Higher-Order Procedure - pair construction (cons, car, cdr)

Question

i need to create this procedures: my-cons, my-car, my-cdr in Scheme.

It should work like this:

(define p1 (my-cons 3 8))

(p1 #t)

3

(p1 #f)

8

(my-car p1)

3

(my-cdr p1)

8

now, I have only this:

(define my-cons

(lambda (x y)

(cons x y)

(let ((a (car (cons x y))))

 (lambda (a)
  (if (equal? a #f) y x)))))

but in this code i can't apply my-cons or my-cdr on defined p1 Can someone help me with this?

Answer 1

First, there's some extraneous code in my-cons which doesn't seem to belong here. This is enough:

(define my-cons
  (lambda (x y)
    (lambda (a)
      (if a x y))))

(Also, you don't need to compare a boolean value with #t or #f — it's usable in if as it is.)

Now you have my-cons that returns a function which returns either x or y depending on its arugment. You can use that when implementing my-car and my-cdr:

(define my-car
  (lambda (c)
    (c #t)))

(define my-cdr
  (lambda (c)
    (c #f)))