我正在尝试编写一个采用多项式并将其简化的程序。每次我运行程序时，通过调用“((evalexp'((x 2)(z 8))p1)”，我都会收到错误消息，“Car:+不是列表”。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))
)
最佳答案
找到了您的错误。声明您的大多数函数都将表达式作为第一个参数，将绑定(bind)作为第二个参数，但是在您的示例中，您正在调用传递绑定(bind)的函数作为第一个参数，将表达式作为第二个参数。您必须调用(evalexp p1 '((x 2) (z 8)))或在函数定义上交换参数顺序(我建议使用此解决方案，因为它看起来更自然，看起来就像链接了@ rainer-joswig的赋值)。
特别是，subst-bindings和evalexp是需要更改的。