清理其渠道地图

Cleansing a Map of Its Channels

假设我们有一个具有以下结构的映射 m

{:a (go "a") 
  :b "b" 
  :c "c" 
  :d (go "d")}

如图所示,m有四个键,其中两个包含频道。

问题:如何编写一个通用函数(或宏?)cleanse-map,它采用像m 并输出其无通道版本(在本例中为 {:a "a" :b "b" :c "c" :d "d"})?

这个问题的一个很好的辅助函数可能如下:

(defn chan? [c]
  (= (type (chan)) (type c)))

如果 cleanse-map 的 return 值(或任何名称)本身就是一个频道,也没有关系。即:

`(cleanse-map m) ;=> (go {:a "a" :b "b" :c "c" :d "d"})
core.async

Limitations 使得 cleanse-map 的实现不是那么简单。但下面的应该有效:

(defn cleanse-map [m]
  (let [entry-chs (map
                   (fn [[k v]]
                     (a/go
                       (if (chan? v)
                         [k (a/<! v)]
                         [k v])))
                   m)]
    (a/into {} (a/merge entry-chs))))

基本上,这里做了什么:

  1. 每个地图条目都被转换为一个包含该地图条目的通道。如果映射条目的值是通道,则在映射函数内的 go-块内提取它。
  2. 带有地图条目的频道 merge-d 变成了一个。完成此步骤后,您将拥有一个包含地图条目集合的频道。
  3. 带有地图条目的频道转换为将包含所需地图的频道(a/into 步骤)。
(ns foo.bar
  (:require
    [clojure.core.async :refer [go go-loop <!]]
    [clojure.core.async.impl.protocols :as p]))

(def m
  {:a (go "a")
   :b "b"
   :c "c"
   :d (go "d")
   :e "e"
   :f "f"
   :g "g"
   :h "h"
   :i "i"
   :j "j"
   :k "k"
   :l "l"
   :m "m"})

(defn readable? [x]
  (satisfies? p/ReadPort x))

(defn cleanse-map
  "Takes from each channel value in m,
   returns a single channel which will supply the fully realized m."
  [m]
  (go-loop [acc {}
            [[k v :as kv] & remaining] (seq m)]
    (if kv
      (recur (assoc acc k (if (readable? v) (<! v) v)) remaining)
      acc)))

(go (prn "***" (<! (cleanse-map m))))

=> "***" {:m "m", :e "e", :l "l", :k "k", :g "g", :c "c", :j "j", :h "h", :b "b", :d "d", :f "f" , :i "i", :a "a"}