活锁的好例子?


141

我知道什么是活锁,但是我想知道是否有人拥有基于代码的良好示例?通过基于代码的方式意思不是指“两个人试图在走廊里互相越过”。如果我再读一遍,我将失去午餐。


96
两个人试图在走廊互相越过的软件模拟怎么样?
1800信息

36
诅咒你!我丢了午餐!
Alex Miller


有关好奇的小伙子的有关笑话:codingarchitect.wordpress.com/2006/01/18/…–
约尔洪

4
两个人试图让过去对方在走廊:gist.github.com/deepankarb/d2dd6f21bc49902376e614d3746b8965:P
冰人

Answers:


119

这是一个非常简单的Java活锁示例,丈夫和妻子试图吃汤,但是中间只有一把汤匙。每个配偶都太客气了,如果对方还没有吃饭,他们会通过汤匙。

public class Livelock {
    static class Spoon {
        private Diner owner;
        public Spoon(Diner d) { owner = d; }
        public Diner getOwner() { return owner; }
        public synchronized void setOwner(Diner d) { owner = d; }
        public synchronized void use() { 
            System.out.printf("%s has eaten!", owner.name); 
        }
    }

    static class Diner {
        private String name;
        private boolean isHungry;

        public Diner(String n) { name = n; isHungry = true; }       
        public String getName() { return name; }
        public boolean isHungry() { return isHungry; }

        public void eatWith(Spoon spoon, Diner spouse) {
            while (isHungry) {
                // Don't have the spoon, so wait patiently for spouse.
                if (spoon.owner != this) {
                    try { Thread.sleep(1); } 
                    catch(InterruptedException e) { continue; }
                    continue;
                }                       

                // If spouse is hungry, insist upon passing the spoon.
                if (spouse.isHungry()) {                    
                    System.out.printf(
                        "%s: You eat first my darling %s!%n", 
                        name, spouse.getName());
                    spoon.setOwner(spouse);
                    continue;
                }

                // Spouse wasn't hungry, so finally eat
                spoon.use();
                isHungry = false;               
                System.out.printf(
                    "%s: I am stuffed, my darling %s!%n", 
                    name, spouse.getName());                
                spoon.setOwner(spouse);
            }
        }
    }

    public static void main(String[] args) {
        final Diner husband = new Diner("Bob");
        final Diner wife = new Diner("Alice");

        final Spoon s = new Spoon(husband);

        new Thread(new Runnable() { 
            public void run() { husband.eatWith(s, wife); }   
        }).start();

        new Thread(new Runnable() { 
            public void run() { wife.eatWith(s, husband); } 
        }).start();
    }
}

6
没有getOwner方法必须同步呢?在有效的Java中,“ 同步没有作用,除非同时进行读写 ”。
Sanghyun Lee

他不应该使用Thread.join()而不是Thread.sleep(),因为他想等待配偶吃饭吗?
安慰

在此特定示例中,我们应该怎么做才能克服活锁问题?
雷神

1
getOwner方法必须同步,因为即使setOwner同步了,也不能保证使用getOwner(或owner直接访问该字段)的线程将看到其他线程执行的更改setOwner。这个视频非常仔细地解释了这一点:youtube.com/watch?
v=WTVooKLLVT8

2
您不需要synchronized setOwner方法使用关键字,因为读写是引用变量的原子动作。
atiqkhaled

75

撇开轻率的注释,已知的一个例子是尝试检测和处理死锁情况的代码。如果两个线程检测到死锁,并尝试彼此“退避”,那么他们将无意中陷入“总是退避”而永远无法前进的循环中。

通过“放开”,我的意思是他们将释放该锁并尝试让另一人获取它。我们可以想象有两个线程执行此操作的情况(伪代码):

// thread 1
getLocks12(lock1, lock2)
{
  lock1.lock();
  while (lock2.locked())
  {
    // attempt to step aside for the other thread
    lock1.unlock();
    wait();
    lock1.lock();
  }
  lock2.lock();
}

// thread 2
getLocks21(lock1, lock2)
{
  lock2.lock();
  while (lock1.locked())
  {
    // attempt to step aside for the other thread
    lock2.unlock();
    wait();
    lock2.lock();
  }
  lock1.lock();
}

抛开竞争条件,这里的情况是两个线程(如果同时进入)将最终在内部循环中运行而不进行处理。显然,这是一个简化的示例。幼稚的解决方法是在线程等待的时间上增加某种随机性。

正确的解决方法是始终尊重锁的层次结构。选择您要获得锁的订单并坚持下去。例如,如果两个线程始终在lock2之前获取lock1,则没有死锁的可能性。


是的,我明白。我正在寻找这样的实际代码示例。问题是“退避”是什么意思,以及它如何产生这种情况。
亚历克斯·米勒

我知道这是一个虚构的示例,但这很可能导致活动锁定?由于线程大声运行的时间和调度的时间不一致,最终是否有可能最终打开一个可以同时捕获两个函数的窗口?
DubiousPusher 2014年

尽管它不是稳定的活锁,因为它们最终显然会脱离它,但我认为它非常符合描述要求
1800 INFORMATION

优秀而有意义的例子。
alecov

7

由于没有标记为已接受答案的答案,因此我尝试创建实时锁定示例。

原始程序我在2012年4月编写了,以学习各种多线程概念。这次我修改了它以创建死锁,竞争条件,活动锁等。

因此,让我们首先了解问题陈述;

Cookie制造商问题

有一些成分容器:ChocoPowederContainerWheatPowderContainerCookieMaker从配料容器中取出一些粉末来烘烤Cookie。如果Cookie制作者发现一个容器为空,它将检查另一个容器以节省时间。然后等待,直到Filler填充所需的容器。有填充物定期检查容器,并在容器需要时填充一定数量。

请检查github上的完整代码;

让我简要解释一下实现。

  • 我开始加油作为守护程序线程。因此,它将保持定期填充容器。首先要填充容器,它会锁定容器->检查是否需要一些粉末->填充它->向所有正在等待它的制造商发出信号->解锁容器。
  • 我创建了CookieMaker,并设置为可以并行烘烤最多8个cookie。然后我开始8个线程来烤饼干。
  • 每个制造商线程创建2个可调用子线程以从容器中取出粉末。
  • 子线程锁定容器并检查其是否有足够的粉末。如果没有,请等待一段时间。填充器填充容器后,它将取粉并解锁容器。
  • 现在,它完成了其他活动,例如:混合和烘烤等。

让我们看一下代码:

CookieMaker.java

private Integer getMaterial(final Ingredient ingredient) throws Exception{
        :
        container.lock();
        while (!container.getIngredient(quantity)) {
            container.empty.await(1000, TimeUnit.MILLISECONDS);
            //Thread.sleep(500); //For deadlock
        }
        container.unlock();
        :
}

IngredientContainer.java

public boolean getIngredient(int n) throws Exception {
    :
    lock();
    if (quantityHeld >= n) {
        TimeUnit.SECONDS.sleep(2);
        quantityHeld -= n;
        unlock();
        return true;
    }
    unlock();
    return false;
}

一切正常,直到Filler填充容器。但是,如果我忘记启动填充器,或者填充器意外休假,子线程会不断更改其状态,以允许其他制造商去检查容器。

我还创建了一个守护进程ThreadTracer,它可以监视线程状态和死锁。这是来自控制台的输出;

2016-09-12 21:31:45.065 :: [Maker_0:WAITING, Maker_1:WAITING, Maker_2:WAITING, Maker_3:WAITING, Maker_4:WAITING, Maker_5:WAITING, Maker_6:WAITING, Maker_7:WAITING, pool-7-thread-1:TIMED_WAITING, pool-7-thread-2:TIMED_WAITING, pool-8-thread-1:TIMED_WAITING, pool-8-thread-2:TIMED_WAITING, pool-6-thread-1:TIMED_WAITING, pool-6-thread-2:TIMED_WAITING, pool-5-thread-1:TIMED_WAITING, pool-5-thread-2:TIMED_WAITING, pool-1-thread-1:TIMED_WAITING, pool-3-thread-1:TIMED_WAITING, pool-2-thread-1:TIMED_WAITING, pool-1-thread-2:TIMED_WAITING, pool-4-thread-1:TIMED_WAITING, pool-4-thread-2:RUNNABLE, pool-3-thread-2:TIMED_WAITING, pool-2-thread-2:TIMED_WAITING]
2016-09-12 21:31:45.065 :: [Maker_0:WAITING, Maker_1:WAITING, Maker_2:WAITING, Maker_3:WAITING, Maker_4:WAITING, Maker_5:WAITING, Maker_6:WAITING, Maker_7:WAITING, pool-7-thread-1:TIMED_WAITING, pool-7-thread-2:TIMED_WAITING, pool-8-thread-1:TIMED_WAITING, pool-8-thread-2:TIMED_WAITING, pool-6-thread-1:TIMED_WAITING, pool-6-thread-2:TIMED_WAITING, pool-5-thread-1:TIMED_WAITING, pool-5-thread-2:TIMED_WAITING, pool-1-thread-1:TIMED_WAITING, pool-3-thread-1:TIMED_WAITING, pool-2-thread-1:TIMED_WAITING, pool-1-thread-2:TIMED_WAITING, pool-4-thread-1:TIMED_WAITING, pool-4-thread-2:TIMED_WAITING, pool-3-thread-2:TIMED_WAITING, pool-2-thread-2:TIMED_WAITING]
WheatPowder Container has 0 only.
2016-09-12 21:31:45.082 :: [Maker_0:WAITING, Maker_1:WAITING, Maker_2:WAITING, Maker_3:WAITING, Maker_4:WAITING, Maker_5:WAITING, Maker_6:WAITING, Maker_7:WAITING, pool-7-thread-1:TIMED_WAITING, pool-7-thread-2:TIMED_WAITING, pool-8-thread-1:TIMED_WAITING, pool-8-thread-2:TIMED_WAITING, pool-6-thread-1:TIMED_WAITING, pool-6-thread-2:TIMED_WAITING, pool-5-thread-1:TIMED_WAITING, pool-5-thread-2:TIMED_WAITING, pool-1-thread-1:TIMED_WAITING, pool-3-thread-1:TIMED_WAITING, pool-2-thread-1:TIMED_WAITING, pool-1-thread-2:TIMED_WAITING, pool-4-thread-1:TIMED_WAITING, pool-4-thread-2:TIMED_WAITING, pool-3-thread-2:TIMED_WAITING, pool-2-thread-2:RUNNABLE]
2016-09-12 21:31:45.082 :: [Maker_0:WAITING, Maker_1:WAITING, Maker_2:WAITING, Maker_3:WAITING, Maker_4:WAITING, Maker_5:WAITING, Maker_6:WAITING, Maker_7:WAITING, pool-7-thread-1:TIMED_WAITING, pool-7-thread-2:TIMED_WAITING, pool-8-thread-1:TIMED_WAITING, pool-8-thread-2:TIMED_WAITING, pool-6-thread-1:TIMED_WAITING, pool-6-thread-2:TIMED_WAITING, pool-5-thread-1:TIMED_WAITING, pool-5-thread-2:TIMED_WAITING, pool-1-thread-1:TIMED_WAITING, pool-3-thread-1:TIMED_WAITING, pool-2-thread-1:TIMED_WAITING, pool-1-thread-2:TIMED_WAITING, pool-4-thread-1:TIMED_WAITING, pool-4-thread-2:TIMED_WAITING, pool-3-thread-2:TIMED_WAITING, pool-2-thread-2:TIMED_WAITING]

您会注意到子线程并更改其状态并等待。


4

一个真实的示例(尽管没有确切的代码)是两个相互竞争的进程实时锁定,以尝试纠正SQL Server死锁,每个进程使用相同的wait-retry算法进行重试。虽然很幸运,但是我已经看到这种情况发生在具有相似性能特征的单独机器上,以响应添加到EMS主题的消息(例如,多次保存单个对象图的更新),并且无法控制锁定顺序。

这种情况下,一个好的解决方案是让竞争的使用者使用(通过将工作划分在不相关的对象上,从而防止重复处理在链中占据较高的位置)。

较不理想的解决方案是提前破解计时厄运(处理过程中的种种力差),或者在死锁后通过使用不同的算法或某种随机性因素来打破它。这仍然可能会出现问题,因为对于每个进程来说,锁定采取的顺序可能是“粘性的”,并且这花费了一定的最短时间,而等待重试中没有考虑到这一点。

另一个解决方案(至少对于SQL Server)是尝试使用不同的隔离级别(例如快照)。


2

我编写了2个人在走廊中经过的示例的代码。一旦两个线程意识到它们的方向相同,它们将彼此避免。

public class LiveLock {
    public static void main(String[] args) throws InterruptedException {
        Object left = new Object();
        Object right = new Object();
        Pedestrian one = new Pedestrian(left, right, 0); //one's left is one's left
        Pedestrian two = new Pedestrian(right, left, 1); //one's left is two's right, so have to swap order
        one.setOther(two);
        two.setOther(one);
        one.start();
        two.start();
    }
}

class Pedestrian extends Thread {
    private Object l;
    private Object r;
    private Pedestrian other;
    private Object current;

    Pedestrian (Object left, Object right, int firstDirection) {
        l = left;
        r = right;
        if (firstDirection==0) {
            current = l;
        }
        else {
            current = r;
        }
    }

    void setOther(Pedestrian otherP) {
        other = otherP;
    }

    Object getDirection() {
        return current;
    }

    Object getOppositeDirection() {
        if (current.equals(l)) {
            return r;
        }
        else {
            return l;
        }
    }

    void switchDirection() throws InterruptedException {
        Thread.sleep(100);
        current = getOppositeDirection();
        System.out.println(Thread.currentThread().getName() + " is stepping aside.");
    }

    public void run() {
        while (getDirection().equals(other.getDirection())) {
            try {
                switchDirection();
                Thread.sleep(100);
            } catch (InterruptedException e) {}
        }
    }
} 

2

Jelbourn代码的C#版本:

using System;
using System.Runtime.CompilerServices;
using System.Threading;
using System.Threading.Tasks;

namespace LiveLockExample
{
    static class Program
    {
        public static void Main(string[] args)
        {
            var husband = new Diner("Bob");
            var wife = new Diner("Alice");

            var s = new Spoon(husband);

            Task.WaitAll(
                Task.Run(() => husband.EatWith(s, wife)),
                Task.Run(() => wife.EatWith(s, husband))
                );
        }

        public class Spoon
        {
            public Spoon(Diner diner)
            {
                Owner = diner;
            }


            public Diner Owner { get; private set; }

            [MethodImpl(MethodImplOptions.Synchronized)]
            public void SetOwner(Diner d) { Owner = d; }

            [MethodImpl(MethodImplOptions.Synchronized)]
            public void Use()
            {
                Console.WriteLine("{0} has eaten!", Owner.Name);
            }
        }

        public class Diner
        {
            public Diner(string n)
            {
                Name = n;
                IsHungry = true;
            }

            public string Name { get; private set; }

            private bool IsHungry { get; set; }

            public void EatWith(Spoon spoon, Diner spouse)
            {
                while (IsHungry)
                {
                    // Don't have the spoon, so wait patiently for spouse.
                    if (spoon.Owner != this)
                    {
                        try
                        {
                            Thread.Sleep(1);
                        }
                        catch (ThreadInterruptedException e)
                        {
                        }

                        continue;
                    }

                    // If spouse is hungry, insist upon passing the spoon.
                    if (spouse.IsHungry)
                    {
                        Console.WriteLine("{0}: You eat first my darling {1}!", Name, spouse.Name);
                        spoon.SetOwner(spouse);
                        continue;
                    }

                    // Spouse wasn't hungry, so finally eat
                    spoon.Use();
                    IsHungry = false;
                    Console.WriteLine("{0}: I am stuffed, my darling {1}!", Name, spouse.Name);
                    spoon.SetOwner(spouse);
                }
            }
        }
    }
}

1

这里的一个示例可能是使用定时tryLock获得多个锁,如果无法完全获得它们,请退回并重试。

boolean tryLockAll(Collection<Lock> locks) {
  boolean grabbedAllLocks = false;
  for(int i=0; i<locks.size(); i++) {
    Lock lock = locks.get(i);
    if(!lock.tryLock(5, TimeUnit.SECONDS)) {
      grabbedAllLocks = false;

      // undo the locks I already took in reverse order
      for(int j=i-1; j >= 0; j--) {
        lock.unlock();
      }
    }
  }
}

我可以想象这样的代码会出现问题,因为您有许多线程在碰撞并等待获取一组锁。但是我不确定这是否对我来说很简单。


1
为了使它成为一个活锁,您将需要另一个线程以不同的顺序来获取那些锁。如果所有线程tryLockAll()locks相同的顺序与锁一起使用,则没有活动锁。
哈维·梅里诺(JaviMerino)2014年

0

Python版本的耶尔本代码:

import threading
import time
lock = threading.Lock()

class Spoon:
    def __init__(self, diner):
        self.owner = diner

    def setOwner(self, diner):
        with lock:
            self.owner = diner

    def use(self):
        with lock:
            "{0} has eaten".format(self.owner)

class Diner:
    def __init__(self, name):
        self.name = name
        self.hungry = True

    def eatsWith(self, spoon, spouse):
        while(self.hungry):
            if self != spoon.owner:
                time.sleep(1) # blocks thread, not process
                continue

            if spouse.hungry:
                print "{0}: you eat first, {1}".format(self.name, spouse.name)
                spoon.setOwner(spouse)
                continue

            # Spouse was not hungry, eat
            spoon.use()
            print "{0}: I'm stuffed, {1}".format(self.name, spouse.name)
            spoon.setOwner(spouse)

def main():
    husband = Diner("Bob")
    wife = Diner("Alice")
    spoon = Spoon(husband)

    t0 = threading.Thread(target=husband.eatsWith, args=(spoon, wife))
    t1 = threading.Thread(target=wife.eatsWith, args=(spoon, husband))
    t0.start()
    t1.start()
    t0.join()
    t1.join()

if __name__ == "__main__":
    main()

错误:在use()中,未使用打印,而且-更重要的是-饥​​饿标志未设置为False。
GDR

0

我修改了@jelbourn的答案。当其中一个人注意到另一个人饿了时,他(她)应该放开汤匙并等待另一个通知,这样就发生了活锁。

public class LiveLock {
    static class Spoon {
        Diner owner;

        public String getOwnerName() {
            return owner.getName();
        }

        public void setOwner(Diner diner) {
            this.owner = diner;
        }

        public Spoon(Diner diner) {
            this.owner = diner;
        }

        public void use() {
            System.out.println(owner.getName() + " use this spoon and finish eat.");
        }
    }

    static class Diner {
        public Diner(boolean isHungry, String name) {
            this.isHungry = isHungry;
            this.name = name;
        }

        private boolean isHungry;
        private String name;


        public String getName() {
            return name;
        }

        public void eatWith(Diner spouse, Spoon sharedSpoon) {
            try {
                synchronized (sharedSpoon) {
                    while (isHungry) {
                        while (!sharedSpoon.getOwnerName().equals(name)) {
                            sharedSpoon.wait();
                            //System.out.println("sharedSpoon belongs to" + sharedSpoon.getOwnerName())
                        }
                        if (spouse.isHungry) {
                            System.out.println(spouse.getName() + "is hungry,I should give it to him(her).");
                            sharedSpoon.setOwner(spouse);
                            sharedSpoon.notifyAll();
                        } else {
                            sharedSpoon.use();
                            sharedSpoon.setOwner(spouse);
                            isHungry = false;
                        }
                        Thread.sleep(500);
                    }
                }
            } catch (InterruptedException e) {
                System.out.println(name + " is interrupted.");
            }
        }
    }

    public static void main(String[] args) {
        final Diner husband = new Diner(true, "husband");
        final Diner wife = new Diner(true, "wife");
        final Spoon sharedSpoon = new Spoon(wife);

        Thread h = new Thread() {
            @Override
            public void run() {
                husband.eatWith(wife, sharedSpoon);
            }
        };
        h.start();

        Thread w = new Thread() {
            @Override
            public void run() {
                wife.eatWith(husband, sharedSpoon);
            }
        };
        w.start();

        try {
            Thread.sleep(10000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        h.interrupt();
        w.interrupt();

        try {
            h.join();
            w.join();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

0
package concurrently.deadlock;

import static java.lang.System.out;


/* This is an example of livelock */
public class Dinner {

    public static void main(String[] args) {
        Spoon spoon = new Spoon();
        Dish dish = new Dish();

        new Thread(new Husband(spoon, dish)).start();
        new Thread(new Wife(spoon, dish)).start();
    }
}


class Spoon {
    boolean isLocked;
}

class Dish {
    boolean isLocked;
}

class Husband implements Runnable {

    Spoon spoon;
    Dish dish;

    Husband(Spoon spoon, Dish dish) {
        this.spoon = spoon;
        this.dish = dish;
    }

    @Override
    public void run() {

        while (true) {
            synchronized (spoon) {
                spoon.isLocked = true;
                out.println("husband get spoon");
                try { Thread.sleep(2000); } catch (InterruptedException e) {}

                if (dish.isLocked == true) {
                    spoon.isLocked = false; // give away spoon
                    out.println("husband pass away spoon");
                    continue;
                }
                synchronized (dish) {
                    dish.isLocked = true;
                    out.println("Husband is eating!");

                }
                dish.isLocked = false;
            }
            spoon.isLocked = false;
        }
    }
}

class Wife implements Runnable {

    Spoon spoon;
    Dish dish;

    Wife(Spoon spoon, Dish dish) {
        this.spoon = spoon;
        this.dish = dish;
    }

    @Override
    public void run() {
        while (true) {
            synchronized (dish) {
                dish.isLocked = true;
                out.println("wife get dish");
                try { Thread.sleep(2000); } catch (InterruptedException e) {}

                if (spoon.isLocked == true) {
                    dish.isLocked = false; // give away dish
                    out.println("wife pass away dish");
                    continue;
                }
                synchronized (spoon) {
                    spoon.isLocked = true;
                    out.println("Wife is eating!");

                }
                spoon.isLocked = false;
            }
            dish.isLocked = false;
        }
    }
}
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