java - 监听器和线程

Question

我觉得这可能是一个愚蠢的问题，但我对听众没有那么丰富的经验......无论如何，我有一个关于线程和听众的问题；考虑这段代码(语法上可能不正确，这是我突然想到的):

public class Stuff {
    private SimpleLongProperty time = new LongProperty(this, "time");
    private Executor execute;

    public Stuff(Clock clock) {
        time.bind(clock.getValue);
        execute = Executors.newFixedThreadPool(5);
    }

    public void someAction() {
        for(int i = 0; i < 5; i++) {
            execute.execute(scheduleTask());
        }
    }

    public Runnable scheduleTask() {
        time.addListener((obs, oldV, newV) -> {
            //Code here
        });
    }
}

当调用 someAction() 并调用 ScheduleTask() 5 次以添加 5 个监听器时，当时间更新时，每个监听器是否会在自己的线程内执行代码？或者它会执行主线程中的代码，因为那是时间所在的地方？

Answer 1

能够在工作站上编写代码并进行测试，找到了我的答案。

public class Stuff {
    private LongProperty time = new SimpleLongProperty(this, "time");
    private Executor execute;

    public Stuff(Clock clock) {
        time.bind(clock.getValue);
        execute = Executors.newCachedThreadPool(runnable -> {
            Thread t = new Thread(runnable);
            t.setDaemon(true);
            return t;
        });
    }

    public void someAction() {
        for(int i = 0; i < 5; i++) {
            execute.execute(scheduleTask(i));
        }
    }

    public Runnable scheduleTask(int i) {
        time.addListener((obs, oldV, newV) -> {
            System.out.println("Task " + i + ": " + Thread.currentThread());
        });
    }
}

上面的代码将打印:

Task 0: Thread[JavaFX Application Thread, 5, main]
Task 1: Thread[JavaFX Application Thread, 5, main]
Task 2: Thread[JavaFX Application Thread, 5, main]
Task 3: Thread[JavaFX Application Thread, 5, main]
Task 4: Thread[JavaFX Application Thread, 5, main]

将 someAction() 和 ScheduleTask() 函数更改为如下所示:

public void someAction() {
    for(int i = 0; i < 5; i++) {
        scheduleTask(i);
    }
}

public void scheduleTask(int i) {
    Runnable test = () -> {
        System.out.println("Task " + i + ": " + Thread.currentThread());
    };
    time.addListener((obs, oldV, newV) -> {
        execute.execute(test);
    });
}

将产生以下结果:

Task 0: Thread[Thread-16, 5, main]
Task 1: Thread[Thread-17, 5, main]
Task 2: Thread[Thread-20, 5, main]
Task 3: Thread[Thread-19, 5, main]
Task 4: Thread[Thread-18, 5, main]