今天的作业题
[Java] 纯文本查看 复制代码 /*1、使用生产者和消费者模式实现,交替输出:
假设只有两个线程,输出以下结果:
t1-->1
t2-->2
t1-->3
t2-->4
t1-->5
t2-->6
....
要求:必须交替,并且t1线程负责输出奇数。t2线程负责输出偶数。
两个线程共享一个数字,每个线程执行时都要对这个数字进行:++
*/
public class Test {
public static void main(String[] args) {
Num num=new Num(1);
Thread t1=new Thread(new OddNumThread(num));
Thread t2=new Thread(new EvenNumThread(num));
t1.setName("奇数线程");
t2.setName("偶数线程");
t1.start();
t2.start();
}
}
class Num{
private int i;
public Num() {
}
Num(int i) {
this.i = i;
}
public int getI() {
return i;
}
public void setI(int i) {
this.i = i;
}
}
class OddNumThread implements Runnable{
private Num num;
public OddNumThread() {
}
public OddNumThread(Num num) {
this.num = num;
}
public void run() {
while (true){
synchronized(num){
if (num.getI() %2 != 0){
System.out.println(Thread.currentThread().getName()+"-->"+num.getI());
int i=num.getI();
num.setI(++i);
num.notify();
}else {
try {
num.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
}
class EvenNumThread implements Runnable{
private Num num;
public EvenNumThread() {
}
public EvenNumThread(Num num) {
this.num = num;
}
public void run() {
while (true){
synchronized(num){
if (num.getI() %2 == 0){
System.out.println(Thread.currentThread().getName()+"-->"+num.getI());
int i=num.getI();
num.setI(++i);
num.notify();
}else {
try {
num.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
} |