为了获得一些好的并行编程实践,我正在尝试使用Clojure的core.async库实现生产者/消费者模式。一切运行良好,但我希望能够在某个时候停止生产者和消费者。
我当前的代码看起来像这样...
(def c (a/chan 5))
(def alive (atom true))
(def producer (a/go-loop []
(Thread/sleep 1000)
(when @alive
(a/>! c 1)
(recur))))
(def consumer (a/go-loop []
(Thread/sleep 3000)
(when @alive
(println (a/<! c))
(recur))))
(do
(reset! alive false)
(a/<!! producer)
(a/<!! consumer))
不幸的是,似乎“ do”块偶尔会无限期地阻塞。我本质上希望能够停止两个继续执行并阻塞的go循环,直到两个循环都退出为止。那里有线程/睡眠代码来模拟执行某些工作单元。
我怀疑停止生产者会使消费者停下来,因此绞死了,尽管我不确定是否有其他方法,但有什么主意吗?
答案 0 :(得分:1)
请see ClojureDocs以获取详细信息。示例:
(let [c (chan 2) ]
(>!! c 1)
(>!! c 2)
(close! c)
(println (<!! c)) ; 1
(println (<!! c)) ; 2
;; since we closed the channel this will return false(we can no longer add values)
(>!! c 1))
对于您的问题,例如:
(let [c (a/chan 5)
producer (a/go-loop [cnt 0]
(Thread/sleep 1000)
(let [put-result (a/>! c cnt)]
(println "put: " cnt put-result)
(when put-result
(recur (inc cnt)))))
consumer (a/go-loop []
(Thread/sleep 3000)
(let [result (a/<! c)]
(when result
(println "take: " result)
(recur))))]
(Thread/sleep 5000)
(println "closing chan...")
(a/close! c))
有结果
put: 0 true
put: 1 true
take: 0
put: 2 true
put: 3 true
closing chan...
put: 4 false
take: 1
take: 2
take: 3