表达式树评估

Question

我试图评估表达式树。这里是我的代码：

(define (eval-tree expr-tree) 
    (eval-treeaux eval-tree expr-tree)) 

(define (eval-treeaux f expr-tree) 
    (if (null? expr-tree) 
     0 
     (cond ((equal? '+ (operator expr-tree)) (+ (left-op expr-tree) (right-op expr-tree))) 
      ((equal? '- (operator expr-tree)) (- (left-op expr-tree) (right-op expr-tree))) 
      ((equal? '* (operator expr-tree)) (* (left-op expr-tree) (right-op expr-tree))) 
      ((equal? '/ (operator expr-tree)) (/ (left-op expr-tree) (right-op expr-tree))) 
      (else (f eval-treeaux expr-tree))))) 

当我运行这个(eval-tree '((6 * 3) + (4 - 2)))，它给了我这个错误：

+: expects type <number> as 1st argument, given: (6 * 3); other arguments 
    were: (4 - 2) 

谁能告诉我是什么问题，我该如何解决？

---

这是我的新代码：

(define (eval-tree expr-tree) 
    (if (null? expr-tree) 
     0 
     ((eval-treeaux eval-tree (car expr-tree)) (eval-tree (cdr expr-tree))))) 

(define (eval-treeaux f expr-tree) 
    (cond ((null? expr-tree) '()) 
     ((not (isExpression expr-tree)) (list expr-tree)) 
     (else (cond ((equal? '+ (operator expr-tree)) (+ (left-op expr-tree) (right-op expr-tree))) 
        ((equal? '- (operator expr-tree)) (- (left-op expr-tree) (right-op expr-tree))) 
        ((equal? '* (operator expr-tree)) (* (left-op expr-tree) (right-op expr-tree))) 
        ((equal? '/ (operator expr-tree)) (/ (left-op expr-tree) (right-op expr-tree))) 
        (else (f eval-treeaux expr-tree)))))) 

但是，当我运行这个(eval-tree '((6 * 3) + (4 - 2)))，我得到这个错误：

procedure application: expected procedure, given: 2; arguments were: 0 

我真的很困惑。任何人都可以帮助我吗？ 感谢

Answer 1

表达((6 * 3) + (4 - 2))似乎是由2子表达式：

(6 * 3) ; e_1 
(4 - 2) ; e_2 

您需要递归评估子表达式，然后才能评估父表达。