java - 多线程我是否正确使用等待和通知

Question

我写了一个哲学家问题的解决方案。它正在运行，我在控制台上得到了正确的输出，但是 wait() 之后的 println() 从未被打印。请告诉我为什么。我在代码中指出了。

该解决方案与http://en.wikipedia.org/wiki/Dining_philosophers_problem#Resource_hierarchy_solution类似

public class Philosopher extends Thread {
    String name;
    // boolean je, czeka;
    int nr;
    Fork left, right;

    public Philosopher(String name, int nr, Fork left, Fork right) {
        this.nr = nr;
        this.name = name;
        this.left = left;
        this.right = right;
        System.out.println("NR " + nr +"  "+ left + " " + right);
    }

    public void run() {
        // while(true){
        try {
            Fork minF = Fork.min(left, right);
            Fork maxF = Fork.max(left, right);

            synchronized (minF) {
                if (!minF.used) {
                    minF.used = true;
                    System.out.println("P" + nr + " took fork " + minF);
                } else {
                    minF.wait();
                    minF.used = true;
                    System.out.println("I waited and took fork " + minF); //why it is never PRINTEDDD???
                }
                synchronized (maxF) {
                    if (!maxF.used) {
                        maxF.used = true;
                        System.out.println("P" + nr + " took fork "
                                + maxF);
                    } else {
                        maxF.wait();
                        maxF.used = true;
                        System.out.println("I waited and took fork "+ maxF); //why it is never PRINTEDDD??
                    }
                    System.out.println("I am eating right now" + this);
                    eating();
                    System.out.println("P" + nr
                            + " I have eaten  I am giving back the forks");
                    minF.used = false;
                    System.out.println("P" + nr +  " NOTIFY fork" + minF);
                    minF.notify();
                    maxF.used = false;
                    System.out.println("P" + nr + " NOTIFY fork" + maxF);
                    maxF.notify();
                }
            }
        } catch (Exception e) {
        }

        // }
    }

    public void eating() throws InterruptedException {
        int time = (int) (Math.random() * 2000);

        for (int i = 0; i < 5; i++) {
            System.out.println("P" + nr + " " + i);
            Thread.sleep(time / 5);
        }
    }

    public String toString() {
        return "Philosopher " + nr;
    }

    public static void startPhilosophers(Philosopher[] f) {
        for (int i = f.length - 1; i >= 0; i--) {
            f[i].start();
        }
    }

    public static void main(String[] args) {
        Fork[] t = Fork.getArrayOfForks();
        Philosopher[] f = { new Philosopher("philosopher 1", 1, t[0], t[4]),
                new Philosopher("philosopher 2", 2, t[1], t[0]),
                new Philosopher("philosopher 3", 3, t[2], t[1]),
                new Philosopher("philosopher 4", 4, t[3], t[2]),
                new Philosopher("philosopher 5", 5, t[4], t[3]), };
        startPhilosophers(f);

    }

}

<小时/>

public class Fork {
    boolean used;
    int nr;

    public Fork(boolean used, int nr) {
        this.used = used;
        this.nr = nr;
    }

    @Override
    public String toString() {
        return "F" + nr;
    }
    public static Fork min(Fork l, Fork p){
        if(l.nr < p.nr)
            return l;
        return p;
    }


    public static Fork max(Fork l, Fork p){
        if(l.nr > p.nr)
            return l;
        return p;
    }
    public static Fork[] getArrayOfForks() {
        Fork[] t = new Fork[5];
        for (int i = 0; i < t.length; i++) {
            t[i] = new Fork(false, (i + 1));
        }
        return t;
    }
}

<小时/>

EXAMPLE output
NR 1  F1 F5
NR 2  F2 F1
NR 3  F3 F2
NR 4  F4 F3
NR 5  F5 F4
P5 took fork F4
P5 took fork F5
I am eating right nowPhilosopher 5
P4 took fork F3
P5 0
P2 took fork F1
P2 took fork F2
I am eating right nowPhilosopher 2
P2 0
P5 1
P2 1
P5 2
P2 2
P5 3
P2 3
P5 4
P2 4
P5 I have eaten  I am giving back the forks
P5 NOTIFY forkF4
P5 NOTIFY forkF5
P4 took fork F4
I am eating right nowPhilosopher 4
P4 0
P2 I have eaten  I am giving back the forks
P2 NOTIFY forkF1
P2 NOTIFY forkF2
P3 took fork F2
P1 took fork F1
P1 took fork F5
I am eating right nowPhilosopher 1
P1 0
P1 1
P4 1
P1 2
P4 2
P1 3
P4 3
P1 4
P4 4
P1 I have eaten  I am giving back the forks
P1 NOTIFY forkF1
P1 NOTIFY forkF5
P4 I have eaten  I am giving back the forks
P4 NOTIFY forkF3
P4 NOTIFY forkF4
P3 took fork F3
I am eating right nowPhilosopher 3
P3 0
P3 1
P3 2
P3 3
P3 4
P3 I have eaten  I am giving back the forks
P3 NOTIFY forkF2
P3 NOTIFY forkF3

Answer 1

synchronized (minF) {
    if (!minF.used) { // always
        minF.used = true;
    }
    ...
    minF.used = false;
    minF.notify();
}

您正在同步 fork ，因此哲学家在检查 fork 是否正在使用时已经锁定了 fork 。哲学家将 fork.used 设置为 true，但在离开同步块(synchronized block)并释放锁之前将其设置回 false。

编辑:根据要求，更新了代码版本。如果您使用同步块(synchronized block)(您已经这样做了)，则无需自己进行管理:

synchronized (minF) {
    synchronized (maxF) {
        System.out.println("I am eating right now" + this);
        eating();
        System.out.println("P" + nr
            + " I have eaten  I am giving back the forks");
    }
}

如果你想明确地写出来，我会使用java.util.concurrent类并让Fork从Semaphore扩展。 。您的代码如下所示:

fork :

public class Fork extends Semaphore {
int nr;

public Fork(int nr) {
    super(1); // can be handed out to only one person at a time
    this.nr = nr;
}
...

还有哲学家:

minF.acquire();
maxF.acquire();
System.out.println("I am eating right now" + this);
eating();
System.out.println("P" + nr
    + " I have eaten  I am giving back the forks");
maxF.release();
minF.release();