Order of execation of commands in Scheme

Question

Hi !

I am writing a program in scheme, that uses recursion to walk through the list, and stop at certain pointer, when counter reaches a certain number N

(define (functX N lst)
  (define counter 1)
  (cond
    [(empty? lst) empty]
    [(negative? N) empty]
    [(< (length lst) N) empty]
    [(<= counter N) ((set! counter (+ counter 1))(cons (first lst) (functX N (rest lst)))))]
    [else empty]))

I don't understand, why the second line from the bottom is giving me trouble: the error I am getting is "procedure application: expected procedure, given: '(1) (no arguments)"

Answer 1

You should consider decrementing N in the recursive call and removing the counter variable altogether.

Answer 2

You have it enclosed in parentheses twice. Expressions in Scheme have the form (func-expr arg-expr ...), so the first expression must evaluate into a function. So if you did:

(define (f n) n)
((f 5))

It would evaluate (f 5) then it would try and evaluate (5) which is an error.

Edit: Some clarification.

You have the following enclosed in brackets twice:

((set! counter (+ counter 1))(cons (first lst) (functX N (rest lst)))))

So first it evaluates set! and reduces down to (where n is a number):

(n (cons ...))

cons is then evaluated along with its arguments (where x is the result):

(n x)

It then tries to apply the argument x to the function n, but since n is a number this results in an error. If you wanted to do two separate computations and only return the value of one, you can use begin.

(begin (set! counter (+ counter 1)) (cons (first lst) (functX N (rest lst))))

Update:

Here is a function that appears to do what you want without voodoo (since mutation is evil).

(define (take n xs)
  (cond
    [(empty? xs) empty]
    [(negative? n) empty]
    [(eq? n 0) empty]
    [else (cons (first xs) (take (- n 1) (rest xs)))]))