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问题的描述
启动3个线程打印递增的数字, 线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到打印到45.
wait+notify实现：
package com.tonyluis;
public class NumberPrintDemo {
static int n = 1;
static int state = 0;
public static void main(String[] args) {
new Thread(new MyThread(0)).start();
new Thread(new MyThread(1)).start();
new Thread(new MyThread(2)).start();
}
}
class MyThread implements Runnable{
private int state;
MyThread(){
this(0);
}
MyThread(int state){
this.state=state;
}
public void run() {
String threadName=Thread.currentThread().getName();
for (int i = 0; i < 3; i++) {
synchronized (MyThread.class) {
while (state != NumberPrintDemo.state)
try {
MyThread.class.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++)
System.out.println(threadName+ ": " + NumberPrintDemo.n++);
System.out.println();
NumberPrintDemo.state++;
NumberPrintDemo.state%=3;
MyThread.class.notifyAll();
}
}
}
}
Lock+condition实现：
package com.tonyluis;
import java.util.concurrent.locks.*;
public class NumberPrint {
static int state = 0;
static int n = 1;
static Lock lock = new ReentrantLock();
static Condition condition[]=new Condition[3];
public static void main(String[] args) {
for(int i=0;i<condition.length;i++)
condition[i]=lock.newCondition();
new Thread(new MyThread(0)).start();
new Thread(new MyThread(1)).start();
new Thread(new MyThread(2)).start();
}
}
class MyThread implements Runnable{
private int state;
MyThread(){
this(0);
}
MyThread(int state){
this.state=state;
}
public void run() {
String threadName=Thread.currentThread().getName();
for (int i = 0; i < 5; i++) {
try {
NumberPrint.lock.lock();
while (state != NumberPrint.state)
try {
NumberPrint.condition[state].await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++)
System.out.println(threadName+ ": " + NumberPrint.n++);
System.out.println();
NumberPrint.state++;
NumberPrint.state%=NumberPrint.condition.length;
NumberPrint.condition[NumberPrint.state].signal();
} finally {
NumberPrint.lock.unlock();
}
}
}
}
使用wait+notify的实现方式，在线程0进入notifyAll()方法之后，会唤醒线程1和线程2，线程1和线程2以及线程0会展开竞争，虽然最终是由线程1获得琐，过程可能比较曲折。
使用lock+condition的实现方法，线程0只会唤醒线程1，这样只有线程1和线程0展开竞争，较为精确，尤其是并发量比较大的情况下。
Lock+Condition 相对于 wait+notify 的一个优势案例分析
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原文地址：http://www.cnblogs.com/tonyluis/p/5581560.html
