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why all producer threads also exit when comsumer threads all exit

I am running the below code, and I have one question:
why do all producer threads also exit when the consumer threads all exit.
Here is the code:
public class NumbersConsumer implements Runnable {
private final BlockingQueue<Integer> queue;
private final int poisonPill;
NumbersConsumer(BlockingQueue<Integer> queue, int poisonPill) {
this.queue = queue;
this.poisonPill = poisonPill;
}
public void run() {
try {
while (true) {
Integer number = queue.take();
if (number.equals(poisonPill)) {
return;
}
String result = number.toString();
System.out.println(Thread.currentThread().getName() + " result: " + result);
}
} catch (InterruptedException e) {
System.out.println(e.getStackTrace());
Thread.currentThread().interrupt();
}
}
}
Producer:
public class NumbersProducer implements Runnable {
private final BlockingQueue<Integer> numbersQueue;
private final int poisonPill;
private final int poisonPillPerProducer;
NumbersProducer(BlockingQueue<Integer> numbersQueue, int poisonPill, int poisonPillPerProducer) {
this.numbersQueue = numbersQueue;
this.poisonPill = poisonPill;
this.poisonPillPerProducer = poisonPillPerProducer;
}
public void run() {
try {
generateNumbers();
} catch (InterruptedException e) {
System.out.println(e.getStackTrace());
Thread.currentThread()
.interrupt();
}
}
private void generateNumbers() throws InterruptedException {
for (int i = 0; i < 100; i++) {
numbersQueue.put(ThreadLocalRandom.current()
.nextInt(100));
}
for (int j = 0; j < poisonPillPerProducer; j++) {
numbersQueue.put(poisonPill);
}
}
}
Tester:
public class BlockingQueueUsage {
public static void main(String[] args) {
int BOUND = 10;
int N_PRODUCERS = 4;
int N_CONSUMERS = Runtime.getRuntime().availableProcessors();
int poisonPill = Integer.MAX_VALUE;
int poisonPillPerProducer =N_CONSUMERS / N_PRODUCERS ;
System.out.printf("poisonPillPerProducer:%s,N_CONSUMERS:%s %n",poisonPillPerProducer,N_CONSUMERS);
BlockingQueue<Integer> queue = new LinkedBlockingQueue<>(BOUND);
for (int i = 0; i < N_PRODUCERS; i++) {
new Thread(new NumbersProducer(queue, poisonPill, poisonPillPerProducer),"Producer"+i).start();
}
for (int j = 0; j < N_CONSUMERS; j++) {
new Thread(new NumbersConsumer(queue, poisonPill),"comsumer"+j).start();
}
}
}

Simple: your producers are missing a loop.
The produces do nothing but produce some numbers, and that poison pill that will end the consumers.
The consumers on the other hand, would be consuming forever ... until that pill comes in.
In other words: your producers "end", because they aren't designed to stay alive.

Java - How to detect deadlocks and recovery from this?

Right now I write a Java program that has as purpose detect deadlocks and recovery from this situation. The program input is two numbers, N = Number of types of resources and M = Number of process.
I wanted to do something like this:
private static void test2() {
final ReentrantLock lock1 = new ReentrantLock();
final ReentrantLock lock2 = new ReentrantLock();
Thread thread1 = new Thread(new Runnable() {
#Override public void run() {
try {
lock1.lock();
System.out.println("Thread1 acquired lock1");
try {
TimeUnit.MILLISECONDS.sleep(50);
} catch (InterruptedException ignore) {}
lock2.lock();
System.out.println("Thread1 acquired lock2");
}
finally {
lock2.unlock();
lock1.unlock();
}
}
});
thread1.start();
Thread thread2 = new Thread(new Runnable() {
#Override public void run() {
try {
lock2.lock();
System.out.println("Thread2 acquired lock2");
try {
TimeUnit.MILLISECONDS.sleep(50);
} catch (InterruptedException ignore) {}
lock1.lock();
System.out.println("Thread2 acquired lock1");
}
finally {
lock1.unlock();
lock2.unlock();
}
}
});
thread2.start();
// Wait a little for threads to deadlock.
try {
TimeUnit.MILLISECONDS.sleep(100);
} catch (InterruptedException ignore) {}
detectDeadlock();
}
But instead of 2, N locks and I have several problems for doing this. Here my code with my attempt:
class Main {
private static int MAX_AVAILABLE = 10;
private static int IDLE = 1000;
public static void main(String[] args) throws java.lang.Exception{
int n, m; //number of resources and process, respectively
ReentrantLock[] resources; // Locks for resources
int[] available; // Number of instances per resource
Process[] processes; // Processes array
DeadlockDetector supervisor; // Deadlock detaction class
n = Integer.valueOf(args[0]);
m = Integer.valueOf(args[1]);
resources = new ReentrantLock[n];
available = new int[n];
processes = new Process[m];
supervisor = new DeadlockDetector();
// Create resources array
for(int i=0; i<n; ++i){
available[i] = (int)(Math.floor(Math.random()*MAX_AVAILABLE + 1));
resources[i] = new ReentrantLock();
System.out.println("R"+String.valueOf(i+1)+"-> instances: "+String.valueOf(available[i]));
}
// Creating processes
for(int i=0; i<m; ++i){
processes[i] = new Process(i, resources, available, n);
System.out.println("P"+String.valueOf(i+1)+"-> requested "+Arrays.toString(processes[i].requested));
processes[i].start();
}
//Run deadlock detection
try {
TimeUnit.MILLISECONDS.sleep(IDLE);
}catch (InterruptedException ignore){}
supervisor.start();
}
}
class Process extends Thread{
public int id;
public int total; // Total of resources instances needed for finished the process
public ReentrantLock[] resources;
public int[] requested; // Number of instances needed per resource type
public boolean[] needed; // Boolean indicating whether the process needs at least one instance of the resource i
public int n;
private static int MIN_TIME = 1000;
private static int MAX_TIME = 3000;
public Process(int index, ReentrantLock[] res, int[] available, int n_resources){
id = index;
n = n_resources;
resources = res;
total = 0;
requested = new int[n];
needed = new boolean[n];
for(int i=0; i<n; ++i){
requested[i] = (int)(Math.floor(Math.random()*available[i]));
needed[i] = requested[i] > 0;
total += requested[i];
}
}
#Override
public void run(){
int resourceT = 0;
int timeToSleep;
System.out.println("P"+String.valueOf(id+1)+" begin running");
try{
while(total > 0){
resourceT = (int)(Math.floor(Math.random()*n));
if(requested[resourceT] < 1){
System.out.println("P"+String.valueOf(id+1)+"-> I do not need more R"+String.valueOf(resourceT+1));
continue;
}
System.out.println("P"+String.valueOf(id+1)+"-> I'll take R"+String.valueOf(resourceT+1));
resources[resourceT].lock();
timeToSleep = (int)(Math.floor(Math.random()*(MAX_TIME - MIN_TIME)) + MIN_TIME);
try{
TimeUnit.MILLISECONDS.sleep(timeToSleep);
}catch (InterruptedException ignore){}
--total;
--requested[resourceT];
}
}finally{
for(int i=0; i<n; ++i){
if(needed[i] && resources[i].isHeldByCurrentThread())
resources[i].unlock();
}
}
System.out.println("P"+String.valueOf(id+1)+"-> Im finished");
}
}
class DeadlockDetector extends Thread{
public ThreadMXBean threadBean;
public long[] threadIds;
public DeadlockDetector(){
}
#Override
public void run(){
Boolean good;
this.threadBean = ManagementFactory.getThreadMXBean();
threadIds = threadBean.findDeadlockedThreads();
int deadlockedThreads = threadIds != null? threadIds.length : 0;
if(deadlockedThreads>1){
good = false;
System.out.println("Number of deadlocked threads: " + deadlockedThreads);
//recoverDeadlock();
//break;
}
}
public void recoverDeadlock(){
}
}
Please, could anyone help me fix this detail? Thanks!

Multi threaded java program to print even and odd numbers alternatively

I was asked to write a two-threaded Java program in an interview. In this program one thread should print even numbers and the other thread should print odd numbers alternatively.
Sample output:
Thread1: 1
Thread2: 2
Thread1: 3
Thread2: 4
... and so on
I wrote the following program. One class Task which contains two methods to print even and odd numbers respectively. From main method, I created two threads to call these two methods. The interviewer asked me to improve it further, but I could not think of any improvement. Is there any better way to write the same program?
class Task
{
boolean flag;
public Task(boolean flag)
{
this.flag = flag;
}
public void printEven()
{
for( int i = 2; i <= 10; i+=2 )
{
synchronized (this)
{
try
{
while( !flag )
wait();
System.out.println(i);
flag = false;
notify();
}
catch (InterruptedException ex)
{
ex.printStackTrace();
}
}
}
}
public void printOdd()
{
for( int i = 1; i < 10; i+=2 )
{
synchronized (this)
{
try
{
while(flag )
wait();
System.out.println(i);
flag = true;
notify();
}
catch(InterruptedException ex)
{
ex.printStackTrace();
}
}
}
}
}
public class App {
public static void main(String [] args)
{
Task t = new Task(false);
Thread t1 = new Thread( new Runnable() {
public void run()
{
t.printOdd();
}
});
Thread t2 = new Thread( new Runnable() {
public void run()
{
t.printEven();
}
});
t1.start();
t2.start();
}
}

I think this should work properly and pretty simple.
package com.simple;
import java.util.concurrent.Semaphore;
/**
* #author Evgeny Zhuravlev
*/
public class ConcurrentPing
{
public static void main(String[] args) throws InterruptedException
{
Semaphore semaphore1 = new Semaphore(0, true);
Semaphore semaphore2 = new Semaphore(0, true);
new Thread(new Task("1", 1, semaphore1, semaphore2)).start();
new Thread(new Task("2", 2, semaphore2, semaphore1)).start();
semaphore1.release();
}
private static class Task implements Runnable
{
private String name;
private long value;
private Semaphore semaphore1;
private Semaphore semaphore2;
public Task(String name, long value, Semaphore semaphore1, Semaphore semaphore2)
{
this.name = name;
this.value = value;
this.semaphore1 = semaphore1;
this.semaphore2 = semaphore2;
}
#Override
public void run()
{
while (true)
{
try
{
semaphore1.acquire();
System.out.println(name + ": " + value);
value += 2;
semaphore2.release();
}
catch (InterruptedException e)
{
throw new RuntimeException(e);
}
}
}
}
}

Well, there are many alternatives. I would probably use a SynchronousQueue instead (I don't like low-level wait/notify and try to use higher-level concurrency primitives instead). Also printOdd and printEven could be merged into single method and no additional flags are necessary:
public class App {
static class OddEven implements Runnable {
private final SynchronousQueue<Integer> queue = new SynchronousQueue<>();
public void start() throws InterruptedException {
Thread oddThread = new Thread(this);
Thread evenThread = new Thread(this);
oddThread.start();
queue.put(1);
evenThread.start();
}
#Override
public void run() {
try {
while (true) {
int i = queue.take();
System.out.println(i + " (" + Thread.currentThread() + ")");
if (i == 10)
break;
queue.put(++i);
if (i == 10)
break;
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
}
}
public static void main(String[] args) throws InterruptedException {
new OddEven().start();
}
}

Is there any better way to write the same program?
Well, the thing is, the only good way to write the program is to use a single thread. If you want a program to do X, Y, and Z in that order, then write a procedure that does X, then Y, then Z. There is no better way than that.
Here's what I would have written after discussing the appropriateness of threads with the interviewer.
import java.util.concurrent.SynchronousQueue;
import java.util.function.Consumer;
public class EvenOdd {
public static void main(String[] args) {
SynchronousQueue<Object> q1 = new SynchronousQueue<>();
SynchronousQueue<Object> q2 = new SynchronousQueue<>();
Consumer<Integer> consumer = (Integer count) -> System.out.println(count);
new Thread(new Counter(q1, q2, 2, 1, consumer)).start();
new Thread(new Counter(q2, q1, 2, 2, consumer)).start();
try {
q1.put(new Object());
} catch (InterruptedException ex) {
throw new RuntimeException(ex);
}
}
private static class Counter implements Runnable {
final SynchronousQueue<Object> qin;
final SynchronousQueue<Object> qout;
final int increment;
final Consumer<Integer> consumer;
int count;
Counter(SynchronousQueue<Object> qin, SynchronousQueue<Object> qout,
int increment, int initial_count,
Consumer<Integer> consumer) {
this.qin = qin;
this.qout = qout;
this.increment = increment;
this.count = initial_count;
this.consumer = consumer;
}
public void run() {
try {
while (true) {
Object token = qin.take();
consumer.accept(count);
qout.put(token);
count += increment;
}
} catch (InterruptedException ex) {
throw new RuntimeException(ex);
}
}
}
}

How about a shorter version like this:
public class OddEven implements Runnable {
private static volatile int n = 1;
public static void main(String [] args) {
new Thread(new OddEven()).start();
new Thread(new OddEven()).start();
}
#Override
public void run() {
synchronized (this.getClass()) {
try {
while (n < 10) {
this.getClass().notify();
this.getClass().wait();
System.out.println(Thread.currentThread().getName() + ": " + (n++));
this.getClass().notify();
}
} catch (InterruptedException ex) {
ex.printStackTrace();
}
}
}
}
There is a bit of a trick to kick-start the threads properly - thus the need to an extra notify() to start the whole thing (instead of have both processes wait, or required the main Thread to call a notify) and also to handle the possibility that a thread starts, does it's work and calls notify before the second thread has started :)

My initial answer was non-functional. Edited:
package test;
public final class App {
private static volatile int counter = 1;
private static final Object lock = new Object();
public static void main(String... args) {
for (int t = 0; t < 2; ++t) {
final int oddOrEven = t;
new Thread(new Runnable() {
#Override public void run() {
while (counter < 100) {
synchronized (lock) {
if (counter % 2 == oddOrEven) {
System.out.println(counter++);
}
}
}
}
}).start();
}
}
}

Multithreading: How to get a thread to execute more frequently than the others?

I've recently been learning about semaphores to specify the ordering of threads, but I'm curious about how to control the frequency as well. Below is a program that prints *, a digit, and then a letter to the screen. Always in that order (e.g. *1A). How can I make it so certain threads print more than once before the others? (e.g. *32A)
import java.lang.Thread;
import java.util.concurrent.*;
public class ThreadSync {
private static boolean runFlag = true;
private static Semaphore canPrintSymbol = new Semaphore(1);
private static Semaphore canPrintDigit = new Semaphore(0);
private static Semaphore canPrintLetter = new Semaphore(0);
public static void main( String[] args ) {
Runnable[] tasks = new Runnable[17];
Thread[] threads = new Thread[17];
// Create 10-digit threads
for (int d = 0; d < 10; d++) {
tasks[d] = new PrintDigit((char)('0' + d));
threads[d] = new Thread(tasks[d]);
threads[d].start();
}
// Create 6-letter threads
for (int d = 0; d < 6; d++) {
tasks[d + 10] = new PrintLetter((char)('A' + d));
threads[d + 10] = new Thread(tasks[d + 10]);
threads[d + 10].start();
}
// Create a thread to print asterisk
tasks[16] = new PrintSymbol('*');
threads[16] = new Thread(tasks[16]);
threads[16].start();
// Let the threads run for a period of time
try { Thread.sleep(500); }
catch (InterruptedException ex) { ex.printStackTrace(); }
runFlag = false;
// Interrupt the threads
for (int i = 0; i < 17; i++) threads[i].interrupt();
}
public static class PrintSymbol implements Runnable {
private char c;
public PrintSymbol(char c) {
this.c = c;
}
public void run() {
while (runFlag) {
try {
canPrintSymbol.acquire();
}
catch (InterruptedException ex) {
ex.printStackTrace();
}
System.out.printf("%c\n", c);
canPrintDigit.release();
}
}
}
public static class PrintDigit implements Runnable {
private char c;
public PrintDigit(char c) { this.c=c; }
public void run() {
while (runFlag) {
try {
canPrintDigit.acquire();
}
catch (InterruptedException ex) {
ex.printStackTrace();
}
System.out.printf("%c\n", c);
canPrintLetter.release();
}
}
}
public static class PrintLetter implements Runnable {
private char c;
public PrintLetter(char c) {
this.c = c;
}
public void run() {
while (runFlag) {
try {
canPrintLetter.acquire();
}
catch (InterruptedException ex) {
ex.printStackTrace();
}
System.out.printf("%c\n", c);
canPrintSymbol.release();
}
}
}
}

Short answer is you can't. At least not to my knowledge. There are hints you can give to the OS like yielding your thread. This means it yields it processing to the next thread. Other then that all you can really do is influence the priority. But all these are just hints to the OS. The OS ultimately determines the order in which the threads are executed. This is one of the main things to keep in mind when working with multiple threads. It is generally not a good idea to have a dependency between separate threads which makes the order of execution important.

How to make sure two threads printing even odd numbers maintain even first then odd order for this implementation?

I have created two runnable jobs: PrintEvenNumbersJob and PrintOddNumbersJob and spawned two threads to execute these jobs. This seems to work perfectly fine! But I smell something suspicious about this implementation. Can I have some comments and advice on this implementation?
The problem that I see with this implementation is that the program terminates only when thread1 gains the lock to the object lock first otherwise it print the odd first even second order and doesn't terminate unless I supply yet another statement "lock.notify" after for statement in PrintEvenNumbersJob (as in this implementation). My question here is how to make sure that thread1 is executed first.
public class PrintEvenNumbersJob implements Runnable {
private Object lock;
public PrintEvenNumbersJob(Object lock) {
this.lock = lock;
}
#Override
public void run() {
synchronized (lock) {
for (int i = 0; i <= 10; i += 2) {
lock.notify();
System.out.println(i);
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify(); // not required if thread1 gains lock first
}
}
}
public class PrintOddNumbersJob implements Runnable {
private Object lock;
public PrintOddNumbersJob(Object lock) {
this.lock = lock;
}
#Override
public void run() {
synchronized (lock) {
for (int i = 1; i < 10; i += 2) {
lock.notify();
System.out.println(i);
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notify();
}
}
}
public class EvenOddManager {
public static void main(String[] args) {
Object lock = new Object();
PrintEvenNumbersJob printEvenNumbersJob = new PrintEvenNumbersJob(lock);
PrintOddNumbersJob printOddNumbersJob = new PrintOddNumbersJob(lock);
Thread thread1 = new Thread(printEvenNumbersJob);
Thread thread2 = new Thread(printOddNumbersJob);
thread2.start();
thread1.start();
}
}

Have you try using Semaphores? It's easier because you don't need to worry about the order that wait and notify are called (if you call notify before the wait, it's "lost")
Sample code:
import java.util.concurrent.*;
public class Test {
private final Semaphore oddJobPermits = new Semaphore(0);
private final Semaphore evenJobPermits = new Semaphore(1);
private class EvenJob implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
try {
evenJobPermits.acquire();
System.out.println(i * 2);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally {
oddJobPermits.release();
}
}
}
}
private class OddJob implements Runnable {
public void run() {
for (int i = 0; i < 10; i++) {
try {
oddJobPermits.acquire();
System.out.println(i * 2 + 1);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} finally {
evenJobPermits.release();
}
}
}
}
public void run() {
new Thread(new EvenJob()).start();
new Thread(new OddJob()).start();
}
public static void main(String[] args) {
new Test().run();
}
}

I believe you will need a referee:
public class Referee {
private boolean evensTurn = true;
public void waitMyTurn(boolean even) {
synchronized(this) {
while (even != evensTurn) {
try {
wait();
} finally {
}
}
}
}
public void done() {
synchronized(this) {
evensTurn = !evensTurn;
notify();
}
}
}
public class PrintEvenNumbersJob implements Runnable {
private Referee referee;
public PrintEvenNumbersJob(Referee referee) {
this.referee = referee;
}
#Override
public void run() {
for (int i = 0; i <= 10; i += 2) {
referee.waitMyTurn(true);
System.out.println(i);
referee.done();
}
}
}
public class PrintOddNumbersJob implements Runnable {
private Referee referee;
public PrintOddNumbersJob(Referee referee) {
this.referee = referee;
}
#Override
public void run() {
for (int i = 0; i <= 10; i += 2) {
referee.waitMyTurn(false);
System.out.println(i);
referee.done();
}
}
}

I tried and tested this code. It works using Semaphore
public class TestSemaphore
{
public static void main(String[] args)
throws Exception
{
AtomicInteger count = new AtomicInteger();
Semaphore s = new Semaphore(1, true);
Semaphore t = new Semaphore(1, true);
OddNumberThread oThread = new OddNumberThread(count, s, t);
EvenNumberThread eThread = new EvenNumberThread(count, s, t);
eThread.start();
oThread.start();
}
static class EvenNumberThread
extends Thread
{
private AtomicInteger count;
private Semaphore s, t;
public EvenNumberThread(AtomicInteger pCount, Semaphore pS, Semaphore pT)
{
super("Even");
count = pCount;
s = pS;
t = pT;
}
#Override
public void run()
{
// Make this thread wait until even thread starts, Order will be incorrect if removed these lines.
s.acquireUninterruptibly();
while (count.intValue() <= 10)
{
try
{
// Double checking to make it work
s.acquireUninterruptibly();
System.out.println(getName() + " " + count.getAndIncrement());
}
finally
{
t.release();
}
}
}
}
static class OddNumberThread
extends Thread
{
private AtomicInteger count;
private Semaphore s, t;
public OddNumberThread(AtomicInteger pCount, Semaphore pS, Semaphore pT)
{
super("Odd");
count = pCount;
s = pS;
t = pT;
}
#Override
public void run()
{
// Start this thread first and start printing, Order will be incorrect if removed these lines.
t.acquireUninterruptibly();
s.release();
while (count.intValue() <= 10)
{
try
{
t.acquireUninterruptibly();
System.out.println(getName() + " " + count.getAndIncrement());
}
finally
{
s.release();
}
}
}
}
}