wait()/notify()编程
wait()/notify()编程
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java并发编程--一道经典多线程题的2种解法
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/**
* Created by emos on 2015/9/13.
*/
public class NumberPrintSimple extends Thread {
static int c = 0;
static volatile int state = 0;
private int id;
@Override
public synchronized void run() {
while (state < 15) {
if (state % 3 == id) {
for (int j = 0; j < 5; j++) {
c++;
System.out.format("Thread %d: %d %n", id, c);
}
state++;
}
}
}
public NumberPrintSimple(int id) {
this.id = id;
}
public static void main(String[] args) {
for (int i = 0; i < 3; i++) {
new NumberPrintSimple(i).start();
}
}
}
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/**
* Created by emos on 2015/9/13.
*/
public class test {
// n为即将打印的数字
private static int n = 1;
// state=1表示将由线程1打印数字, state=2表示将由线程2打印数字, state=3表示将由线程3打印数字
private static int state = 1;
public static void main(String[] args) {
final test pn = new test();
new Thread(new Runnable() {
public void run() {
// 3个线程打印75个数字, 单个线程每次打印5个连续数字, 因此每个线程只需执行5次打印任务. 3*5*5=75
for (int i = 0; i < 5; i++) {
// 3个线程都使用pn对象做锁, 以保证每个交替期间只有一个线程在打印
synchronized (pn) {
// 如果state!=1, 说明此时尚未轮到线程1打印, 线程1将调用pn的wait()方法, 直到下次被唤醒
while (state != 1)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
// 当state=1时, 轮到线程1打印5次数字
for (int j = 0; j < 5; j++) {
// 打印一次后n自增
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
// 线程1打印完成后, 将state赋值为2, 表示接下来将轮到线程2打印
state = 2;
// notifyAll()方法唤醒在pn上wait的线程2和线程3, 同时线程1将退出同步代码块, 释放pn锁.
// 因此3个线程将再次竞争pn锁
// 假如线程1或线程3竞争到资源, 由于state不为1或3, 线程1或线程3将很快再次wait, 释放出刚到手的pn锁.
// 只有线程2可以通过state判定, 所以线程2一定是执行下次打印任务的线程.
// 对于线程2来说, 获得锁的道路也许是曲折的, 但前途一定是光明的.
pn.notifyAll();
}
}
}
}, "线程1").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
synchronized (pn) {
while (state != 2)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 3;
pn.notifyAll();
}
}
}
}, "线程2").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
synchronized (pn) {
while (state != 3)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 1;
pn.notifyAll();
}
}
}
}, "线程3").start();
}
}
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import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;
/**
* Created by emos on 2015/9/13.
*/
public class NumberPrintByLock implements Runnable {
private int state = 1;
private int n = 1;
// 使用lock做锁
private ReentrantLock lock = new ReentrantLock();
// 获得lock锁的3个分支条件
private Condition c1 = lock.newCondition();
private Condition c2 = lock.newCondition();
private Condition c3 = lock.newCondition();
@Override
public void run() {
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
// 线程1获得lock锁后, 其他线程将无法进入需要lock锁的代码块.
// 在lock.lock()和lock.unlock()之间的代码相当于使用了synchronized(lock){}
lock.lock();
while (state != 1)
try {
// 线程1竞争到了lock, 但是发现state不为1, 说明此时还未轮到线程1打印.
// 因此线程1将在c1上wait
// 与解法一不同的是, 三个线程并非在同一个对象上wait, 也不由同一个对象唤醒
c1.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
// 如果线程1竞争到了lock, 也通过了state判定, 将执行打印任务
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
// 打印完成后将state赋值为2, 表示下一次的打印任务将由线程2执行
state = 2;
// 唤醒在c2分支上wait的线程2
c2.signal();
} finally {
// 打印任务执行完成后需要确保锁被释放, 因此将释放锁的代码放在finally中
lock.unlock();
}
}
}
}, "线程1").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
lock.lock();
while (state != 2)
try {
c2.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 3;
c3.signal();
} finally {
lock.unlock();
}
}
}
}, "线程2").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
lock.lock();
while (state != 3)
try {
c3.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 1;
c1.signal();
} finally {
lock.unlock();
}
}
}
}, "线程3").start();
}
public static void main(String[] args) {
new NumberPrintByLock().run();
}
}
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