我在http://www.learningclojure.com/2010/11/yet-another-way-to-write-factorial.html上找到了这段代码,但我不明白弹出任务是否应该是线程安全的,或者是如何实现的。难道不允许返回两次相同的头吗?
(def to-do-list (atom '()))
(defn add-task![t] (swap! to-do-list #(cons t %)))
(defn pop-task![] (let [h (first @to-do-list)] (swap! to-do-list rest) h))如果是这样的话,有没有可能继续使用atom并编写peek和交换!原子操作,或者这是ref机制的工作?
发布于 2011-04-20 18:12:46
或者你降到一个更低的级别。
(def to-do-list (atom nil))
(defn add-task!
[t]
(swap! to-do-list conj t))
(defn pop-task!
[]
(let [[h & r :as l] @to-do-list]
(if (compare-and-set! to-do-list l r)
h
(recur))))发布于 2011-04-20 17:20:01
是的,这段代码不是线程安全的。你可以让它成为线程安全的,通过利用交换!返回原子的新值,这意味着您需要将队列与"popped“值组合在一起。
(def to-do-list
(atom {}))
(defn add-task!
[t]
(swap! to-do-list
(fn [tl]
{:queue (cons t (:queue tl))})))
(defn pop-task!
[]
(let [tl (swap! to-do-list
(fn [old]
{:val (first (:queue old))
:queue (rest (:queue old))}))]
(:val tl)))https://stackoverflow.com/questions/5727611
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