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For below program, when threads finish their work, why doesn't main thread stops?

This program works fine by printing alternate numbers via different threads but when all the numbers from 0-9 are printed Why does this program not stop? I have to manually stop my application.
public class EvenOddPrinter implements Runnable{
private AtomicInteger num = new AtomicInteger(0);
private Object lock = new Object();
#Override
public void run() {
synchronized (lock){
while (num.get()<10){
System.out.println(num.getAndAdd(1) + " - "+Thread.currentThread().getName());
lock.notify();
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
public class Executor {
public static void main(String[] args) throws InterruptedException {
EvenOddPrinter eop = new EvenOddPrinter();
Thread t1 = new Thread(eop);
Thread t2 = new Thread(eop);
t1.start();
t2.start();
}
}

that's because in the last Thread getting stuck at wait. notifyAll will notify all waiting thread if any there and release lock.
while (num.get()<10){
// existing implementation
}
lock.notifyAll();

The second thread t2 keeps waiting on the lock in the end, and t1 doesnt do the notify() anymore because the while condition becomes false. You must put a lock.notify(); statement outside of the while loop.

As soon as the number reached 8 the first thread calls notify() and goes to wait(). Second thread then makes the number 9 and calls notify() and goes to wait(). First thread is then not able to go inside the loop as specified in the condition, therefore, it exits the synchronized and block and finishes but second thread is still waiting. There has to be a mechanism to notifyAll() as soon as one of the threads exits the synchronized block which is exactly what I did.
#Override
public void run() {
synchronized (lock){
while (num.get()<10){
System.out.println(num.getAndAdd(1) + " - "+Thread.currentThread().getName());
lock.notify();
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
lock.notifyAll();
}
Also, lock is pointless when I am using AtomicInteger (or the other way round).

Why does interrupt() interrupt two threads in quick succession?

I want to start a thread, interrupt it and start a new thread. The Problem is, that this doesn't really work. The first thread starts and gets interrupted, but the following thread gets interrupted before it even can start. So interrupt() interrupts the old and the new thread. The output looks like this:
run()-method starts
Thread Counter:0
Thread Counter:1
Thread Counter:2
Thread Counter:3
Thread Counter:4
Thread Counter:5
Thread Counter:6
8 seconds are over
Thread is not null
Thread will be interrupted now
catch
run()-method starts
catch
8 seconds are over
Thread is not null
Thread will be interrupted now
....
You can see that the thread starts the first time. Then the thread gets interrupted and 'catch' is called. So far, so good. After this, the next thread is going to start, but this time the thread gets interrupted immediately and 'catch' is called right after 'run()-method starts'.
So, I can't figure out why this is happening. I don't want two threads being interrupted in quick succession.
Here is my code:
public class MyRunnable {
static Thread myThread;
static boolean stop;
static Runnable myRunny = new Runnable() {
#Override
public void run() {
System.out.println("run()-method starts");
try {
int j = 0;
while (!stop) {
Thread.sleep(1000);
System.out.println("Thread Counter:"+j);
j++;
}
} catch (InterruptedException e) {
System.out.println("catch");
myThread.interrupt();
}
};
};
public static void main(String[] args){
myThread = null;
while(true) {
stop = false;
if(myThread != null) {
System.out.println("Thread is not null ");
System.out.println("Thread will be interrupted now");
myThread.interrupt();
}
myThread = new Thread(myRunny);
myThread.start();
try {
Thread.sleep(8000);
System.out.print("8 seconds are over "+ "\n");
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}

catch (InterruptedException e) {
System.out.println("catch");
myThread.interrupt();
}
myThread field is replaced with new reference before calling myThread.interrupt(), so you interrupt the new thread!

So interrupt() interrupts the old and the new thread.
Your diagnosis is incorrect. A call to Thread.interrupt will interrupt one thread once.
What your example is doing is interrupting one thread, and that thread is catching InterruptedException and interrupting a second thread in the exception handler. Two calls to interrupt are being made in quick succession on different threads.
I don't want two threads being interrupted in quick succession.
Well change
} catch (InterruptedException e) {
System.out.println("catch");
myThread.interrupt();
}
to
} catch (InterruptedException e) {
System.out.println("catch");
}

How can I start a thread from another and restart a thread after execution?

I have 2 threads, the "main" thread which starts a secondary thread to run a little process.
The "main" thread must wait for the secondary thread for a few of seconds to complete the process, after that time, the "main" thread must start again no matter what happened with the process of the secondary thread.
If the secondary process ended earlier, the "main" thread must start to work again.
How can I start a thread from another, wait for the end of execution, and restart the thread after?
I have a code here, but the ExampleRun class, must wait, for example, 10 sec and start again, no matter what happend with MyProcess
public class ExampleRun {
public static void main(String[] args) {
MyProcess t = new MyProcess();
t.start();
synchronized (t) {
try {
t.wait();
} catch (InterruptedException e) {
System.out.println("Error");
}
}
}
}
public class MyProcess extends Thread {
public void run() {
System.out.println("start");
synchronized (this) {
for (int i = 0; i < 5; i++) {
try {
System.out.println("I sleep");
sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
flag = true;
System.out.println("Wake up");
notify();
}
}
}

The simplest way to achieve what you want is to use Thread.join(timeout).
Also, do not use synchronized, wait, or notify on Thread objects. This will interfere with the Thread.join implementation. See the documentation for details.
Here's what your main program would look like:
public static void main(String[] args) {
MyProcess t = new MyProcess();
t.start();
try {
t.join(10000L);
} catch (InterruptedException ie) {
System.out.println("interrupted");
}
System.out.println("Main thread resumes");
}
Note that when the main thread resumes after the join() call, it can't tell whether the child thread completed or whether the call timed out. To test this, call t.isAlive().
Your child thread of course could do anything, but it's important for it not to use synchronized, wait, or notify on itself. For example, here's a rewrite that avoids using these calls:
class MyProcess extends Thread {
public void run() {
System.out.println("MyProcess starts");
for (int i = 0; i < 5; i++) {
try {
System.out.println("MyProcess sleeps");
sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("MyProcess finishes");
}
}

You can do this with a simple lock method:
public static void main (String[] args)
{
// create new lock object
Object lock = new Object();
// create and start thread
Thread t = new Thread(() ->
{
// try to sleep 1 sec
try { Thread.sleep(1000); }
catch (InterruptedException e) { /* do something */ }
// notify main thread
synchronized (lock) { lock.notifyAll(); }
};
t.start();
// wait for second thread to finish
synchronized (lock)
{
while (t.isAlive())
lock.wait();
}
// second thread finished
System.out.println("second thread finished :)");
}

You could call Thread.join() on the Thread you want to wait for, per the Javadoc,
Waits for this thread to die.
Alternatively, you could use a Future and simply call get(), from its' Javadoc,
Waits if necessary for the computation to complete, and then retrieves its result.

Getting a thread to pause - Thread.wait()/Thread.notify()

I'm trying to understand how threads work, and I wrote a simple example where I want to create and start a new thread, the thread, display the numbers from 1 to 1000 in the main thread, resume the secondary thread, and display the numbers from 1 to 1000 in the secondary thread. When I leave out the Thread.wait()/Thread.notify() it behaves as expected, both threads display a few numbers at a time. When I add those functions in, for some reason the main thread's numbers are printed second instead of first. What am I doing wrong?
public class Main {
public class ExampleThread extends Thread {
public ExampleThread() {
System.out.println("ExampleThread's name is: " + this.getName());
}
#Override
public void run() {
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
}
}
public static void main(String[] args) {
new Main().go();
}
public void go() {
Thread t = new ExampleThread();
t.start();
synchronized(t) {
try {
t.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
synchronized(t) {
t.notify();
}
}
}

You misunderstand how wait/notify works. wait does not block the thread on which it is called; it blocks the current thread until notify is called on the same object (so if you have threads A and B and, while in thread A, called B.wait(), this will stop thread A and not thread B - for as long as B.notify() is not called).
So, in your specific example, if you want main thread to execute first, you need to put wait() inside the secondary thread. Like this:
public class Main {
public class ExampleThread extends Thread {
public ExampleThread() {
System.out.println("ExampleThread's name is: " + this.getName());
}
#Override
public void run() {
synchronized (this) {
try {
wait();
} catch (InterruptedException e) {
}
}
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
}
}
public static void main(String[] args) {
new Main().go();
}
public void go() {
Thread t = new ExampleThread();
t.start();
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
synchronized(t) {
t.notify();
}
}
}
However, even this code may not work like you want. In a scenario where the main thread gets to the notify() part before the secondary thread had a chance to get to the wait() part (unlikely in your case, but still possible - you can observe it if you put Thread.sleep at the beginning of the secondary thread), the secondary thread will never be waken up. Therefore, the safest method would be something similar to this:
public class Main {
public class ExampleThread extends Thread {
public ExampleThread() {
System.out.println("ExampleThread's name is: " + this.getName());
}
#Override
public void run() {
synchronized (this) {
try {
notify();
wait();
} catch (InterruptedException e) {
}
}
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
}
}
public static void main(String[] args) {
new Main().go();
}
public void go() {
Thread t = new ExampleThread();
synchronized (t) {
t.start();
try {
t.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
for(int i = 1; i < 1000; i++) {
System.out.println(Thread.currentThread().getName());
System.out.println(i);
}
synchronized(t) {
t.notify();
}
}
}
In this example the output is completely deterministic. Here's what happens:
Main thread creates a new t object.
Main thread gets a lock on the t monitor.
Main thread starts the t thread.
(these can happen in any order)
Secondary thread starts, but since main thread still owns the t monitor, the secondary thread cannot proceed and must wait (because its first statement is synchronized (this), not because it happens to be the t object - all the locks, notifies and waits could as well be done on an object completely unrelated to any of the 2 threads with the same result.
Primary thread continues, gets to the t.wait() part and suspends its execution, releasing the t monitor that it synchronized on.
Secondary thread gains ownership of t monitor.
Secondary thread calls t.notify(), waking the main thread. The main thread cannot continue just yet though, since the secondary thread still holds ownership of the t monitor.
Secondary thread calls t.wait(), suspends its execution and releases the t monitor.
Primary thread can finally continue, since the t monitor is now available.
Primary thread gains ownership of the t monitor but releases it right away.
Primary thread does its number counting thing.
Primary thread again gains ownership of the t monitor.
Primary thread calls t.notify(), waking the secondary thread. The secondary thread cannot continue just yet, because the primary thread still holds the t monitor.
Primary thread releases the t monitor and terminates.
Secondary thread gains ownership of the t monitor, but releases it right away.
Secondary thread does its number counting thing and then terminates.
The entire application terminates.
As you can see, even in such a deceptively simple scenario there is a lot going on.

You are lucky that your program terminates at all.
When you call t.wait() your main threads stops and waits indefinitely on a notification.
It never gets it, but I believe is awaken by spurious wakeup when the secondary thread finishes. (Read here on what a spurious wakeup is).

ExampleThread doesn't wait() or notify(), and isn't synchronized on anything. So it will run whenever it can without any coordination with other threads.
The main thread is waiting for a notification which never comes (this notification should be sent by another thread). My guess is that when the ExampleThread dies, the main thread is woken "spuriously," and completes.
The thread that should wait for another to complete must perform the call to wait() inside a loop that checks for a condition:
class ExampleThread extends Thread {
private boolean ready = false;
synchronized void ready() {
ready = true;
notifyAll();
}
#Override
public void run() {
/* Wait to for readiness to be signaled. */
synchronized (this) {
while (!ready)
try {
wait();
} catch(InterruptedException ex) {
ex.printStackTrace();
return; /* Interruption means abort. */
}
}
/* Now do your work. */
...
Then in your main thread:
ExampleThread t = new ExampleThread();
t.start();
/* Do your work. */
...
/* Then signal the other thread. */
t.ready();

Simultaneous access to a Java synchronized block using threads?

How can two threads access a synchronized block simultaneously? That is, how can I make one thread give the chance for the other thread to execute a synchronized block, even before this thread finishes the execution of the same synchronized block?

See wait(), notify(), and notifyAll().
Edit: The edit to your question is incorrect. The sleep() method does not release the monitor.
For example:
private static final Object lock = new Object();
public static void main(String[] args) {
ExecutorService executorService = Executors.newFixedThreadPool(2);
executorService.execute(new One());
executorService.execute(new Two());
}
static class One implements Runnable {
#Override
public void run() {
synchronized (lock) {
System.out.println("(One) I own the lock");
System.out.println("(One) Giving up the lock and waiting");
try {
lock.wait();
} catch (InterruptedException e) {
System.err.println("(One) I shouldn't have been interrupted");
}
System.out.println("(One) I have the lock back now");
}
}
}
static class Two implements Runnable {
#Override
public void run() {
try {
Thread.sleep(100);
} catch (InterruptedException e) {
System.err.println("(Two) I shouldn't have been interrupted");
}
synchronized (lock) {
System.out.println("(Two) Now I own the lock (Two)");
System.out.println("(Two) Giving up the lock using notify()");
lock.notify();
}
}
}

It sounds like you might want to consider using more than one synchronized block, particularly if there's a blocking operation that one thread is getting caught on and thus blocking another thread that wants to execute something else in the block.

A synchronized block is a block of code which can (by definition) only be accessed by one thread at a time.
Saying that you want another thread to enter this block while another thread also currently processes it, does make the synchronized block scheme useless.
You probably want to split the synchronized block into many other ones.

The only way I can see if one thread calls wait() on monitor object. Then it will release monitor and wait for notification while other thread can execute synchronized block. Then other thread will have to call notify()/notifyAll() so first thread gets monitor back and continue.

A thread can release its monitor using lock.wait(). Another thread can then pick up the monitor and enter the synchronized block.
Example:
public class MultipleThreadsInSynchronizedBlock {
public static void main(String... args) {
final Object lock = new Object();
Runnable runnable = new Runnable() {
public void run() {
synchronized (lock) {
System.out.println("Before wait");
try {
lock.wait();
} catch (InterruptedException e) {
}
System.out.println("After wait");
}
}
};
new Thread(runnable).start();
new Thread(runnable).start();
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
}
synchronized (lock) {
lock.notifyAll();
}
}
}
This prints:
Before wait
Before wait
After wait
After wait
However it's not a "hack" to allow a mutually exclusive block to be run non-atomically. If you're going to use very low-level synchronization primitives like this you need to know what you're doing.