Understanding Clojure concurrency example

Question

I just go through various documentation on Clojure concurrency and came accross the example on the website (http://clojure.org/concurrent_programming).

(import '(java.util.concurrent Executors))
(defn test-stm [nitems nthreads niters]
(let [refs  (map ref (replicate nitems 0))
      pool  (Executors/newFixedThreadPool nthreads)
      tasks (map (fn [t]
                   (fn []
                     (dotimes [n niters]
                       (dosync
                         (doseq [r refs]
                           (alter r + 1 t))))))
                (range nthreads))]
(doseq [future (.invokeAll pool tasks)]
  (.get future))
(.shutdown pool)
(map deref refs)))

I understand what it does and how it works, but I don't get why the second anonymous function fn[] is needed?

Many thanks,

dusha.

P.S. Without this second fn [] I get NullPointerException.

Answer 1

Here is a classic example of using higher-order functions:

;; a function returns another function
(defn make-multiplyer [times]
  (fn [x]
    (* x times)))

;; now we bind returned function to a symbol to use it later
(def multiply-by-two (make-multiplyer 2))

;; let's use it
(multiply-by-two 100)   ; => 200

---

In that code sample fn inside fn works the same way. When map invokes (fn [t] (fn [] ...)) it gets inner fn.

(def list-of-funcs (map (fn [t]
                          (fn [] (* t 10)))   ; main part
                        (range 5)))
;; Nearly same as
;; (def list-of-funcs (list (fn [] (* 0 10))
;;                          (fn [] (* 1 10))
;;                          ...
;;                          (fn [] (* 4 10))))


(for [i list-of-funcs]
  (i))
; => (0 10 20 30 40)

Update: And as Alex said tasks in the code sample is bound to list of callables which is passed then to .invokeAll().