1 死锁定义
死锁是指,两个或两个以上的线程在执行过程中,由于竞争资源而造成的一种阻塞的现象,若无外力作用,它们都将无法推进下去。简化一点说就是:一组相互竞争资源的线程因为互相等待,导致“永久”阻塞的现象。
例如下面案例:
public class DeadLockTest {
private static String A = "A";
private static String B = "B";
public static void main(String[] args) {
new DeadLockTest().deadLock();
}
private void deadLock() {
Thread t1 = new Thread(new Runnable() {
@Override
public void run() {
synchronized (A) {
System.out.println("t1成功获取A锁");
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
e.printStackTrace();
}
synchronized (B) {
System.out.println("t1成功获取B锁");
}
}
}
});
Thread t2 = new Thread(new Runnable() {
@Override
public void run() {
synchronized (B) {
System.out.println("t2成功获取B锁");
synchronized (A) {
System.out.println("t2成功获取A锁");
}
}
}
});
t1.start();
t2.start();
}
}
t1锁住了A,然后尝试对B进行加锁,同时t2已经锁住了B,接着尝试对A进行加锁,这时死锁就发生了。t1永远得不到B,t2也永远得不到A,它们将永远阻塞下去。
Thread 1 locks A, waits for B
Thread 2 locks B, waits for A
可以使用 jps 定位进程 id,再用 jstack 定位死锁。
D:\workplace\GiteeProjects\java-learn\concurrent-learn\src\main\java\com\kai\demo\basic>jps
Picked up JAVA_TOOL_OPTIONS: -Dfile.encoding=gb2312
21812 Jps
15848 RemoteMavenServer36
25132 KotlinCompileDaemon
25580
27516 DeadLockTest
29052 Launcher
D:\workplace\GiteeProjects\java-learn\concurrent-learn\src\main\java\com\kai\demo\basic>jstack 27516
Picked up JAVA_TOOL_OPTIONS: -Dfile.encoding=gb2312
2020-10-31 17:27:38
Full thread dump Java HotSpot(TM) 64-Bit Server VM (25.261-b12 mixed mode):
"DestroyJavaVM" #14 prio=5 os_prio=0 tid=0x0000016945db5000 nid=0xfd0 waiting on condition [0x0000000000000000]
java.lang.Thread.State: RUNNABLE
"Thread-1" #13 prio=5 os_prio=0 tid=0x0000016962b9b800 nid=0x1814 waiting for monitor entry [0x00000005fc8ff000]
java.lang.Thread.State: BLOCKED (on object monitor)
at com.kai.demo.basic.DeadLockTest$2.run(DeadLockTest.java:47)
- waiting to lock <0x000000076bf40af8> (a java.lang.String)
- locked <0x000000076bf40b28> (a java.lang.String)
at java.lang.Thread.run(Thread.java:748)
"Thread-0" #12 prio=5 os_prio=0 tid=0x0000016962b03000 nid=0x6490 waiting for monitor entry [0x00000005fc7ff000]
java.lang.Thread.State: BLOCKED (on object monitor)
at com.kai.demo.basic.DeadLockTest$1.run(DeadLockTest.java:35)
- waiting to lock <0x000000076bf40b28> (a java.lang.String)
- locked <0x000000076bf40af8> (a java.lang.String)
at java.lang.Thread.run(Thread.java:748)
--- omit ---
Found one Java-level deadlock:
=============================
"Thread-1":
waiting to lock monitor 0x0000016960a81c38 (object 0x000000076bf40af8, a java.lang.String),
which is held by "Thread-0"
"Thread-0":
waiting to lock monitor 0x0000016960a84578 (object 0x000000076bf40b28, a java.lang.String),
which is held by "Thread-1"
Java stack information for the threads listed above:
===================================================
"Thread-1":
at com.kai.demo.basic.DeadLockTest$2.run(DeadLockTest.java:47)
- waiting to lock <0x000000076bf40af8> (a java.lang.String)
- locked <0x000000076bf40b28> (a java.lang.String)
at java.lang.Thread.run(Thread.java:748)
"Thread-0":
at com.kai.demo.basic.DeadLockTest$1.run(DeadLockTest.java:35)
- waiting to lock <0x000000076bf40b28> (a java.lang.String)
- locked <0x000000076bf40af8> (a java.lang.String)
at java.lang.Thread.run(Thread.java:748)
Found 1 deadlock.
2 避免死锁
产生死锁的四个必要条件:
(1) 互斥:一个资源每次只能被一个线程使用。
(2) 请求与保持:一个线程因请求资源而阻塞时,不释放已获得的资源。
(3) 不剥夺: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:若干线程之间形成一种头尾相接的循环等待资源关系。例如上述案例中,线程 t1 等待线程t2占有的资源,线程t2等待线程t1占有的资源。
只有这四个条件都发生时才会出现死锁,那么反过来,也就是说只要我们破坏其中一个,就可以成功预防死锁的发生。
编写程序时避免死锁的常见方法:
(1)避免一个线程同时获取多个锁。
(2)避免一个线程在锁内同时占用多个资源,尽量保证每个锁只占用一个资源。
(3)尝试使用定时锁,使用lock.tryLock(timeout)来替代使用内部锁机制。
(4) 对于数据库锁,加锁和解锁必须在一个数据库连接里,否则会出现解锁失败的情况。
3 哲学家就餐问题
哲学家就餐问题是1965年由Dijkstra提出的一种线程同步的问题。
问题描述:一圆桌前坐着5位哲学家,每两个人中间有一只筷子,桌子中央有面条。哲学家思考问题,当饿了的时候拿起左右两只筷子吃饭,必须拿到两只筷子才能吃饭。上述问题会产生死锁的情况,当5个哲学家都拿起自己右手边的筷子,准备拿左手边的筷子时产生死锁现象。
用代码描述就是:
public class DeadLockTest2 {
public static void main(String[] args) {
ExecutorService executorService = 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++) {
executorService.execute(new Philosopher(chopsticks[i], chopsticks[(i + 1) % sum]));
}
}
}
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) {// 获取左手边筷子
synchronized (right) {// 获取右手边筷子
Thread.sleep(1000);// 进餐
}
}
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
// 筷子
class Chopstick {
public Chopstick() {
}
}
3.1 破坏循环等待条件
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 DeadLockTest2 extends Thread {
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]));
}
}
}
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() < right.getId()) {
this.left = left;
this.right = right;
} else {
this.left = right;
this.right = left;
}
}
@Override
public void run() {
try {
while (true) {
Thread.sleep(1000);
System.out.println(Thread.currentThread().getName()+"思考一段时间");
synchronized (left) {
System.out.println(Thread.currentThread().getName()+"获取到第一只筷子");
synchronized (right) {
System.out.println(Thread.currentThread().getName()+"获取到第二只筷子");
Thread.sleep(1000);
System.out.println(Thread.currentThread().getName()+"进餐一段时间");
}
}
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
执行结果:
pool-1-thread-4思考一段时间
pool-1-thread-4获取到第一只筷子
pool-1-thread-4获取到第二只筷子
pool-1-thread-5思考一段时间
pool-1-thread-5获取到第一只筷子
pool-1-thread-1思考一段时间
... ...
3.2 破坏请求与保持条件
使用可重入锁ReentrantLock来避免死锁。使用内置锁的线程获取不到锁时会被阻塞,而ReentrantLock是可重入的,也可以指定一个超时时限(调用tryLock(long timeout, TimeUnit unit)
方法),在等待超过该时间后tryLock就会返回一个失败信息,释放所持有的资源。
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) {
}~~~~
}
}
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