var*_*nie 2 asynchronous channel clojure clojurescript
我有以下场景:
有一些服务我用来检索一些数据传递给我输入.
有一些输入参数我需要针对上述服务执行N个请求,收集输出并为每个输出执行一些CPU繁重的任务.
我试图使用核心/异步通道实现这一目标.
这是我的尝试(示意图)哪种有效,但它不表现我想要它.将对如何改进它的任何提示表示感谢.
(defn produce-inputs
[in-chan inputs]
(let input-names-seq (map #(:name %) inputs)]
(doseq [input-name input-names-seq]
(async/go
(async/>! in-chan input-name)))))
(defn consume
[inputs]
(let [in-chan (async/chan 1)
out-chan (async/chan 1)]
(do
(produce-inputs in-chan inputs)
(async/go-loop []
(let [input-name (async/<! in-chan)]
(do
(retrieve-resource-from-service input-name
; response handler
(fn [resp]
(async/go
(let [result (:result resp)]
(async/>! out-chan result)))))
(when input-name
(recur)))))
; read from out-chan and do some heavy work for each entry
(async/go-loop []
(let [result (async/<! out-chan)]
(do-some-cpu-heavy-work result))))))
; entry point
(defn run
[inputs]
(consume inputs))
Run Code Online (Sandbox Code Playgroud)
有没有办法更新它,以便每次不会有超过五个service(retrieve-resource-from-service)活动请求?
如果我的解释不明确,请提出问题,我会更新.
您可以创建另一个通道作为令牌桶,以限制请求的速率.
要限制同时请求的数量,您可以执行以下操作:
(defn consume [inputs]
(let [in-chan (async/chan 1)
out-chan (async/chan 1)
bucket (async/chan 5)]
;; ...
(dotimes [_ 5] (async/put! bucket :token))
(async/go-loop []
(let [input-name (async/<! in-chan)
token (async/<! bucket)]
(retrieve-resource-from-service
input-name
; response handler
(fn [resp]
(async/go
(let [result (:result resp)]
(async/>! out-chan result)
(async/>! bucket token)))))
(when input-name
(recur))))
;; ...
))
Run Code Online (Sandbox Code Playgroud)
bucket创建一个新频道,并将五个项目放入其中.在触发请求之前,我们从存储桶中获取令牌,并在请求完成后将其放回.如果bucket频道中没有令牌,我们必须等到其中一个请求完成.
注意:这只是代码的草图,您可能需要更正它.特别是,如果您的retrieve-resource-from-service函数中有任何错误处理程序,则应该在发生错误时放回令牌以避免最终的死锁.