我了解什么是活锁，但我想知道是否有人有一个很好的基于代码的示例？而基于代码的，我确实不是的意思是“两个人试图在走廊里超越对方”。如果我再读一遍，我会失去午餐。
【问题讨论】：
标签： concurrency livelock
这是一个非常简单的 Java 活锁示例，其中一对夫妻试图喝汤，但他们之间只有一个勺子。夫妻双方都太客气了，对方还没吃饭就递勺子。
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();
}
}
运行程序，你会得到：
Bob: You eat first my darling Alice!
Alice: You eat first my darling Bob!
Bob: You eat first my darling Alice!
Alice: You eat first my darling Bob!
Bob: You eat first my darling Alice!
Alice: You eat first my darling Bob!
...
如果不间断，这将永远持续下去。这是一个活锁，因为 Alice 和 Bob 在无限循环中反复要求对方先走（因此 live）。在陷入僵局的情况下，Alice 和 Bob 都会被冻结，等待对方先走——除了等待（因此 dead），他们不会做任何事情。
【讨论】：
getOwner 方法不是也必须同步吗？来自 Effective Java “除非读写同步，否则同步无效”。
Thread.join()而不是Thread.sleep()，因为他想等配偶吃饭吗？
getOwner 方法必须同步，因为即使 setOwner 已同步，这并不能保证使用 getOwner 的线程（或直接访问字段 owner）会看到更改由执行setOwner 的另一个线程创建。该视频非常仔细地解释了这一点：youtube.com/watch?v=WTVooKLLVT8
setOwner 方法不需要使用synchronized 关键字，因为读取和写入是引用变量的原子操作。
考虑一个有 50 个进程槽的 UNIX 系统。
十个程序正在运行，每个程序必须创建 6 个（子）进程。
每个进程创建了4个进程后，原来的10个进程和40个新进程都用完了表。 10 个原始进程中的每一个现在都处于无限循环中，分叉和失败——这恰如其分地出现了活锁的情况。这种情况发生的可能性很小，但有可能发生。
【讨论】：
由于没有答案标记为已接受答案，我试图创建活锁示例；
Original program 是我在 2012 年 4 月写的，用来学习多线程的各种概念。这次我修改了它以创建死锁、竞争条件、活锁等。
那么我们先来了解一下问题陈述；
Cookie 制作问题
有一些配料容器：ChocoPowderContainer、WheatPowderContainer。 CookieMaker 从配料容器中取出一定量的粉末来烘烤 Cookie。如果饼干制造商发现一个容器是空的，它会检查另一个容器以节省时间。并等待 Filler 填充所需的容器。有一个Filler定期检查容器并在容器需要时填充一些数量。
请查看github上的完整代码；
让我简要解释一下实现。
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;
}
在 Filler 填充容器之前一切正常。但是如果我忘记启动填充器，或者填充器意外离开，子线程会不断改变它们的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，它监视线程状态和死锁。这是控制台的输出；
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]
你会注意到子线程和改变它们的状态和等待。
【讨论】：
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;
}
}
}
【讨论】：
我修改了@jelbourn 的答案。 当其中一个注意到另一个饿了时，他（她）应该释放勺子并等待另一个通知，这样就会发生活锁。
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();
}
}
}
【讨论】：
jelbourn 的 C# 版本代码：
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);
}
}
}
}
}
【讨论】：
我编写了 2 个人通过走廊的示例。两条线一旦意识到它们的方向相同就会相互避开。
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) {}
}
}
}
【讨论】：
抛开轻率的做法不谈，一个已知的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，并试图互相“让开”，那么他们将最终陷入循环，总是“让开”并且永远无法继续前进。
“让步”我的意思是他们会释放锁并试图让另一个人获得它。我们可以想象两个线程这样做的情况（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();
}
抛开竞争条件不谈，我们这里的情况是，如果两个线程同时进入，它们最终会在内部循环中运行而不再继续。显然这是一个简化的例子。一个天真的解决方法是在线程等待的时间中加入某种随机性。
正确的解决方法是始终尊重lock heirarchy。选择一个获得锁的顺序并坚持下去。例如，如果两个线程总是在 lock2 之前获得 lock1，那么就没有死锁的可能性。
【讨论】：
jelbourn 代码的 Python 版本：
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()
【讨论】：
一个真实的（尽管没有确切的代码）示例是两个竞争进程实时锁定以尝试纠正 SQL 服务器死锁，每个进程使用相同的等待重试算法进行重试。虽然这是时机的运气，但我已经看到这种情况发生在具有相似性能特征的不同机器上，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。
在这种的情况下，一个好的解决方案是让相互竞争的消费者（通过划分无关对象的工作来防止重复处理在链中尽可能高的位置）。
一个不太理想的（好吧，肮脏的黑客）解决方案是提前打破计时坏运气（处理中的一种强制差异），或者在死锁后通过使用不同的算法或一些随机元素来打破它。这可能仍然存在问题，因为每个进程的锁定顺序可能是“粘性的”，并且这需要一定的最短时间，而不会在等待重试中考虑。
另一个解决方案（至少对于 SQL Server）是尝试不同的隔离级别（例如快照）。
【讨论】：
这里的一个例子可能是使用定时 tryLock 来获得多个锁，如果您无法获得所有锁，请退出并重试。
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有问题，因为你有很多线程冲突并等待获得一组锁。但作为一个简单的例子，我不确定这对我来说是否非常引人注目。
【讨论】：
tryLockAll() 和locks 中的锁，则没有活锁。