# [racket] taking from a generator, and a possible bug

I want to,
(take k (in-generator (infinite-generator (yield (random n)))))
where k and n are positive integers, k<n.
Unfortunately, 'in-generator' returns a sequence while 'take' needs a
list and calling 'sequence->list' results in an error. So, I thought
maybe I needed 'take' from 'lazy'. But it also complains about the
index being too large.
That's when I encountered a possible (minor) bug with 'take' vs
'lazy/take'. They appear to have their arguments swapped:
>* (take '(a b) 1)
*'(a)
>* (require (only-in lazy [take lazy-take]))
*>* (lazy-take '(a b) 1)
*. . take: expects type <non-negative exact integer> as 1st argument,
given: '(a b); other arguments were: 1
My current approach aside, is there a better way?
Here's my workaround,
(define (call-n-times f n)
(define (helper f n lst)
(if (< n 1)
lst
(helper f (- n 1) (cons (f) lst))))
(helper f n '()))
(define (take-randoms k n)
(call-n-times (curry random n) k))
--
Ian Tegebo