[求助]子函数能否调用自身来进行循环?

gisshow

[求助]子函数能否调用自身来进行循环?

 

Andyhon

前辈的一些讨论
Ref: http://autocad.xarch.at/stdlib/archive/10/msg00054.html

3. ;;; by Randy Richardson
4. (defun subst1 (index new lst / counter lst1)
5.   (setq counter 0)
6.   (repeat index
7.     (setq lst1 (cons (nth counter lst) lst1))
8.     (setq counter (1+ counter))
9.   )
10.   (setq lst1 (cons new lst1))
11.   (setq counter (1+ counter))
12.   (repeat (- (length lst) (length lst1))
13.     (setq lst1 (cons (nth counter lst) lst1))
14.     (setq counter (1+ counter))
15.   )
16.   (reverse lst1)
17. )
18. ;;; by Frank Oquendo
19. (defun replace-element (index newelement lst / tmp)
20.   (repeat index
21.     (setq tmp (cons (car lst) tmp)
22.    lst (cdr lst)
23.     )
24.   )
25.   (append (reverse tmp) (list newelement) (cdr lst))
26. )
28. ;;; by Tony Tanzillo
29. (defun SetNth (index newitem lst / i)
30.    (setq i -1)
31.    (mapcar
32.      '(lambda (elem)
33.          (if (eq index (setq i (1+ i)))
34.             newitem
35.             elem
36.          )
37.       )
38.       lst
39.    )
40. )
41. There's a number of posts here offering solutions, but
42. if you need to perform multiple replacements of elements
43. in the same list, by position, then none of the solutions
44. you've seen so far, including the one Reini posts a link
45. to (with a password), are very effective.
46. If you have to do many replacements, you should
47. index your list, by cons'ing each element with a
48. unique integer value. This allows you to use (SUBST)
49. to quickly replace elements:
50. (defun index-list (lst / i)
51.    (setq i -1)
52.    (mapcar '(lambda (elem) (cons (setq i (1+ i)) elem)) lst)
53. )
54. Given the list (A B C D E F), the above function
55. returns the list:
56.     ((0 . A) (1 . B) (2 . C) (3 . D) ...)
57. With this resulting list, you can now use (subst)
58. to replace only one element by position. This is
59. much more effecient if you have many replacements
60. to perform on the same list.
61. To replace an element at a given position:
62.    (setq indexed-list
63.       (subst (cons <index> <newvalue>)
64.              (nth <index> indexed-list)
65.              indexed-list
66.       )
67.    )
68. You can still quickly reference elements in the list
69. by position, using (cdr (nth <index> <indexed-list>))
70. Or, once you're finished replacing elements, you can
71. just do (mapcar 'cdr <indexed-list>) to remove the
72. indexing.
73. ;;; by Tony Tanzillo ===============================
75. ;;; by Reini Urban
76. (defun std-%setnth (new i lst / fst len)
77.   (cond
78.     ((minusp i) lst)
79.     ((> i (setq len (length lst))) lst)
80.     ((> i (/ len 2))
81.       (reverse (std-%setnth new (1- (- len i)) (reverse lst))))
82.     (T
83.       (append
84.         (progn
85.           (setq fst nil)        ; ; possible VL lsa compiler bug
86.           (repeat (rem i 4)
87.             (setq fst (cons (car lst) fst)
88.                   lst (cdr lst)))
89.           (repeat (/ i 4)
90.             (setq fst (cons (cadddr lst)
91.                             (cons (caddr lst)
92.                                   (cons (cadr lst)
93.                                         (cons (car lst) fst))))
94.                   lst (cddddr lst)))
95.           (reverse fst)
96.         )
97.         (if (listp new) new (list new)) ; v0.4001
98.         (cdr lst)))))
99. ;|
100. ;;; old slower versions:
101. (defun STD-RPLACE (lst i new)
102.   (if (< -1 i (length lst))     ; fixed v0.3003
103.     (append (std-firstn i lst)
104.             (list new)
105.             (std-nthcdr (1+ i) lst))
106.     lst))
107. |;
109. ;;; by Marc'Antonio Alessi
110. ;;; newitem /= nil
111. (defun ALE_SUBST_NTH (index newitem lst / len r_lst olditem )
112.   (if (and newitem (> (setq len (- (length lst) (1+ index))) -1))
113.     (progn
114.       (setq
115.         r_lst   (reverse lst)
116.         olditem (nth index lst)
117.       )
118.       (while (/= index (length (setq r_lst (cdr (member olditem r_lst))))))
119.       (while (/= len   (length (setq lst   (cdr (member olditem   lst))))))
120.       (append (reverse r_lst)  (list newitem) lst)
121.     )
122.     lst
123.   )
124. )

;;; by Randy Richardson (defun subst1 (index new lst / counter lst1)   (setq counter 0)   (repeat index     (setq lst1 (cons (nth counter lst) lst1))     (setq counter (1+ counter))   )   (setq lst1 (cons new lst1))   (setq counter (1+ counter))   (repeat (- (length lst) (length lst1))     (setq lst1 (cons (nth counter lst) lst1))     (setq counter (1+ counter))   )   (reverse lst1) ) ;;; by Frank Oquendo (defun replace-element (index newelement lst / tmp)   (repeat index     (setq tmp (cons (car lst) tmp)    lst (cdr lst)     )   )   (append (reverse tmp) (list newelement) (cdr lst)) ) ;;; by Tony Tanzillo (defun SetNth (index newitem lst / i)    (setq i -1)    (mapcar      '(lambda (elem)          (if (eq index (setq i (1+ i)))             newitem             elem          )       )       lst    ) ) There's a number of posts here offering solutions, but if you need to perform multiple replacements of elements in the same list, by position, then none of the solutions you've seen so far, including the one Reini posts a link to (with a password), are very effective. If you have to do many replacements, you should index your list, by cons'ing each element with a unique integer value. This allows you to use (SUBST) to quickly replace elements: (defun index-list (lst / i)    (setq i -1)    (mapcar '(lambda (elem) (cons (setq i (1+ i)) elem)) lst) ) Given the list (A B C D E F), the above function returns the list:     ((0 . A) (1 . B) (2 . C) (3 . D) ...) With this resulting list, you can now use (subst) to replace only one element by position. This is much more effecient if you have many replacements to perform on the same list. To replace an element at a given position:    (setq indexed-list       (subst (cons <index> <newvalue>)              (nth <index> indexed-list)              indexed-list       )    ) You can still quickly reference elements in the list by position, using (cdr (nth <index> <indexed-list>)) Or, once you're finished replacing elements, you can just do (mapcar 'cdr <indexed-list>) to remove the indexing. ;;; by Tony Tanzillo =============================== ;;; by Reini Urban (defun std-%setnth (new i lst / fst len)   (cond     ((minusp i) lst)     ((> i (setq len (length lst))) lst)     ((> i (/ len 2))       (reverse (std-%setnth new (1- (- len i)) (reverse lst))))     (T       (append         (progn           (setq fst nil)        ; ; possible VL lsa compiler bug           (repeat (rem i 4)             (setq fst (cons (car lst) fst)                   lst (cdr lst)))           (repeat (/ i 4)             (setq fst (cons (cadddr lst)                             (cons (caddr lst)                                   (cons (cadr lst)                                         (cons (car lst) fst))))                   lst (cddddr lst)))           (reverse fst)         )         (if (listp new) new (list new)) ; v0.4001         (cdr lst))))) ;| ;;; old slower versions: (defun STD-RPLACE (lst i new)   (if (< -1 i (length lst))     ; fixed v0.3003     (append (std-firstn i lst)             (list new)             (std-nthcdr (1+ i) lst))     lst)) |; ;;; by Marc'Antonio Alessi ;;; newitem /= nil (defun ALE_SUBST_NTH (index newitem lst / len r_lst olditem )   (if (and newitem (> (setq len (- (length lst) (1+ index))) -1))     (progn       (setq         r_lst   (reverse lst)         olditem (nth index lst)       )       (while (/= index (length (setq r_lst (cdr (member olditem r_lst))))))       (while (/= len   (length (setq lst   (cdr (member olditem   lst))))))       (append (reverse r_lst)  (list newitem) lst)     )     lst   ) )

ZZXXQQ

函数自身调用就是“递归”。只要有限次的“递归”（确保堆栈不溢出），且函数结束条件设定合理就行。