[CareerCup] 16.4 A Lock Without Deadlocks 无死锁的锁
16.4 Design a class which provides a lock only if there are no possible deadlocks.
有很多方法可以避免死锁的发生,一个常用的方法是列出所需要的锁,然后判断锁上这些锁后会不会发生死锁,比如有如下的锁的顺序:
A = {1, 2, 3, 4}
B = {1, 3, 5}
C = {7, 5, 9, 2}
这是有可能产生死锁的,比如当A锁上2等待3,当B锁上3等待5,当C锁上5等待2,我们可以将其看做图,2连上3,3连上5,5连上2,那么就会有环。一条边(w,v)表示锁上v后马上锁w,那么上述里子的图中的边为(1, 2), (2, 3), (3, 4), (1, 3), (3, 5), (7, 5), (5, 9), (9, 2)。那么我们在检测是否有死锁的情况就是要找图中是否存在环,我们可以用DFS来搜索所有的相连的部分,我们用DFS需要标记点的状态,我们首先定义个LockFactory类,用来保存锁的序列,比如上面的A,B,C,然后对于每一个锁序列,我们先将所有的锁都链起来,然后标记为false,然后用DFS判断是否有环,如果有环,则断开当前锁序列之间的所有连接,参见代码如下;
import java.util.Hashtable;
import java.util.LinkedList;
import java.util.concurrent.locks.Lock; public class LockFactory {
private static LockFactory instance;
private int numberOfLocks = 5;
private LockNode[] locks;
private Hashtable<Integer, LinkedList<LockNode>> lockOrder; private LockFactory(int count) {
numberOfLocks = count;
locks = new LockNode[numberOfLocks];
lockOrder = new Hashtable<Integer, LinkedList<LockNode>>();
for (int i = 0; i < numberOfLocks; ++i) {
locks[i] = new LockNode(i, count);
}
} public static LockFactory getInstance() {
return instance;
} public static LockFactory initialize(int count) {
if (instance == null) {
instance = new LockFactory(count);
}
return instance;
} public boolean hasCycle(Hashtable<Integer, Boolean> touchedNodes, int[] resourcesInOrder) {
for (int resource : resourcesInOrder) {
if (touchedNodes.get(resource) == false) {
LockNode n = locks[resource];
if (n.hasCycle(touchedNodes)) {
return true;
}
}
}
return false;
} public boolean declare(int ownerId, int[] resourcesInOrder) {
Hashtable<Integer, Boolean> touchedNodes = new Hashtable<Integer, Boolean>();
int index = 1;
touchedNodes.put(resourcesInOrder[0], false);
for (index = 1; index < resourcesInOrder.length; ++index) {
LockNode pre = locks[resourcesInOrder[index - 1]];
LockNode cur = locks[resourcesInOrder[index]];
pre.joinTo(cur);
touchedNodes.put(resourcesInOrder[index], false);
}
if (hasCycle(touchedNodes, resourcesInOrder)) {
for (int j = 1; j < resourcesInOrder.length; ++j) {
LockNode p = locks[resourcesInOrder[j - 1]];
LockNode c = locks[resourcesInOrder[j]];
p.remove(c);
}
return false;
}
LinkedList<LockNode> list = new LinkedList<LockNode>();
for (int i = 0; i < resourcesInOrder.length; ++i) {
LockNode resource = locks[resourcesInOrder[i]];
list.add(resource);
}
lockOrder.put(ownerId, list);
return true;
} public Lock getLock(int ownerId, int resourceId) {
LinkedList<LockNode> list = lockOrder.get(ownerId);
if (list == null) return null;
LockNode head = list.getFirst();
if (head.getId() == resourceId) {
list.removeFirst();
return head.getLock();
}
return null;
}
} import java.util.ArrayList;
import java.util.Hashtable;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock; public class LockNode {
public enum VisitState {
FRESH, VISITING, VISITED
};
private ArrayList<LockNode> children;
private int lockId;
private Lock lock;
private int maxLocks; public LockNode(int id, int max) {
lockId = id;
children = new ArrayList<LockNode>();
maxLocks = max;
} public void joinTo(LockNode node) {
children.add(node);
} public void remove(LockNode node) {
children.remove(node);
} public boolean hasCycle(Hashtable<Integer, Boolean> touchedNodes) {
VisitState[] visited = new VisitState[maxLocks];
for (int i = 0; i < maxLocks; ++i) {
visited[i] = VisitState.FRESH;
}
return hasCycle(visited, touchedNodes);
} public boolean hasCycle(VisitState[] visited, Hashtable<Integer, Boolean> touchedNodes) {
if (touchedNodes.containsKey(lockId)) {
touchedNodes.put(lockId, true);
}
if (visited[lockId] == VisitState.VISITING) {
return true;
} else if (visited[lockId] == VisitState.FRESH) {
visited[lockId] = VisitState.VISITING;
for (LockNode n : children) {
if (n.hasCycle(visited, touchedNodes)) {
return true;
}
}
visited[lockId] = VisitState.VISITED;
}
return false;
} public Lock getLock() {
if (lock == null) {
lock = new ReentrantLock();
}
return lock;
} public int getId() {
return lockId;
}
} public class j {
public static void main(String[] args) {
int[] res1 = {1, 2, 3, 4};
int[] res2 = {1, 5, 4, 1};
int[] res3 = {1, 4, 5};
LockFactory.initialize(10); LockFactory lf = LockFactory.getInstance();
System.out.println(lf.declare(1, res1));
System.out.println(lf.declare(2, res2));
System.out.println(lf.declare(3, res3)); System.out.println(lf.getLock(1, 1));
System.out.println(lf.getLock(1, 2));
System.out.println(lf.getLock(1, 3));
System.out.println(lf.getLock(1, 4));
System.out.println(lf.getLock(2, 1));
System.out.println(lf.getLock(2, 5));
System.out.println(lf.getLock(2, 4));
System.out.println(lf.getLock(3, 1));
System.out.println(lf.getLock(3, 4));
System.out.println(lf.getLock(3, 5));
}
}
[CareerCup] 16.4 A Lock Without Deadlocks 无死锁的锁的更多相关文章
- 转:synchronized和LOCK的实现原理---深入JVM锁机制
JVM底层又是如何实现synchronized的? 目前在Java中存在两种锁机制:synchronized和Lock,Lock接口及其实现类是JDK5增加的内容,其作者是大名鼎鼎的并发专家Doug ...
- 015-线程同步-synchronized几种加锁方式、Java对象头和Monitor、Mutex Lock、JDK1.6对synchronized锁的优化实现
一.synchronized概述基本使用 为确保共享变量不会出现并发问题,通常会对修改共享变量的代码块用synchronized加锁,确保同一时刻只有一个线程在修改共享变量,从而避免并发问题. syn ...
- 16、Java并发性和多线程-死锁
以下内容转自http://ifeve.com/deadlock/: 死锁是两个或更多线程阻塞着等待其它处于死锁状态的线程所持有的锁.死锁通常发生在多个线程同时但以不同的顺序请求同一组锁的时候. 例如, ...
- Lock方法是用于数据库的锁机制,
Lock方法是用于数据库的锁机制,如果在查询或者执行操作的时候使用: lock(true); 复制代码 就会自动在生成的SQL语句最后加上 FOR UPDATE或者FOR UPDATE NOWAI ...
- [CareerCup] 16.3 Dining Philosophers 哲学家聚餐问题
16.3 In the famous dining philosophers problem, a bunch of philosophers are sitting around a circula ...
- [CareerCup] 16.5 Semphore 信号旗
16.5 Suppose we have the following code:public class Foo { public Foo() { . . . } public void first( ...
- [CareerCup] 16.6 Synchronized Method 同步方法
16.6 You are given a class with synchronized method A and a normal method B. If you have two threads ...
- [CareerCup] 16.2 Measure Time in a Context Switch 测量上下文转换的时间
16.2 How would you measure the time spent in a context switch? 上下文转换发生在两个进程之间,比如让一个等待进程进入执行和让一个运行进程进 ...
- [CareerCup] 16.1 Thread and Process 线程和进程
16.1 What's the difference between a thread and a process? 进程Process是程序执行时的一个实例.一个进程是被分配系统资源的独立单元,每个 ...
随机推荐
- C++模板【转】
1. 模板的概念. 我们已经学过重载(Overloading),对重载函数而言,C++的检查机制能通过函数参数的不同及所属类的不同.正确的调用重载函数.例如,为求两个数的最大值,我们定义MAX()函数 ...
- 【T_SQL】 基础 事务
1.使用 T-SQL 语句来管理事务 开始事务:BEGIN TRANSACTION 提交事务:COMMIT TRANSACTION 回滚(撤销)事务:ROLLBAC ...
- 编译报错dereferencing pointer to incomplete type
关于编译报错“dereferencing pointer to incomplete type... 多是没找到结构体的定义,可以在本地复制其定义试试. 参考: http://my.oschina.n ...
- windows下自动关机
定时:at 00:00 shutdown -s //在00:00时关机 倒计时:shutdown -s -t 3600 //3600s后关机 取消:shutdown -a
- Spring常用工具类
Spring框架下自带了丰富的工具类,在我们开发时可以简化很多工作: 1.Resource访问文件资源: 具体有: ResourceUtils.getFile(url); FileSystemReso ...
- iOS10 UI教程视图调试
iOS10 UI教程视图调试 iOS10 UI教程视图调试,当视图很复杂的时候,层次结构就不会很简单了.Xcode可以通过视图(View)调试帮助开发者解决层次结构复杂的问题.视图调试是在Xcode ...
- 17996 Daily Cool Run (dp)
时间限制:1000MS 内存限制:65535K 提交次数:0 通过次数:0 题型: 编程题 语言: 不限定 Description Daily Cool Run is a popular gam ...
- 1.ARC和非ARC文件共存
1.ARC和非ARC文件共存 项目->Build Parses->对应的类 1.1.新项目兼容老的非ARC:-fno-objc-arc 1.2.老项目兼容ARC:-fobjc-arc
- stack
stack介绍:先进后出 实现C++STL,栈有两个参数:template<class T,class Container = deque<T>> class stack: 参 ...
- Codeforces Round #347 (Div. 2)
unrating的一场CF A - Complicated GCD #include <bits/stdc++.h> const int N = 1e5 + 5; char a[105], ...