Java-JUC（十）：线程按序交替执行

问题：

有a、b、c三个线程，使得它们按照abc依次执行10次。

实现：

package com.dx.juc.test;import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;public class ABC {public static void main(String[] args) {final AlternateDemo alternateDemo=new AlternateDemo();new Thread(new Runnable() {public void run() {for(int i=1;i<10;i++){alternateDemo.loopA(i);}}}, "A").start();new Thread(new Runnable() {public void run() {for(int i=1;i<10;i++){alternateDemo.loopB(i);}}}, "B").start();new Thread(new Runnable() {public void run() {for(int i=1;i<10;i++){alternateDemo.loopC(i);System.out.println("---------------------------------------");}}}, "C").start();}
}class AlternateDemo {private Lock lock = new ReentrantLock();private Condition conditionA = lock.newCondition();private Condition conditionB = lock.newCondition();private Condition conditionC = lock.newCondition();// 一个标号，标记当前可以执行的线程编号.1-a线程可以执行，2-b线程可以执行，3-c线程可以执行。private int flag = 1;public void loopA(int loopNum) {lock.lock();try {// 1)等待喚醒// 如果flag标记值不是1的话，就让线程处于等待状态，直到其他线程唤醒它。if (flag != 1) {conditionA.await();}// 2)被喚醒后，開始執行。for (int i = 1; i <= 1; i++) {System.out.println(Thread.currentThread().getName() + "-" + loopNum + "-" + i);}// 3)修改標記flag，并喚醒下一個線程。flag = 2;conditionB.signal();} catch (InterruptedException e) {e.printStackTrace();} finally {lock.unlock();}}public void loopB(int loopNum) {lock.lock();try {// 1)等待喚醒// 如果flag标记值不是2的话，就让线程处于等待状态，直到其他线程唤醒它。if (flag != 2) {conditionB.await();}// 2)被喚醒后，開始執行。for (int i = 1; i <= 1; i++) {System.out.println(Thread.currentThread().getName() + "-" + loopNum + "-" + i);}// 3)修改標記flag，并喚醒下一個線程。flag = 3;conditionC.signal();} catch (InterruptedException e) {e.printStackTrace();} finally {lock.unlock();}}public void loopC(int loopNum) {lock.lock();try {// 1)等待喚醒// 如果flag标记值不是3的话，就让线程处于等待状态，直到其他线程唤醒它。if (flag != 3) {conditionC.await();}// 2)被喚醒后，開始執行。for (int i = 1; i <= 1; i++) {System.out.println(Thread.currentThread().getName() + "-" + loopNum + "-" + i);}// 3)修改標記flag，并喚醒下一個線程。flag = 1;conditionA.signal();} catch (InterruptedException e) {e.printStackTrace();} finally {lock.unlock();}}
}

测试打印结果：

A-1-1
B-1-1
C-1-1
---------------------------------------
A-2-1
B-2-1
C-2-1
---------------------------------------
A-3-1
B-3-1
C-3-1
---------------------------------------
A-4-1
B-4-1
C-4-1
---------------------------------------
A-5-1
B-5-1
C-5-1
---------------------------------------
A-6-1
B-6-1
C-6-1
---------------------------------------
A-7-1
B-7-1
C-7-1
---------------------------------------
A-8-1
B-8-1
C-8-1
---------------------------------------
A-9-1
B-9-1
C-9-1
---------------------------------------