CODEWITHSUNDEEP
scheme
chendoy
chendoy

Reputation: 275

car and cdr in Scheme won't give expected output

I'm learning scheme and when trying to use car and cdr in a very simply function I don't get the expected results:

(define test
  (lambda (lst)
    '((car lst).cdr (lst))))

And when running 

(test '(a.b.c))

I get output:

'((car lst) .cdr (lst))

How do I make it use the car and cdr functions and not treat them like strings?

Upvotes: 0

Views: 130

Answers (2)

Sylwester
Sylwester

Reputation: 48745

When you quote something, like '(1 2 3 4) the result will be the data structure created at read time without any evaluation. 

'((car lst) .cdr (lst))))
; ==> ((car lst) .cdr (lst)))

It does exactly what you tell it to. It's like complaining that evaluating the string "5 + 5" doesn't give you 10.

If you actually want the car of something you do this:

(define (test lst)
  (car lst))

(test '(1 2 3)) ; ==> 1

You can make a shallow copy of a pair like this:

(define (shallow-copy pair)
  (cons (car pair) (cdr pair)))

(define test-value (list 1 2 3))
(define shallow-value (shallow-copy test-value)))

(eq? test-value shallow-value)            ; ==> #f
(eq? (cdr test-value) (cdr shallow-value)); ==> #t

I have no idea what your procedure supposed to do so cannot help you further.

Upvotes: 3

Omri Attiya
Omri Attiya

Reputation: 4037

The symbol ' or quote is a procedure that creates constant. That's why you got '((car lst) .cdr (lst)) as output, You basically calculated the string expression of ((car lst) .cdr (lst)). Read here about it. The correct form of what you wanted to do should be written like this:

(define test
    (lambda (lst)
        (list (car lst) (cdr lst))))

(test '(a b c))

The output will be a list in which the first item will be the car of '(a b c) which is a and the second item will be the cdr of '(a b c) which is the list (b c). The output will be: (a (b c))

Upvotes: 2

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