死锁的四个条件：
(1) 互斥条件：一个资源每次只能被一个进程使用。
(2) 请求与保持条件：一个进程因请求资源而阻塞时，对已获得的资源保持不放。
(3) 不剥夺条件:进程已获得的资源，在末使用完之前，不能强行剥夺。
(4) 循环等待条件:若干进程之间形成一种头尾相接的循环等待资源关系
先写一个会造成死锁的哲学家问题。当所有哲学家同时决定进餐，拿起左边筷子时候，就发生了死锁。
public class test {
public static void main(String[] args) {
ExecutorService exec = Executors.newCachedThreadPool();
int sum = 5;
Chopstick[] chopsticks = new Chopstick[sum];
for (int i = 0; i < sum; i++) {
chopsticks[i] = new Chopstick();
}
for (int i = 0; i < sum; i++) {
exec.execute(new Philosopher(chopsticks[i], chopsticks[(i + 1) % sum]));
}
}
}
// 筷子
class Chopstick {
public Chopstick() {
}
}
class Philosopher implements Runnable {
private Chopstick left;
private Chopstick right;
public Philosopher(Chopstick left, Chopstick right) {
this.left = left;
this.right = right;
}
@Override
public void run() {
try {
while (true) {
Thread.sleep(1000);//思考一段时间
synchronized (left) {
synchronized (right) {
Thread.sleep(1000);//进餐一段时间
}
}
}
}
catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
解决方案一：破坏死锁的循环等待条件。
不再按左手边右手边顺序拿起筷子。选择一个固定的全局顺序获取，此处给筷子添加id，根据id从小到大获取，(不用关心编号的具体规则，只要保证编号是全局唯一并且有序的)，不会出现死锁情况。
public class test {
public static void main(String[] args) {
ExecutorService exec = Executors.newCachedThreadPool();
int sum = 5;
Chopstick[] chopsticks = new Chopstick[sum];
for (int i = 0; i < sum; i++) {
chopsticks[i] = new Chopstick(i);
}
for (int i = 0; i < sum; i++) {
exec.execute(new Philosopher(chopsticks[i], chopsticks[(i + 1) % sum]));
}
}
}
// 筷子
class Chopstick {
//状态
private int id;
public Chopstick(int id) {
this.id=id;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
}
// 哲学家
class Philosopher implements Runnable {
private Chopstick left;
private Chopstick right;
public Philosopher(Chopstick left, Chopstick right) {
if(left.getId()
this.left = left;this.right = right;
}else{
this.left=right;this.right=left;
}
}
@Override
public void run() {
try {
while (true) {
Thread.sleep(1000);//思考一段时间
synchronized (left) {
synchronized (right) {
Thread.sleep(1000);//进餐一段时间
}
}
}
}
catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
方法二：破坏死锁的请求与保持条件，使用lock的特性，为获取锁操作设置超时时间。这样不会死锁(至少不会无尽的死锁)
class Philosopher extends Thread{
private ReentrantLock left,right;
public Philosopher(ReentrantLock left, ReentrantLock right) {
super();
this.left = left;
this.right = right;
}
public void run(){
try {
while(true){
Thread.sleep(1000);//思考一段时间
left.lock();
try{
if(right.tryLock(1000,TimeUnit.MILLISECONDS)){
try{
Thread.sleep(1000);//进餐一段时间
}finally {
right.unlock();
}
}else{
//没有获取到右手的筷子，放弃并继续思考
}
}finally {
left.unlock();
}
}
} catch (InterruptedException e) {
}
}
}
方法三：设置一个条件遍历与一个锁关联。该方法只用一把锁，没有chopstick类，将竞争从对筷子的争夺转换成了对状态的判断。仅当左右邻座都没有进餐时才可以进餐。提升了并发度。前面的方法出现情况是：只有一个哲学家进餐，其他人持有一根筷子在等待另外一根。这个方案中，当一个哲学家理论上可以进餐(邻座没有进餐)，他肯定可以进餐。
public class Philosopher extends Thread{
private boolean eating;
private Philosopher left;
private Philosopher right;
private ReentrantLock lock;
private Condition condition;
public Philosopher(ReentrantLock lock){
eating =false;
this.lock=lock;
condition=lock.newCondition();
}
public void setLeft(Philosopher left) {
this.left = left;
}
public void setRight(Philosopher right) {
this.right = right;
}
public void think() throws InterruptedException{
lock.lock();
try {
eating=false;
System.out.println(Thread.currentThread().getName()+"开始思考");
left.condition.signal();
right.condition.signal();
} finally{
lock.unlock();
}
Thread.sleep(1000);
}
public void eat() throws InterruptedException{
lock.lock();
try {
while(left.eating||right.eating)
condition.await();
System.out.println(Thread.currentThread().getName()+"开始吃饭");
eating=true;
} finally {
// TODO: handle finally clause
lock.unlock();
}
Thread.sleep(1000);
}
public void run(){
try{
while(true){
think();
eat();
}
}catch(InterruptedException e){}
}
}
 - How to use it in a guest)
