unfold in Clojure

unfold.clj

(use 'clojure.test)

(defn unfold
 "A variant of 'iterate' which accepts a stopping condition,
  having the same syntax as 'unfold' of scheme.
  Supposed
    (x1 x2 x3 ... xk-1 xk ... ) == (s (g s) (g (g s)) ... )
  and '(p xi)' returns true first at i == k,
    ((f x1) (f x2) (f x3) ... (f xk-1))
  is returned."
  [p f g s]
  (loop [c s rvsd '()]
    (if (p c)
        (loop [acc '() [r & rs] rvsd]
          (if r (recur (cons (f r) acc) rs) acc))
        (recur (g c) (cons c rvsd)))))

(deftest test-unfold
  (are [expected result] (= expected result)
    (unfold #(<=  12 %) #(* 2 %) #(+ 2 %) 3) '(6 10 14 18 22)
    (unfold #(<= 100 %) #(+ 2 %) #(* 2 %) 1) '(3 4 6 10 18 34 66)
    ))

'cont-unfold' consumes large stack.
(cont-unfold #(<= 10000 %) identity inc 0) ; -> (0 1 2 ... 9999)
(cont-unfold #(<= 100000 %) identity inc 0) ; -> StackOverflowError clojure.lang.RT.boundedLength (RT.java:1607)
What is a better implementation of unfold?

Using a list which contains what calculations should be done, finally 'reverse' it and apply.
It became not to eat stack, but still using large heap. Hum...

'comp' calls 'reverse' internally.
So I changed not to use 'reverse' and 'comp' but to 'comp'oseing manually.
Still eat large heap.

The operation '#(cons (f %) %2)' doesn't vary in the chain of calculations.
We only have to consing results of '#(g %)' in a list and return with reversed order.