synchronized 关键字给出了预期的输出，但对不同线程调用该方法的顺序不满意

Question

下面是我的 3 classes(threads) 实现 Runnable 接口：

public class Thread1 implements Runnable {
Shared s;

public Thread1(Shared s){
    this.s = s;
}

@Override   
public void run() {
    System.out.println("Sum of 10 and 20 by " + Thread.currentThread().getName() + " is "+ s.add(10, 20));
}

}

public class Thread2 implements Runnable {
Shared s;

public Thread2(Shared s){
    this.s = s;
}
    
@Override
public void run() {
    System.out.println("Sum of 100 and 200 by " + Thread.currentThread().getName() + " is " + s.add(100, 200));
}

}

public class Thread3 implements Runnable {
Shared s;

public Thread3(Shared s){
    this.s = s;
}
    
@Override
public void run() {
    System.out.println("Sum of 1000 and 2000 by " + Thread.currentThread().getName() + " is " + s.add(1000, 2000));
}

}

下面是class，其对象在这些线程之间共享：

public class Shared {
private int x;
private int y;

synchronized public int add(int a, int b) {
    x = a;
    y = b;

    try {
        System.out.println(Thread.currentThread().getName() + "is going into sleep state");
        Thread.sleep(1000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    return x + y;
}

}

最后，这就是我启动线程和传递共享对象的方式：

    public static void main(String[] args) {
    Shared s = new Shared();
    Thread t1 = new Thread(new Thread1(s), "Thread-1");
    Thread t2 = new Thread(new Thread2(s), "Thread-2");
    Thread t3 = new Thread(new Thread3(s), "Thread-3");
    
    t1.start();
    t2.start();
    t3.start();
}

这是我得到的正确输出：

      Thread-2is going into sleep state
      Thread-1is going into sleep state
      Sum of 100 and 200 by Thread-2 is 300
      Thread-3is going into sleep state
      Sum of 10 and 20 by Thread-1 is 30
      Sum of 1000 and 2000 by Thread-3 is 3000

但是如果你看到这里，Thread-1 在 Thread-2 完成它的工作之前开始执行 add 方法（这是同步的）。由于 Thread-2 进入休眠状态，因此它自己也获得了锁，并且不应允许其他线程进入 add(...) 方法。 Thread-1 或 Thread-3 只能在 Thread-2 完成后才开始执行 add(...) 方法。所以，我期待的输出是：

      Thread-2is going into sleep state
      Sum of 100 and 200 by Thread-2 is 300
      Thread-1is going into sleep state
      Sum of 10 and 20 by Thread-1 is 30
      Thread-3is going into sleep state
      Sum of 1000 and 2000 by Thread-3 is 3000

请告诉我哪里做错了或者输出结果是这样的，如果是，请告诉原因。

Answer 1

原因是System.out.println()速度很慢。

您可以使用它从运行() 方法内部和 add() 内部发送消息。无法保证 sysout 中的消息一致。

我重写了Shared如下：

public class Shared {
    private int x;
    private int y;

    synchronized public int add(int a, int b) {
        x = a;
        y = b;

        try {
            long now = System.currentTimeMillis() - start;

            Thread.sleep(1000);

            // notice FIRST sleep and THEN sysout
            System.out.println(Thread.currentThread().getName() + " calculation has taken place at time " + now);

        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        return x + y;
    }

    public static long start;

    public static void main(String[] args) {

        start = System.currentTimeMillis();

        Shared s = new Shared();
        Thread t1 = new Thread(new Thread1(s), "Thread-1");
        Thread t2 = new Thread(new Thread2(s), "Thread-2");
        Thread t3 = new Thread(new Thread3(s), "Thread-3");

        t1.start();
        t2.start();
        t3.start();
    }
}

每个线程现在看起来像：

public class Thread1 implements Runnable {
    Shared s;

    public Thread1(Shared s) {
        this.s = s;
    }

    @Override
    public void run() {
        int result = s.add(10, 20);
        long now= System.currentTimeMillis()-Shared.start;
        System.out.println("Sum of 10 and 20 by " + Thread.currentThread().getName() + " is " + result+ " found at "+ now);
    }
}

生成以下输出：

Thread-1 calculation has taken place at time 2
Sum of 10 and 20 by Thread-1 is 30 found at 1005
Thread-2 calculation has taken place at time 1005
Sum of 100 and 200 by Thread-2 is 300 found at 2006
Thread-3 calculation has taken place at time 2006
Sum of 1000 and 2000 by Thread-3 is 3000 found at 3007

每个线程都在 add() 中阻塞。

等到后面显示结果的时候，下一个线程已经开始计算了。

注意 add() 现在首先是 sysout（很慢），它发生在睡眠期间，这给了它足够的时间。

Answer 2

您需要同步s如下图：

@Override
public void run() {
    synchronized (s) {
        System.out.println("Sum of 10 and 20 by " + Thread.currentThread().getName() + " is " + s.add(10, 20));
    }
}

在类、Thread2 和 Thread3 中也这样做。检查 this 以获得解释。

synchronized 关键字给出了预期的输出，但对不同线程调用该方法的顺序不满意

synchronized keyword is giving the expected output but not satisfied with the order in which the method is called by different threads

java

multithreading

synchronization

synchronized