曾经有面试官问过这样一个问题，如何让两个线程交替打印奇数和偶数？如何让多个线程交替执行？
回答这个问题之前需要先了解java多线程的运行机制，线程间通信机制，线程同步问题。
这个问题我知道的实现方案有两种，一种是基于synchronized和wait/notify，另外一种是基于Lock和Condition.
1.基于synchronized和wait/notify
package org.stathry.jdkdeep.concurrent;
/**
* 线程间通信-基于Object的wait/notify
* @date 2017年12月21日
*/
public class ThreadTurnTest1 {
private Object lock = new Object();
private boolean RUN0 = true;
private static final int LIMIT = 1000;
public static void main(String[] args) throws InterruptedException {
final ThreadTurnTest1 o = new ThreadTurnTest1();
new Thread(new Runnable() {
@Override
public void run() {
try {
o.m0();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t0").start();
new Thread(new Runnable() {
@Override
public void run() {
try {
o.m1();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t1").start();
Thread.sleep(10 * 1000);
}
private void m1() throws InterruptedException {
for (int i = 1; i < LIMIT; i += 2) {
synchronized (lock) {
if (RUN0) {
lock.wait();
}
System.out.println(Thread.currentThread().getName() + "___" + i);
RUN0 = true;
lock.notify();
}
}
}
private void m0() throws InterruptedException {
for (int i = 0; i < LIMIT; i += 2) {
synchronized (lock) {
if (!RUN0) {
lock.wait();
}
System.out.println(Thread.currentThread().getName() + "___" + i);
RUN0 = false;
lock.notify();
}
}
}
}
执行结果如下：
t0___0
t1___1
t0___2
t1___3
t0___4
t1___5
t0___6
t1___7
t0___8
t1___9
t0___10
...
2.基于Lock和Condition
基于Lock和Condition可以实现两个以上线程的通信。
package org.stathry.jdkdeep.concurrent;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.LinkedBlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
/**
* 线程间通信-基于lock的condition
* @date 2017年12月21日
*/
public class ThreadTurnTest2 {
private Lock lock = new ReentrantLock();
private Condition c0 = lock.newCondition();
private Condition c1 = lock.newCondition();
private Condition c2 = lock.newCondition();
private int status = 0;
private static final int LIMIT = 10000;
public static void main(String[] args) throws InterruptedException {
ExecutorService exec = new ThreadPoolExecutor(3, 30, 60, TimeUnit.SECONDS, new LinkedBlockingQueue(300),
Executors.defaultThreadFactory(), new ThreadPoolExecutor.AbortPolicy());
final ThreadTurnTest2 o = new ThreadTurnTest2();
exec.submit(o.createTask(0));
exec.submit(o.createTask(1));
exec.submit(o.createTask(2));
exec.shutdown();
exec.awaitTermination(5, TimeUnit.MINUTES);
}
/**
* @param name
* @param i
*/
private Runnable createTask(final int i) {
return new Runnable() {
@Override
public void run() {
try {
m(i);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
};
}
private void m(final int n) throws InterruptedException {
for (int i = n; i < LIMIT; i += 3) {
lock.lock();
try {
switch (n) {
case 0:
if (status != 0) {
c0.await();
}
System.out.println(Thread.currentThread().getName() + "___" + i);
status = 1;
c1.signal();
break;
case 1:
if (status != 1) {
c1.await();
}
System.out.println(Thread.currentThread().getName() + "___" + i);
status = 2;
c2.signal();
break;
case 2:
if (status != 2) {
c2.await();
}
System.out.println(Thread.currentThread().getName() + "___" + i);
status = 0;
c0.signal();
break;
default:
break;
}
} finally {
lock.unlock();
}
}
}
}
执行结果如下：
pool-1-thread-1___0
pool-1-thread-2___1
pool-1-thread-3___2
pool-1-thread-1___3
pool-1-thread-2___4
pool-1-thread-3___5
pool-1-thread-1___6
pool-1-thread-2___7
pool-1-thread-3___8
pool-1-thread-1___9
pool-1-thread-2___10
pool-1-thread-3___11
pool-1-thread-1___12
pool-1-thread-2___13
pool-1-thread-3___14
...
看完上面两种方式你学会了吗？
其实这两种方式思路都是类似的，大致可以分三步：
循环内加锁
判断共享状态变量，如果状态值表示还没轮到当前线程执行，则调用调用锁对象的wait方法
等待状态变化的线程被唤醒，执行任务，然后改变状态，然后调用锁对象的notify方法
再多问几个问题，为什么要加锁呢？因为只有加锁才能保证多线程交替有序执行，否则线程的执行是由操作系统随机调度的，那么执行顺序自然是乱序的。为什么要对同一个对象加锁呢？因为只有对同一对象加锁，才能保证访问该锁的两个线程之间相互通信(不同的对象调用notify不会唤醒处于wait状态的线程)，否则即使线程已经获取到锁了，但可能会因为没轮到自己执行而一直处于wait状态。
如果觉得写的不错，记得点击关注，如果写的不好欢迎批评指正，让我们一起进步！
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