来自池的CompletableFuture重用线程 [英] CompletableFuture reuse thread from pool

问题描述

我正在测试可实现的未来。
如此处所述
我以为线程将从公共池中重用，但是此摘要显示了奇怪的行为

I'm testing Completable Future. As described here I thought thread would be reused from common pool but this snippet shows strange behaviour

for (int i = 0; i < 10000; i++) {
            final int counter = i;
            CompletableFuture.supplyAsync(() -> {

                System.out.println("Looking up " + counter + " on thread " + Thread.currentThread().getName());
                return null;
            });
}

我有这样的输出：

Looking up 0 on thread Thread-2
Looking up 1 on thread Thread-3
Looking up 2 on thread Thread-4
...
Looking up 10000 on thread Thread-10002

线程中查找10000为每个任务创建线程。
为什么我的所有completableFuture都不会重用公共池中的线程？

it looks like a new thread is created for each task. Why all of my completableFuture do not reuse thread from common pool?

MoreOver我已经用RxJava测试过了，并且可以在以下代码中正常工作：

MoreOver I've tested with RxJava and it works as excpected with this code:

for (int i = 0; i < 10000; i++) {
            rxJobExecute(i).subscribeOn(Schedulers.io()).subscribe();
        }

private Observable<String> rxJobExecute(int i) {
        return Observable.fromCallable(() -> {

            System.out.println("emission " + i + " on thread " + Thread.currentThread().getName());
            return "tata";

        });
    }

输出

emission 8212 on thread RxIoScheduler-120
emission 8214 on thread RxIoScheduler-120
emission 8216 on thread RxIoScheduler-120
emission 8218 on thread RxIoScheduler-120
emission 8220 on thread RxIoScheduler-120
emission 7983 on thread RxIoScheduler-275
emission 1954 on thread RxIoScheduler-261
emission 1833 on thread RxIoScheduler-449
emission 1890 on thread RxIoScheduler-227

推荐答案

由于您只有2个处理器，则在应用程序启动时，值 Runtime.getRuntime（）。availableProcessors（）仅观察到 1 处理器（

Chances are that since you only have 2 processors then the value Runtime.getRuntime().availableProcessors() on starting of the application only observed 1 processor (the avialableProcessors will return any number between 1 and the number of processors on your machine and isn't quite deterministic).

ForkJoin公共池将每线程使用一个线程，而avialableProcessors将返回介于1和计算机上处​​理器数量之间的任意数字，并且不确定性）。任务线程池（如果并行度为1）。

The ForkJoin common pool will use a thread-per-task thread pool if the parallelism is 1.

要强制系统以特定的并行度加载（至少为此），请定义系统属性 -Djava.util.concurrent.ForkJoinPool.common.parallelism = 2 作为运行时参数

To force the system to load with a particular parallelism (for this at least) define the system property -Djava.util.concurrent.ForkJoinPool.common.parallelism=2 as a runtime argument

编辑：

我再次查看了内部逻辑。由于您有 2 个内核，因此并行性始终是使用每个任务线程。该逻辑期望大于或等于3，因此您需要将并行度更新为3

I looked again at the internal logic. Since you have 2 cores, the parallelism will always be to use thread-per-task. The logic expects greater than or equal to 3, so you will need to update parallelism to 3

-Djava.util.concurrent.ForkJoinPool.common .parallelism = 3 。

另一种方法是定义自己的ThreadPool

The alternative is to define your own ThreadPool

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