背景
用两个线程交替输出A-Z和1-26，即一个线程输出A-Z，另一个线程输出1-26
而且是交替形式
- 线程1输出A——线程二输出1
- 线程1输出B——线程二输出2
- 线程1输出C——线程二输出3
以此类推
分析
主要考察线程之间的通信，思路就是创建两个线程
在一个线程输出一个内容之后，自己进入阻塞，去唤醒另一个线程
另一个线程同样，输出一个内容之后，自己进入阻塞，去唤醒另一个线程
代码实现（一）
public class AlternateCover {
public static void main(String[] args) {
final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};
threadRun(arrLetter, arrNumber);
}
private static void threadRun(char[] arrLetter,String[] arrNumber){
final Object lock = new Object();// 设置一个锁对象
// print arrNumber
new Thread(() -> {
synchronized (lock) {
for (String a : arrNumber) {
System.out.print( a);
try {
lock.notify();// 唤醒其他等待的线程 此处唤醒 arrLetter
lock.wait();// arrNumber自己进入等待 让出CPU资源和锁资源
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();
}
}, "arrNumber ").start();
// print arrLetter
new Thread(() -> {
synchronized (lock) {// 获取对象锁
for (char a : arrLetter) {
System.out.print(a);
try {
lock.notify();// 唤醒其他等待的线程 此处唤醒 arrNumber
lock.wait();// arrLetter自己进入等待 让出CPU资源和锁资源
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
}
}, "arrLetter ").start();
}
}
运行一下，确实实现了交替输出，但是多运行几次，就会发现问题
有时候是数字先输出，有时候是字母先输出
即两个线程谁先启动的顺序是不固定的
倘若试题中再加一句，必须要字母先输出，怎么办？
代码实现（二）
public class AlternateCover {
public static volatile Boolean flg = false;// 谁先开始的标志 volatile修饰目的是让该值修改对所有线程可见，且防止指令重排序
public static void main(String[] args) {
final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};
threadRun(arrLetter, arrNumber);
}
private static void threadRun(char[] arrLetter,String[] arrNumber){
final Object lock = new Object();// 锁对象
// print arrLetter
new Thread(() -> {
synchronized (lock) {
if (!flg){ // 如果flg是false 就将值设为true
flg = true;
}
for (char a : arrLetter) {
System.out.print(a);// 输出内容
try {
lock.notify();// 唤醒在等待的其他线程中的一个（此处也只有另一个）
lock.wait();// 自己进入等待 让出CPU资源和锁资源
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
}
}, "arrLetter").start();
// print arrNumber
new Thread(() -> {
synchronized (lock) {
if (!flg){// 倘若是该线程先执行，那么flg次数还是false 就先等着
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
for (String a : arrNumber) {
System.out.print( a);
try {
lock.notify();
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();
}
}, "arrNumber").start();
}
}
如此问题可以得到解决，但有更优(装)雅（B）的解决办法
CountDownLatch实现
public class AlternateCover {
private static CountDownLatch count = new CountDownLatch(1);// 计数器容量为1
public static void main(String[] args) {
final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};
threadRun(arrLetter, arrNumber);
}
private static void threadRun(char[] arrLetter,String[] arrNumber){
final Object lock = new Object();
// print arrLetter
new Thread(() -> {
synchronized (lock) {// 获取对象锁
count.countDown();// 对计数器进行递减1操作，当计数器递减至0时，当前线程会去唤醒阻塞队列里的所有线程（只针对count）
for (char a : arrLetter) {
System.out.print(a);
try {
lock.notify();// 唤醒其他等待的线程 此处唤醒 arrNumber
lock.wait();// arrLetter自己进入等待 让出CPU资源和锁资源
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
}
}, "arrLetter ").start();
// print arrNumber
new Thread(() -> {
synchronized (lock) {
try {
count.await();// 如果该线程先执行 阻塞当前线程，将当前线程加入阻塞队列
} catch (InterruptedException e) {
e.printStackTrace();
}
for (String a : arrNumber) {
System.out.print( a);
try {
lock.notify();// 唤醒其他等待的线程 此处唤醒 arrLetter
lock.wait();// arrNumber自己进入等待 让出CPU资源和锁资源
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();
}
}, "arrNumber ").start();
}
}
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