Clojure 的 pmap 函数为 URL 获取操作生成了多少个线程? [英] How many threads does Clojure's pmap function spawn for URL-fetching operations?
问题描述
关于 pmap
函数的文档让我想知道对于通过网络获取 XML 提要集合之类的事情会有多高效.我不知道 pmap 会产生多少并发提取操作以及最大值是多少.
The documentation on the pmap
function leaves me wondering how efficient it would be for something like fetching a collection of XML feeds over the web. I have no idea how many concurrent fetch operations pmap would spawn and what the maximum would be.
推荐答案
如果你查看源码你会看到:
If you check the source you see:
> (use 'clojure.repl)
> (source pmap)
(defn pmap
"Like map, except f is applied in parallel. Semi-lazy in that the
parallel computation stays ahead of the consumption, but doesn't
realize the entire result unless required. Only useful for
computationally intensive functions where the time of f dominates
the coordination overhead."
{:added "1.0"}
([f coll]
(let [n (+ 2 (.. Runtime getRuntime availableProcessors))
rets (map #(future (f %)) coll)
step (fn step [[x & xs :as vs] fs]
(lazy-seq
(if-let [s (seq fs)]
(cons (deref x) (step xs (rest s)))
(map deref vs))))]
(step rets (drop n rets))))
([f coll & colls]
(let [step (fn step [cs]
(lazy-seq
(let [ss (map seq cs)]
(when (every? identity ss)
(cons (map first ss) (step (map rest ss)))))))]
(pmap #(apply f %) (step (cons coll colls))))))
(+ 2 (.. Runtime getRuntime availableProcessors))
是一个重要的线索.pmap 将获取第一个 (+ 2 个处理器)
的工作片段并通过 future
异步运行它们.因此,如果您有 2 个内核,它将一次启动 4 个工作,试图保持领先于您,但最大值应该是 2+n.
The (+ 2 (.. Runtime getRuntime availableProcessors))
is a big clue there. pmap will grab the first (+ 2 processors)
pieces of work and run them asynchronously via future
. So if you have 2 cores, it's going to launch 4 pieces of work at a time, trying to keep a bit ahead of you but the max should be 2+n.
future
最终使用支持无限数量线程的代理 I/O 线程池.它会随着工作的增加而增长,如果线程未使用则收缩.
future
ultimately uses the agent I/O thread pool which supports an unbounded number of threads. It will grow as work is thrown at it and shrink if threads are unused.
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