Question

我需要编写可以将列表对中的变量替换为列表的函数。例如(subsitute-var '((p #t) (Q #f)) '(P and Q or Q))

我写了一些代码

(define substitute   
  (lambda (A B list)     
    (cond      
     ((null? list) '())      
     ((list? (car list))
      (cons (substitute A B (car list)) (substitute A B (cdr list))))
     ((eq? (car list) A) (cons B ( substitute A B (cdr list))))      
     (else       
      (cons (car list) (substitute A B (cdr list)))))))

(define substitute-var
  (lambda (list var)
   (cond
     ((null? list) '())
     ((null? var) '())
     ((substitute (caar var) (car (cdr (car var))) list))       
      (substitute-var list (cdr var)))))

但问题是它只替换了第一对(p #t)并将列表的其余部分保留为相同。我尝试递归地调用substitute-var，但它也无法正常工作。所以我需要帮助。请帮帮我谢谢

Answer 1

我认为你的var和list混淆了

Answer 2

试试这个：

(define (substitute-var var lst)
  (if (or (null? var) (null? lst))
      '()
      (substitute (car var) (cadr var) lst)))

(define (substitute a b lst)
  (cond ((null? lst) '())
        ((eq? (car lst) (car a))
         (cons (cadr a) (substitute a b (cdr lst))))
        ((eq? (car lst) (car b))
         (cons (cadr b) (substitute a b (cdr lst))))
        (else (cons (car lst) (substitute a b (cdr lst))))))

现在，在使用您的示例进行测试时：

(substitute-var '((P #t) (Q #f)) '(P and Q or Q))

该程序返回预期答案：

(#t and #f or #f)

在方案中需要帮助替代功能

2 个答案: