Java - 使用同步方法进行多线程练习

Question

我有这个练习，一种医院模拟，其中我必须控制每个房间的访问。医生一次只能进入一个房间，并且只有在没有访客进入的情况下才能进入。相反，访客只能在没有医生在场且最多还有 4 名访客进入的情况下才能进入该房间。这是我的代码:

public class Room {

public Room(){
}

public synchronized void friendVisit() throws InterruptedException{
    if(visitors>4 || doctors>0)
        wait();
    visitors++;
}

public synchronized void exitFriend(){
    visitors--;
    notify();
}

public synchronized void doctorVisit() throws InterruptedException{
    if(doctors>0 || visitors>0)
        wait();
    doctors++;
}

public synchronized void exitDoctor(){
    --doctors;
    notify();
}

public int getVisitors(){
    return visitors;
}

public int getDoctors(){
    return doctors;
}

int visitors=0; //number of visitors in the room
int doctors=0; //number of doctors in the room

医生和访客(称为 Friend 的类)是线程

public class Friend extends Thread{
public Friend(Room room_reference){
    room=room_reference;
}

public void run(){
        try {
            sleep(500);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        try {
            room.friendVisit();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        try {
            sleep(500);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        room.exitFriend();
}

private Room room; //reference to the room

这是医生线程:

public class Doctor extends Thread{
public Doctor(Room room_reference){
    room=room_reference;
}

public void run(){
        try {
            sleep(500);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        try {
            room.doctorVisit();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        try {
            sleep(500);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        room.exitDoctor();
}

private Room room; //reference to the room

这是一个显示线程，用于跟踪访客和医生的数量:

public class Display extends Thread{
public Display(Room room_reference){
    room=room_reference;
}

public void run(){
    while(true)
    {
    try {
        sleep(300);
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    System.out.println("The room contains "+room.getDoctors()+
                       " doctors and "+room.getVisitors()+" visitors.");
    }
}

private Room room;

这是我的主要内容:

public class Demo {
public static void main(String[]args){
    Room room=new Room();
    Friend friend=new Friend(room);
    Doctor doctor=new Doctor(room);
    Display display=new Display(room);
    display.start();
    while(true){
        if(new Random().nextBoolean()==true){
            friend=new Friend(room);
            friend.start();
        }
        if(new Random().nextInt(5)==3){
            doctor=new Doctor(room);
            doctor.start();
        }   
    }
}

问题是不止一名医生可以访问该房间，我不明白为什么，因为 Room 类中的方法适用于访客。提前致谢。

Answer 1

我认为您的错误之一是假设 wait() 仅当满足上面的条件时才会返回:

if(doctors>0 || visitors>0)
        wait();

您可以在 if 语句中的条件为 false 的情况下从此调用返回 wait()。也许尝试一下 while 循环:

while (doctors>0 || visitors>0) {
        wait();
}

(当然要添加括号，因为你知道缺少括号是邪恶的......)

可能还有其他问题 - 我还没有启动您的代码。