我想用 T1-1、T2-2、T3-3、T1-4、T2-5、T3-6 等线程按顺序打印数字
public class NumberGame {
static int a=1;
public static void main(String args[]) throws InterruptedException
{
PrintSequenceRunnable C1=new PrintSequenceRunnable("T1",a);
PrintSequenceRunnable C2=new PrintSequenceRunnable("T2",a);
PrintSequenceRunnable C3=new PrintSequenceRunnable("T3",a);
Thread t1 = new Thread(C1);
Thread t2 = new Thread(C2);
Thread t3 = new Thread(C3);
t1.start();
t2.start();
t3.start();
}
}
public class PrintSequenceRunnable implements Runnable {
String tname;
int a;
PrintSequenceRunnable(String tname, int a )
{
this.tname = tname;
this.a = a;
}
@Override
public void run() {
synchronized (this) {
for(int i=0; i<10;i++)
{
System.out.println(tname+" "+a);
a++;
try {
this.wait(1000);
this.notify();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
// TODO Auto-generated method stub
}
}
但我的输出就像
T1-1 T2-1 T3-1 T1-2 T3-2 T2-2 T3-3 T1-3 T2-3 T3-4 T1-4 T2-4 T3-5 T1-5 T2-5 T3-6 T1-6 T2-6 T1-7 T2-7 T3-7 T2-8 T3-8 T1-8 T2-9 T3-9 T1-9 T2-10 T1-10 T3-10
谁能帮帮我。
最佳答案
问题是:
- 设计存在线程之间引发条件的问题,需要同步它们。
- 使用构造函数时
PrintSequenceRunnable(String tname, int a )
您正在发送原始变量a
的副本,这是一个静态成员NumberGame
。所以,每个PrintSequenceRunnable
有自己的变量a
.
我的建议是使用方法 wait
同步每个线程和notify
。我拿了你的代码并做了一些修改:
数字游戏
public class NumberGame {
public static void main(String args[]) throws InterruptedException
{
PrintSequenceRunnable C1=new PrintSequenceRunnable("T1");
PrintSequenceRunnable C2=new PrintSequenceRunnable("T2");
PrintSequenceRunnable C3=new PrintSequenceRunnable("T3");
Thread t1 = new Thread(C1);
Thread t2 = new Thread(C2);
Thread t3 = new Thread(C3);
t1.start();
t2.start();
t3.start();
Thread.sleep(1);//Wait 1 ms to avoid a raise condition
PrintSequenceRunnable.activateNextItem(); //Start sequence.
t1.join();
t2.join();
t3.join();
System.out.println("--END--");
}
}
PrintSequenceRunnable
import java.util.Vector;
public class PrintSequenceRunnable implements Runnable {
static private int a = 0;
private static Vector<PrintSequenceRunnable> items = new Vector<PrintSequenceRunnable>();
/**
* Method to select the next Thread which will be activate to continue its thread.
*/
public static synchronized void activateNextItem() {
int index = a % items.size();
items.get(index).activate();
}
private String tname;
private Object sempahoro = new Object(); //Object to sinchrony the thread
public PrintSequenceRunnable(String tname)
{
this.tname = tname;
items.add(this);
}
public void activate()
{
synchronized (sempahoro) {
sempahoro.notify();
}
}
@Override
public void run() {
for(int i=0; i<10;i++)
{
synchronized (sempahoro) {
try {
sempahoro.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
a++;
System.out.println(tname+" "+a);
activateNextItem(); //Raise the next thread.
}
// TODO Auto-generated method stub
}
}
在此示例中,方法 activateNextItem, from
PrintSequenceRunnable`,将决定通知哪个实例执行其线程。
重要,我需要设置一个 sleep
初始化每个线程后一秒以避免引发条件,我的意思是:WAITING所有线程启动并使所有线程处于等待状态。
输出:
T1 1
T2 2
T3 3
T1 4
T2 5
T3 6
T1 7
T2 8
T3 9
T1 10
T2 11
T3 12
T1 13
T2 14
T3 15
T1 16
T2 17
T3 18
T1 19
T2 20
T3 21
T1 22
T2 23
T3 24
T1 25
T2 26
T3 27
T1 28
T2 29
T3 30
--END--
关于java - 如何使用多线程打印序列中的数字,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57875711/