解决死锁问题-open call

当调用一个外部的接口时,如果持有锁,由于不知道外部接口做了什么操作,此时,可能由于资源依赖形成锁环路,造成死锁。解决办法是,使用开放调用,调用外部接口时,不持有锁。

先看看错误的调用:如果有两个线程同时调用Taxi的setLocation和Dispatcher的printLocation,则可能因为错误的执行时序造成两个线程都只获得了一个锁,等待另一个永远无法获得的锁,从而造成死锁。

public class Taxi {

    private Point location, destination;

    private final Dispatcher dispatcher;

    public Taxi(Dispatcher dispatcher) {
        this.dispatcher = dispatcher;
    }

    public synchronized Point getLocation() {
        return location;
    }

    public synchronized void setLocation(Point location) {
        this.location = location;
        if(location.equals(destination)) {
            dispatcher.notifyAvailable(this);
        }
    }

}
public class Dispatcher {

    private final Set<Taxi> taxis;
    private final Set<Taxi> availableTaxis;

    public Dispatcher() {
        taxis = new HashSet<>();
        availableTaxis = new HashSet<>();
    }

    public synchronized void notifyAvailable(Taxi taxi) {
        availableTaxis.add(taxi);
    }

    public synchronized void printLocation() {
        for(Taxi taxi :taxis) {
            System.out.println(taxi.getLocation());
        }
    }
}

开放调用:在调用某个方法时不需要持有锁。

修改Taxi类

public void setLocation(Point location) {
    this.location = location;
    boolean reachedDestination;
    synchronized (this) {
        this.location = location;
        reachedDestination = location.equals(destination);
    }
    if(reachedDestination) {
        dispatcher.notifyAvailable(this);
    }
}

修改Dispatcher类:

public void printLocation() {
    Set<Taxi> copy;
    synchronized (this) {
        copy = new HashSet<>(taxis);
    }
    for(Taxi taxi :taxis) {
        System.out.println(taxi.getLocation());
    }
}

但是,使用开放调用技术改造接口会失去原子性,在大多数情况下,非原子性是可以接受的。如果必须保证原子性,则需要设计一些协议,而不是使用加锁,来补偿原子性。例如,使用单线程执行所有任务。

猜你喜欢

转载自blog.csdn.net/Wengzhengcun/article/details/87708645