Java - Multithreading exercise with synchronized methods

Question

I have this exercise, a kind of hospital simulation in which I have to control the accesses to each singular room. Doctors can enter the room one at the time, and can enter only if no visitors are in. A visitor, instead, can only access the room if no doctors are in it and a max of 4 more visitors are in. Here's my code:

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

Doctors and Visitors(the class it's called Friend) are threads

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

Here's the doctor thread:

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

Here's a Display thread to keep trace of the number of visitors and doctors:

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;

And here's my main:

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();
        }   
    }
}

The problem is that more than one doctor can access the room and I don't understand why, since the methods in the Room class work for the Visitors. Thanks in advance.

have you tried debugging? Also, you never start the initial friend and doctor you instantiate. — Taylor
– Taylor, Commented Nov 4, 2013 at 14:19

Duncan Jones · Accepted Answer · 2013-11-04 14:19:21Z

3

I think one of your mistakes is assuming that wait() will return only when the condition above it is satisfied:

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

You may return from this call to wait() with the condition in your if statement false. Perhaps try a while loop:

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

(adding brackets, of course, because you know a lack of brackets is evillll.....)

There may be other problems - I've not yet fired up your code.

answered Nov 4, 2013 at 14:19

Duncan Jones

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1 Comment

Andrew Over a year ago

Yep, it seems you're completely right, and it makes sense too. Many thanks man.

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