Empty and '() in function not equal

Question

i am learning racket for my course and while going through solutions to an assignment, I found that a function doesn't return true, when comparing empty and '()

Here is the output I'm getting:

(lst-contains? '(1 2 3 ()) '())      ; output: #t
(lst-contains? '(1 2 3 ()) empty)    ; output: #t
(lst-contains? '(1 2 3 empty) empty) ; output: #f (unexpected!)
(lst-contains? '(1 2 3 empty) '())   ; output: #f (unexpected!)

Here's my definition of lst-contains?:

(define (lst-contains? lst element)
  (cond
    [(empty? lst) #f]
    [(equal? (first lst) (element)) #t]
    [else (lst-contains? (rest lst) element)]))

Thanks : )

is it against the forum guidelines to post code screenshots ? (I am new to stackoverflow) ? — Mek, Jan 04 '21 at 14:59
No it isn't but you can't eaasily copy the code and test it for finding a solution. — marsh-wiggle, Jan 04 '21 at 15:10
this question/answer is related: [_What is the difference between quote and list?_](https://stackoverflow.com/questions/34984552/what-is-the-difference-between-quote-and-list) — Alex Knauth, Jan 04 '21 at 18:56

score 2 · Answer 1 · answered Jan 02 '21 at 14:52

This is due to the fact that quote (or ') "distributes" across a list. So, when you write '(1 2 3 empty), you're actually creating something that's equivalent to (list 1 2 3 'empty) not (list 1 2 3 empty).

You can check for yourself that this is the case by looking at the result of:

(list-ref '(1 2 3 empty) 3)      ; output: 'empty
(list-ref (list 1 2 3 empty) 3)  ; output: '()

score 0 · Answer 2 · answered Jan 02 '21 at 14:51

In racket they are equal:

(define (lst-contains lst element)
  (cond
    ((empty? lst) #f)
    ((equal? (first lst) element) #t)
    (else (lst-contains (rest lst) element))))

(lst-contains '(1 2 3 ()) empty) ; -> #t

Tested in DrRacket.

Note that '(1 2 3 empty) is different from '(1 2 3 ()) since in the first case empty is taken as a symbol.

Empty and '() in function not equal

2 Answers2