You don't need to recur to another function, just recur to a loop:
(defn factorial [n]
(loop [x n acc 1]
(if (zero? x) acc (recur (dec x) (* acc x)))))
(println (factorial 5))
--
Science answers questions; philosophy questions answers.
On Jul 15, 2010, at 6:38 AM, Brisance wrote:
> Here's a factorial function as found in the WikiBook "Learning
> Clojure" <http://en.wikibooks.org/wiki/Learning_Clojure>:
>
> (defn factorial [n]
> (defn fac [n acc]
> (if (zero? n)
> acc
> (recur (- n 1) (* acc n)))) ; recursive call to fac, but reuses
> the stack; n will be (- n 1), and acc will be (* acc n)
> (fac n 1))
>
> Question: how would I go about writing idiomatic Clojure to return
> factorials of n, for large values of n. e.g. 1e6 or more? Preferably
> without having to create another function.
>
> Thanks in advance for any insight.
>
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