I've started learning Clojure a few days ago and I wrote this code that factorises numbers. How do I make it better? Is there any way to avoid doing it with a loop? Is that how I am supposed to code in Clojure?

(defn classify [n]
  (loop [n n
         i 2
         f []]
      (= n i) (conj f i)
      (= (mod n i) 0) (recur (/ n i) 2 (conj f i))
      :else (recur n (inc i) f))))
  • 1
    \$\begingroup\$ This is functional. loop is the functional way to "loop" in Clojure. doseq for example would be the "bad", imperative way of doing things. \$\endgroup\$ May 2, 2021 at 17:35

1 Answer 1


Maybe a slightly more functional way to express your algorithm is ...

(defn classify [n]
  (letfn [(factors-from [i from]
              (= i 1) '()
              (zero? (mod i from)) (cons from (factors-from (quot i from) from))
              :else (recur i (inc from))))]
   (factors-from n 2)))
  • The local function factors-from plays the role of the loop.
  • The recursive call when we have found a factor saves us carrying the sequence of factors as a parameter.
  • We use quot instead of / for the division, since it is simpler.
  • The terminal case is 1instead of from. This gives the correct answer of an empty factor list when n is 1.
  • The normal case, when we don't find a factor, is a straight tail recursion, hence can be a recur.

There is no danger of running out of stack space, since the smallest number that could do so is about 2^10000, far beyond anything long arithmetic could deal with.

  • \$\begingroup\$ Thank you! I can learn a lot by reading this code and trying out the functions you used. \$\endgroup\$ May 4, 2021 at 9:52

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