如果一个线程完成了它的最后一条语句的执行，那么它是否保证像显式调用return一样死亡？

Question

例如，以下面的代码段为例，

static volatile boolean isDone = false;
public static void main(String[] args) throws InterruptedException {
    // I know ArrayBlockingQueue cannot take 0 as the constructor arg, but please for the sake of example, let's pretend this was legal
    final ExecutorService exec = new ThreadPoolExecutor(1, 1, 0L, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<>(0));
    exec.execute(() -> {
        isDone = true;
        // does the compiler add a return statement here?
    });
    while (!isDone){
    }
    try {
        exec.execute(() -> {});
    } catch (RejectedExecutionException e) {
        System.out.println("What if this might actually happen in a super bad timing by the scheduler??");
    }
    exec.shutdown();
}

ExecutorService创建的第一个线程似乎已经死亡，但真的是这样吗？

exec.execute(() -> {
    isDone = true;
    return;
});

但是，在实际执行return之前可能会有一些延迟，因为这取决于调度器，所以如果调度器决定在执行exec.execute(()->{}）之前不运行return语句，下一个提交给executor的runnable可能会被拒绝；在尝试区。

Answer 1

是的，编译器的行为就像任何方法的结尾都包含return；语句一样。

不，你的代码不正确。

问题是两个线程可以以任何可能的方式交织。正如墨菲所说，能发生的事情（最终）会发生。

public static void main(String[] args) throws InterruptedException {
    final ExecutorService exec = new ThreadPoolExecutor(1, 1, 0L, TimeUnit.SECONDS, new SynchronousQueue<Runnable>());
    Future<?> future = exec.submit(() -> {
        // Do work
    });

    try {
        future.get(); //blocking call
    } catch (ExecutionException e) {
        e.getCause().printStackTrace(); // The RuntimeException thrown in the lambda.
    }

    try {
        exec.submit(() -> {});
    } catch (RejectedExecutionException e) {
        // Shouldn't happen
    }
    exec.shutdown();
}

而队列长度为0的执行器实际上是可能的：

new ThreadPoolExecutor(1, 1, 0L, TimeUnit.SECONDS, new SynchronousQueue<Runnable>());