Question

我的MultiThread有问题，因为它无法独立工作，但是，有2个不同的线程。等待并通知

我的主要课程：

    Mechanics mechanics = new Mechanics(busShop, "Mechanic 1");
    Mechanics mechanics2 = new Mechanics(busShop, "Mechanic 2");

    Thread thMechanic = new Thread(mechanics);
    Thread thMehanic2 = new Thread(mechanics2);

    thMechanic.start();
    thMehanic2.start();

创建两个不同的独立线程，现在我这样做：

我的力学课程：

@Override
public void run() {
    fixEngine();
}

private void fixEngine() {
    while (true) {
        busShop.FixEngine(MechanicsName);
    }
}

运行时，只需执行此功能：

public void FixEngine(String mechanicsName) {
    //Call mechanics to fix engine
    Bus bus;
    synchronized (EntryRamp.ListBusDepotWaiting) {
        while (EntryRamp.ListBusDepotWaiting.size() == 0) {
            try {
                EntryRamp.ListBusDepotWaiting.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }

    try {
        Thread.sleep(5000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    synchronized (ListBusEngineFix) {
        //Fix the mechanics


        bus = (Bus) ((LinkedList<?>) EntryRamp.ListBusDepotWaiting).poll();
        ((LinkedList<Bus>) ListBusEngineFix).offer(bus);
        System.out.println("Bus: " + bus.getBusName() + "is being fixed by" + mechanicsName);



        ListBusEngineFix.notify();
        //Done fix
    }


}

通知来自何处。

    synchronized (ListBusDepotWaiting) {
        //If depot is not 2
        if (ListBusDepotWaiting.size() < 2 && ListBusRamp.size() == 0) {
            //If empty we can start moving
            bus = (Bus) ((LinkedList<?>) ListBusWaiting).poll();
            ((LinkedList<Bus>) ListBusRamp).offer(bus);
            System.out.println("RAMP: " + bus.getBusName() + "Enter the RAMP");

            try {
                Thread.sleep(2000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

            bus = (Bus) ((LinkedList<?>) ListBusRamp).poll();
            ((LinkedList<Bus>) ListBusDepotWaiting).offer(bus);
            System.out.println("DEPOT: " + bus.getBusName() + "Enter the DEPOT");
            ListBusDepotWaiting.notify();
        }
    }

运行时显示：

RAMP: Bus 1Enter the RAMP
DEPOT: Bus 1Enter the DEPOT
RAMP: Bus 2Enter the RAMP
DEPOT: Bus 2Enter the DEPOT
Bus: Bus 1is being fixed byMechanic 1
RAMP: Bus 3Enter the RAMP
Bus: Bus 1being cleaned
Bus: Bus 2is being fixed byMechanic 2

总线1不会立即处理，但它等待总线2到来，并与另一个线程一起工作。这意味着线程1无法自行完成，您可能在这里看不到，但是机制1和机制2同时完成，而机制1首先完成任务。任何想法的人？

处理多个线程（相同功能）等待通知

0 个答案: