return range of list in scheme

in scheme,

list-ref returns only one element.

but I want to do like

(my-list-operation 0 4 '(1 2 3 4 5 6 7 8 9开发者_JAVA百科 10 11 12))

=> '(1 2 3 4)

could somebody tell me how to do this?

I believe in using existing quality libraries where available. So, this answer uses SRFI 1 (if you're using Racket, load it using (require srfi/1)):

(define (list-range lst start end)
  (take (drop lst start) (- end start)))

Example:

(list-range (iota 12 1) 0 4)  ; => (1 2 3 4)

(define get-n-items
    (lambda (lst num)
        (if (> num 0)
            (cons (car lst) (get-n-items (cdr lst) (- num 1)))
            '()))) ;'

(define slice
    (lambda (lst start count)
        (if (> start 1)
            (slice (cdr lst) (- start 1) count)
            (get-n-items lst count))))

get-n-items is a helper, which appends n items together. slice takes a list, offset and count and again loops until it finds the start index--then returns n items from there.

(From How do I take a slice of a list (A sublist) in scheme?)

(define my-list-operation
   (lambda (start end lst)
      (cond ((= 0 end) '())
            ((= 0 start) (cons (car lst) (my-list-operation 0 (- end 1) (cdr lst))))
            (else
               (my-list-operation (- start 1) (- end 1) (cdr lst))))))

That should do it.

A better solution is:

(define (slice lst start [end #f])
  (let ([lst0 (drop lst start)])
    (take lst0 (or end (length lst0)))))

Since that will pass the following tests:

 (require rackunit)
 (test-case "slice list"
  (check-equal? (slice '(1 2 3 4) 1 1) '(2))
  (check-equal? (slice '(1 2 3 4) 0 1) '(1))
  (check-equal? (slice '(1 2 3 4) 0 2) '(1 2))
  (check-equal? (slice '(1 2 3 4) 0) '(1 2 3 4))
  (check-equal? (slice '(1 2 3 4) 1 2) '(2 3))
  (check-equal? (slice '(1 2 3 4) 1) '(2 3 4)))

Chris' solution doesn't cover all of these scenarios.