java - 如何防止 CompletableFuture#whenComplete 在上下文线程中执行

Question

我有以下代码:

ConcurrentHashMap taskMap= new ConcurrentHashMap();
....
taskMap.compute(key, (k, queue) -> {
        CompletableFuture<Void> future = (queue == null)
                ? CompletableFuture.runAsync(myTask, poolExecutor)
                : queue.whenCompleteAsync((r, e) -> myTask.run(), poolExecutor);
        //to prevent OutOfMemoryError in case if we will have too much keys
        future.whenComplete((r, e) -> taskMap.remove(key, future));            
        return future;
    });

此代码的问题是，如果 future 已经完成 whenComplete 函数参数在与 compute 调用相同的线程中调用。在此方法的主体中，我们从 map 中删除条目。但计算方法文档禁止这样做，应用程序会卡住。

如何解决这个问题？

Answer 1

最明显的解决方案是使用 whenCompleteAsync 而不是 whenComplete，因为前者保证使用提供的 Executor 执行操作，而不是调用线程。可以通过以下方式演示

Executor ex = r -> { System.out.println("job scheduled"); new Thread(r).start(); };
for(int run = 0; run<2; run++) {
    boolean completed = run==0;
    System.out.println("*** "+(completed? "with already completed": "with async"));
    CompletableFuture<String> source = completed?
        CompletableFuture.completedFuture("created   in "+Thread.currentThread()):
        CompletableFuture.supplyAsync(() -> {
            LockSupport.parkNanos(TimeUnit.SECONDS.toNanos(1));
            return "created   in "+Thread.currentThread();
        }, ex);

    source.thenApplyAsync(s -> s+"\nprocessed in "+Thread.currentThread(), ex)
          .whenCompleteAsync((s,t) -> {
                if(t!=null) t.printStackTrace(); else System.out.println(s);
                System.out.println("consumed  in "+Thread.currentThread());
            }, ex)
          .join();
}

这将打印类似的内容

*** with already completed
job scheduled
job scheduled
created   in Thread[main,5,main]
processed in Thread[Thread-0,5,main]
consumed  in Thread[Thread-1,5,main]
*** with async
job scheduled
job scheduled
job scheduled
created   in Thread[Thread-2,5,main]
processed in Thread[Thread-3,5,main]
consumed  in Thread[Thread-4,5,main]

所以你可以使用

taskMap.compute(key, (k, queue) -> {
        CompletableFuture<Void> future = (queue == null)
                ? CompletableFuture.runAsync(myTask, poolExecutor)
                : queue.whenCompleteAsync((r, e) -> myTask.run(), poolExecutor);
        //to prevent OutOfMemoryError in case if we will have too much keys
        future.whenCompleteAsync((r, e) -> taskMap.remove(key, future), poolExecutor);
        return future;
    });

如果提前完成的可能性很大，您可以使用以下方法减少开销

taskMap.compute(key, (k, queue) -> {
        CompletableFuture<Void> future = (queue == null)
                ? CompletableFuture.runAsync(myTask, poolExecutor)
                : queue.whenCompleteAsync((r, e) -> myTask.run(), poolExecutor);
        //to prevent OutOfMemoryError in case if we will have too much keys
        if(future.isDone()) future = null;
        else future.whenCompleteAsync((r, e) -> taskMap.remove(key, future), poolExecutor);
        return future;
    });

也许，您没有找到这个明显的解决方案，因为您不喜欢依赖操作始终被安排为池中的新任务，即使完成已经发生在不同的任务中。您可以使用专门的执行程序来解决此问题，该执行程序只会在必要时重新安排任务:

Executor inPlace = Runnable::run;
Thread forbidden = Thread.currentThread();
Executor forceBackground
       = r -> (Thread.currentThread()==forbidden? poolExecutor: inPlace).execute(r);

…

future.whenCompleteAsync((r, e) -> taskMap.remove(key, future), forceBackground);

---

但是您可能会重新考虑这种复杂的每个映射清理逻辑是否真的需要。它不仅很复杂，而且可能会产生显着的开销，可能会安排大量清理操作，而这些操作在执行时已经过时了，而这些操作并不是真正需要的。

执行起来可能会更简单、更高效

taskMap.values().removeIf(CompletableFuture::isDone);

不时清理整个 map 。