syntax-error - 汽车 : + is not a list, lisp 错误？

Question

我正在尝试编写一个采用多项式并对其进行化简的程序。每次我运行该程序时，通过调用“(evalexp '( (x 2) (z 8) ) p1)”我都会收到错误消息“Car: + is not a list”。它应该返回“(+ 2 (* 2 (- y 4)))” 这是我的代码:

(setq p1 '(+ x (* x (- y (/ z 2)))))
(setq p2 '(+ (- z 2) (* x 5)))
(setq p3 '(+ 1 a))
(setq p4 '(+ x (+ x (* x 0))))

(defun addexp (e1 e2) (list '+ e1 e2))
(defun subexp (e1 e2) (list '- e1 e2))
(defun mulexp (e1 e2) (list '* e1 e2))
(defun divexp (e1 e2) (list '/ e1 e2))

(defun deep-subst (old new expr)
  (cond
   ((null expr) 
     nil
   )
   ((listp (car expr))
    (cons (deep-subst old new (car expr)) (deep-subst old new (cdr expr)))
   )
   ((eq old (car expr)) 
    (cons new (deep-subst old new (cdr expr)))
   )
   (T   
    (cons (car expr) (deep-subst old new (cdr expr)))
    )
  )
)

(defun subst-bindings (expr bindinglist)
  (cond   
    ((null bindinglist) 
       expr )
    (T 
       (deep-subst (car (car bindinglist)) (car (cdr (car bindinglist))) 
        (subst-bindings expr (cdr bindinglist))
    ))))

(defun simplify-triple(op left-arg right-arg)
  (cond
    ((and (numberp left-arg) (numberp right-arg))
      (eval (list op left-arg right-arg))
    )
    ((and (eq op '+) (eql right-arg 0))
      left-arg
    )
    ((and (eq op '+) (eql left-arg 0))
      right-arg
    )
    ((and (eq op '-) (eql right-arg 0))
      left-arg
    )
    ((and (eq op '-) (eql right-arg left-arg))
      0
    )
    ((and (eq op '*) (eql right-arg 0))
      0
    )
    ((and (eq op '*) (eql left-arg 0))
      0
    )
    ((and (eq op '*) (eql right-arg 1))
      left-arg
    )
    ((and (eq op '*) (eql left-arg 1))
      right-arg
    )
    ((and (eq op '/) (eql left-arg 0))
      0
    )
    ((and (eq op '/) (eql right-arg 1))
      left-arg
    )
    ((and (eq op '/) (eql right-arg left-arg))
      1
    )

    (T
        (list op left-arg (simplify right-arg))
        ;(list op right-arg (simplify left-arg))
    )
  )
)

(defun simplify (exp)
  (cond
    ( (listp exp)
        (simplify-triple (car exp) (simplify (car (cdr exp))) (simplify (car (cdr (cdr exp)))))
    )   
    (T 
        exp)))


(defun evalexp (exp binding-list)
  (simplify (subst-bindings exp binding-list))
)

Answer 1

找到你的错误。您的大多数函数都被声明为接收表达式作为第一个参数，绑定(bind)作为第二个参数，但是在您的示例中，您调用的函数将绑定(bind)作为第一个参数传递，将表达式作为第二个参数传递。您必须调用 (evalexp p1 '((x 2) (z 8))) 或交换函数定义的参数顺序(我建议使用此解决方案，因为它看起来更自然且看起来就像@rainer-joswig 链接的作业一样)。

特别是，subst-bindings 和 evalexp 是需要更改的。