lisp - 从未调用过 Lisp 函数

Question

   (defun func (in s f l)
      (cond 
       ((null in) (append l (list (list 'end (+ 1 f)))))
       ((eq (car in) 'foo) (foo-asd (cdr in) s f l)) 
       ((atom (car in))(atom-asd in (+ 1 s) (+ 1 f) l))
       ))

.

(defun atom-asd (in s f l)
  (cond ((eql in nil) ())
    (append l (list (list 'frob s (car in) (+ 1 f))))))

.

(defun foo-asd (in s f l)
  (cond 
   ((eql in nil) (append l (list (list 'frob s 'myst f))))
   ((func in s f (append l (list (list 'frob s 'myst (+ 1 f))))))
   ((foo-asd (cdr in) s f l))
  ))

. 关于这段代码，如果调用(func '(foo x y) 0 0 ()) 函数foo-asd 将被调用，那么func 是再次调用会进入函数atom-asd，当atom-asd执行结束，所有程序结束，不调用递归调用foo-asd 。我需要调用 foo-asd，但我不明白为什么在 atom-asd 结束后它没有被调用。

Answer 1

[4]> (trace func)
;; Tracing function func.
(func)
[5]> (trace atom-asd)
;; Tracing function atom-asd.
(atom-asd)
[6]> (trace foo-asd)
;; Tracing function foo-asd.
(foo-asd)
[7]> (func '(foo x y) 0 0 ())
1. Trace: (func '(foo x y) '0 '0 'nil)
2. Trace: (foo-asd '(x y) '0 '0 'nil)
3. Trace: (func '(x y) '0 '0 '((frob 0 myst 1)))
4. Trace: (atom-asd '(x y) '1 '1 '((frob 0 myst 1)))
*** - cond: variable append has no value
The following restarts are available:
USE-VALUE      :R1      Input a value to be used instead of append.
STORE-VALUE    :R2      Input a new value for append.
ABORT          :R3      Abort main loop
Break 1 [8]>

所以在atom-asd中:

(defun atom-asd (in s f l)
  (cond
    ((eql in nil)
     ())
    (append
     l (list (list 'frob s (car in) (+ 1 f))))))

您有两个谓词。一个测试 (eql in nil) 显然是 nil，然后它检查变量 append 是否有一个非 nil 值(value)。问题是 append 不是绑定(bind)变量。它是函数命名空间中的绑定(bind)，但这里每个术语都包含在一组括号中，因此 append 本身就是要测试的表达式。你可能意味着它应该在第一个学期没有开始时执行 (append ...) 并且你应该这样写:

(defun atom-asd (in s f l)
  (cond
    ((eql in nil)
     ())
    (t
     (append l (list (list 'frob s (car in) (+ 1 f)))))))

在这个版本中我们得到了一个结果:

[8]> (func '(foo x y) 0 0 ())
5. Trace: (func '(foo x y) '0 '0 'nil)
6. Trace: (foo-asd '(x y) '0 '0 'nil)
7. Trace: (func '(x y) '0 '0 '((frob 0 myst 1)))
8. Trace: (atom-asd '(x y) '1 '1 '((frob 0 myst 1)))
8. Trace: atom-asd ==> ((frob 0 myst 1) (frob 1 x 2))
7. Trace: func ==> ((frob 0 myst 1) (frob 1 x 2))
6. Trace: foo-asd ==> ((frob 0 myst 1) (frob 1 x 2))
5. Trace: func ==> ((frob 0 myst 1) (frob 1 x 2))
((frob 0 myst 1) (frob 1 x 2))