Function which returns function scheme

Question

Ok so I've spent quite a lot of time on this, I can't seem to grasp how to do this. I do understand it quite a bit when it's a simple variable but can't seem to grasp a little more complicated problem with the same concept.

This is the simple one I understand quite a bit:

(define (add n)
   (lambda (x) (+ x n)))

(define total (add 5))
(total 12) => **17**

This is what I'm trying to achieve, when an integer is entered it should find the value of that position, I have the function that finds the position already but not sure how to implement this into the function which returns a function way:

(define (position N L)
  (cond ((null? L) L)
        ((= N 1) (car L))
        (else (position (- N 1) (cdr L)))))

For example if I enter (define X (position '(1 5 8 2 7))) and then input (X 4) it should output 2 which is at position #4. I'm sure it's something simple but I've been sitting here for a while trying to put it together but I'm not doing well. Any help is appreciated!

Answer 1

It's easiest to create a second procedure:

(define (position L N)
  (cond ((null? L) L)
        ((= N 1) (car L))
        (else (position (cdr L) (- N 1)))))

(define (position2 L)
  (lambda (N)
    (position L N)))

then

> (define X (position2 '(1 5 8 2 7)))
> (X 4)
2

In Racket, you can also use curry:

> (define x (curry position '(1 5 8 2 7)))
> (x 4)
2

Answer 2

As uselpa said, you could do this by defining a second procedure, but this is a fairly straightforward implementation so you could do something like this:

(define (position L)
  (lambda (N) 
    (cond ((null? L) '())
          ((= N 1) (car L))
          (else ((position (cdr L)) (- N 1))))))

(define X (position '(1 5 8 2 7)))

Here, X is defined as a #<procedure> bound to the function position with the list L. In the else statement, you see (position (cdr L)) which is equivalent to X, which we follow with (- N 1).