Reputation: 77
I find this exercise is interesting. Here's my solution:
(define (my-equal? a b)
(cond ((eq? a b) #t)
((and (pair? a) (pair? b))
(and (my-equal? (car a) (car b)) (my-equal? (cdr a) (cdr b))))
(else #f)))
Is it right ? I wonder if (eq? a b) is true, (equal? a b) should be always true.
Upvotes: 2
Views: 250
Reputation: 235984
I think we can give a more accurate answer by considering other data types, and recursively testing the elements in proper/improper lists. Here's my shot:
(define (same-type? a b)
(or (and (number? a) (number? b))
(and (symbol? a) (symbol? b))
(and (string? a) (string? b))
(and (list? a) (list? b))
(and (pair? a) (pair? b))))
(define (my-equal? a b)
(cond ((not (same-type? a b)) #f)
((or (symbol? a) (null? a) (null? b))
(eq? a b))
((list? a)
(if (not (= (length a) (length b)))
#f
(and (my-equal? (car a) (car b))
(my-equal? (cdr a) (cdr b)))))
((pair? a)
(and (my-equal? (car a) (car b))
(my-equal? (cdr a) (cdr b))))
((string? a) (string=? a b))
((number? a) (= a b))))
For the last part of your question, I suggest you take a look at this very detailed answer.
Upvotes: 2