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Main thread to wait two parallel threads children java

first what i am trying to do:
During the main thread execution i want to pause the main thread and start two parallel threads. As soon as both this parallel threads terminate, i'd like to start again with the main thread.
What i tried:
...
...
main thread is executing
...
...
CyclicBarrier barrier = new CyclicBarrier(2);
Thread child1 = new Thread(new ThreadBuilderTask(barrier,0));
Thread child2 = new Thread(new ThreadBuilderTask(barrier,1));
child1.start();
child2.start();
/* Now i'm expecting that child1 and child2 are running in parallel calling their fooFunction */
child1.join();
child2.join();
/*Now i'm expecting that main thread will wait for child1and also for child2 (that are running in parallel).*/
... main thread starts again after both child1 and child2 finished (reached the await of the barrier)
... (break point set here, never reached)
...
Thread builder custom class
public class ThreadBuilderTask implements Runnable{
private CyclicBarrier barrier;
private int index;
...setters and getters..
#Override
public void run() {
fooFunction(this.getIndex());
try {
this.getBarrier().await();
} catch (InterruptedException | BrokenBarrierException e) {
return;
}
}
public ThreadBuilderTask(CyclicBarrier barrier,int index){
this.barrier = barrier;
this.index = index;
}
public fooFunction(int index){
//Something taking some seconds to execute
}
It's not clear what is happening here but it is definetely not working. As soon as i call join everything stops and the main thread never restart. (I put a breakpoint after the joins to see when the main thread restarts).
Maybe there is a bit of confusion with these concepts and also i'm not sure if i need to use both the barrier and the joins or simply one of those techniques.
Thanks
Davide

As mentioned in the comments I'd also suggest to use CompletableFuture. A very basic example of your described requirements could look like this:
final Runnable runnable1 = ...;
final Runnable runnable2 = ...;
CompletableFuture<Void> future1 = CompletableFuture.runAsync(runnable1);
CompletableFuture<Void> future2 = CompletableFuture.runAsync(runnable2);
CompletableFuture.allOf(future1, future2).get(); // waits for both runnables to finish
You might want to add more/some exception handling to this example. But it should give an idea how this might work.

You may consider to use Java CompletableFuture to achieve the objective.
Using its functions like supplyAsync or runAsync you may start child threads and join their respective result in the end. Or you can simply let the main thread wait until the subsequent threads completes.
Recently I managed to implement a sample scatter-gather function using the same class.
Check Java Doc for more offerings and to find best available function: https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/CompletableFuture.html

can multiple threads with different objects of same runnable class overlap

Does creating a new object everytime of runnable class and passing it to Thread create multithreading problems.
For example:
class exmpl implements Runnable {
int a = 0;
exmpl(int x) {
a = x;
}
}
public class toRun {
public static void main(String[] args) {
Thread t = new Thread(new exmpl(5));
Thread t1 = new Thread(new exmpl(6));
t.start();
t1.start();
}
}
Will the two thread overlap ? ie. will the value of a for t be 5 and for t1 be 6 always?

The value of a in the exmpl instance that you construct for the t thread is initially set to 5, and the value of a in the other exmpl instance is initially set to 6, but a is not a final variable, and you have not shown us the run() method. Whether or not the two a fields will ever change depends on what the run() method does with them.

Both t and t1 are totally different objects, the value of a in each instance will be kept unless you modified it.

Yes, of course! The whole point of threads is that is so that multiple threads (even if they're different objects of class thread will run at once.
Think about it:
If it were impossible to run multiple threads at once, then wouldn't you just use individual methods? That way, the code would run in a single order, going from 1 to 2 to 3.
The thing is, java has something called a thread. The whole point of this thread is so that instead of going from 1 to 2 to 3, the compiler can execute 1 2 and 3 all at the same time.

Creating and naming multiple, simultaneous threads with a for loop

Is there a way to create multiple threads that run simultaneously with a for loop? Consider this example:
for(int i = 1; i<=36; i++) {
if(new Random().nextInt(2)==0){
ActionThread nr = new ActionThread();
}
}
I don't want the threads to be killed after completion of the if statement. The end of each thread is randomly determined in the ActionThread class itself. Also, how do I name the threads automatically? For example, instead of nr, the first thread should be named nr1, the second nr2, the third nr3, and so on.

I'm assuming that ActionThread is some custom class that you have created that extends Thread.
I don't want the threads to be killed after completion of the if statement.
They won't be. However, it doesn't look like you have started them yet. Read the javadocs for Thread. Read the material at the top, then look at the start() and run() methods.
If you don't start a thread ... nothing happens.
Also, if you want some other part of your application to be able to "do things" to the threads once they have been created, you should replace the nr local variable with a data structure that the the rest of the application can get at; e.g. a list or an array.
(It is also possible to find extant threads via the ThreadGroup tree, but it is complicated.)
Also, how do I name the threads automatically?
Call Thread.setName(), or pass the thread name to the (relevant) Thread constructor. For example:
nr.setName("thr" + i);
Or you could even make your ActionThread set its own name in the constructor.
I should also point out that is is generally considered to be a bad idea to create subclasses of Thread. It is better to put your thread logic into a custom Runnable class, then create and pass a Runnable instance as a Thread construct argument. Like this:
public class MyRunnable implements Runnable {
#Override
public void run() {
// thread logic goes here
}
}
Thread th = new Thread(new MyRunnable());
th.start();
If you want to pass parameters to the thread logic, add a constructor to your runnable class with some arguments, and provide them when you instantiate the runnable.
Why do it this way? Because it allows you to easily change your code to use a thread loop or executor or some such.

public static void main(String[] a) {
List<ActionThread> threads = new ArrayList<>();
for (int i = 1; i <= 36; i++) {
if (new Random().nextInt(2) == 0) { // no idea why you have put this
// but seems unecessary
ActionThread thread = new ActionThread();
threads.add(thread);
thread.start();
}
}
}
class ActionThread extends Thread {
#Override
public void run() {
// Write what to do in Thread here
}
}
Once the list of ActionThread is there you have handle to all the Threads that you have created. using threads.get(index). From question its appears that by name you meant handle to Thread instance

For automatic naming, may be use static field (counter) in ActionThread and increment him in the constructor, before generate thread name.
class ActionThread extend Thread {
private static int id = 0;
ActionThread() {
setName(String.format("n%d", ++id);
}
}

Java synchronized keyword issue

I am trying to understand the keyword synchronized from the following example
Java Main Method -->
public int methodA(){
Hello h = new Hello();
h.callSomeSynchronizedMethod();
sysout("Main");
return 0;
}
In the Hello Class-->
public synchronized void callSomeSynchronizedMethod(){
Hi h = new Hi();
h.someMethod();
sysout("Hello");
}
In the Hi class
public void someMethod(){
sysout("Hi");
}
So what would be the list of outputs that i will get;
1.) Is it in the order of Hi, Hello and Main ?
2.) What i understand about the synchronized keyword is that it will only execute 1 method and then execute the other, without multi-threading. Is this correct ?

To really understand what synchronized does you need to run the program twice, once synchronized and once not. Also your program should use multiple threads. So here is an example of such a test.
public class Synchro {
public static void main(String args[]){
new Synchro();
}
public Synchro(){
final Moo moo = new Moo();
Thread t = new Thread(new Runnable(){
public void run(){
moo.aMethod("Second");
}
});
t.start();//calling the method in a thread
moo.aMethod("First");//calling the same method from the same object in the main thread
}
class Moo{
public Moo(){
}
public void aMethod(String name){
//this loop just prints slowly so you can see the execution
for(int i = 1; i <= 100; i++){
System.out.println(String.format("%s : %d", name, i));
try{
Thread.sleep(50);
}catch(InterruptedException e){}
}
}
}
}
Now, if you run the above code, noticing that the method is not synchronized, you will see the printout from the two executions of the method interleaved. That is you will see First 1 then Second 1 then First 2 etc.
Now, add the synchronized keyword to the method making it:
public synchronized void aMethod(String name){ ....
and run the code again. This time, one execution of the method completes before the other begins.
The synchronized keyword is only necessary when multiple threads are accessing the very same object.

You would get "Hi", then "Hello", then "Main", yes. The synchronized modifier has nothing to do with the order the methods are called in; and, other than adding a bit of overhead, it does nothing at all when running the code in a single thread. You could run this same test without synchronized and get the same result.
Now, if you ran a similar test where multiple threads were calling these methods, your results would be less determinate.
Synchronized is meant to allow for the more safe execution of code and management of resources in a multi-threaded environment.
http://docs.oracle.com/javase/tutorial/essential/concurrency/syncmeth.html
Hope this helps.

all these methods will be executed in one thread so the answer for the first question is "yes".
synchronized keyword emans that the method can be executed in only one thread at every moment of time. So if you call it from another thread - it will wait till the execution is finished in the first thread.

In Java there is no automatic multithreading: you must explicitly start a thread and pass it a run method that it will execute. Only in that case will the synchronized keyword start to matter, but its meaning is not quite as you understand it: the methods will execute in whatever thread calls them, but while one is executing, another thread will block before it is able to execute a method guarded by the same lock.

Using ThreadLocal in tandem with Volatile gives unpredictable results

I was reading through Java Memory model and was playing with volatile. I wanted to check how Volatile will work in tandem with ThreadLocal. As per definition ThreadLocal has its own, independently initialized copy of the variable whereas when you use volatile keyword then JVM guarantees that all writes and subsequent reads are done directly from the memory. Based on the high level definitions i knew what i was trying to do will give unpredictable results. But just out of curiosity wanted to ask if someone can explain in more details as if what is going on in the background. Here is my code for your reference...
public class MyMainClass {
public static void main(String[] args) throws InterruptedException {
ThreadLocal<MyClass> local = new ThreadLocal<>();
local.set(new MyClass());
for(int i=0;i<5; i++){
Thread thread = new Thread(local.get());
thread.start();
}
}
}
public class MyClass implements Runnable {
private volatile boolean flag = false;
public void printNameTillFlagIsSet(){
if(!flag)
System.out.println("Flag is on for : " + Thread.currentThread().getName());
else
System.out.println("Flag is off for : " + Thread.currentThread().getName());
}
#Override
public void run() {
printNameTillFlagIsSet();
this.flag = true;
}
}

In your code you create a ThreadLocal reference as a local variable of your main method. You then store an instance of MyClass in it and then give that same reference of MyClass to 5 threads created in the main method.
The resulting output of the program is unpredictable since the threads are not synchronized against each other. At least one thread will see the flag as false the other four could see the flag as either true or false depending on how the thread execution is scheduled by the OS. It is possible that all 5 threads could see the flag as false, or 1 could see it false and 4 see it true or anything in between.
The use of a ThreadLocal has no impact on this run at all based on the way you are using it.

As most have pointed out you have deeply misunderstood ThreadLocal. This is how I would write it to be more accurate.
public class MyMainClass {
private static final ThreadLocal<MyClass> local = new ThreadLocal<>(){
public MyClass initialValue(){
return new MyClass();
}
}
public static void main(String[] args) throws InterruptedException {
local.set(new MyClass());
for(int i=0;i<5; i++){
Thread thread = new Thread(new Runnable(){
public void run(){
local.get().printNameTillFlagIsSet();
local.get().run();
local.get().printNameTillFlagIsSet();
}
});
thread.start();
}
}
}
So here five different instances of MyClass are created. Each thread will have their own accessible copy of each MyClass. That is Thread created at i = 0 will always have a different instance of MyClass then i = 1,2,3,4 despite how many local.get() are done.
The inner workings are a bit complicated but it can be done similar to
ConcurrentMap<Long,Thread> threadLocalMap =...;
public MyClass get(){
long id = Thread.currentThread().getId();
MyClass value = threadLocalMap.get(id);
if(value == null){
value = initialValue();
threadLocalMap.put(id,value);
}
return value;
}
To further answer your question about the volatile field. It is in essence useless here. Since the field itself is 'thread-local' there will be no ordering/memory issues that can occur.

Just don't divinize the JVM. ThreadLocal is a regular class. Inside it uses a map from current thread ID into an object instance. So that the same ThreadLocal variable could have its own value for each thread. That's all. Your variable exists only in the main thread, so it doesn't make any sence.
The volatile is something about java code optimization, It just stops all possible optimizations which allow avoid redundant memory reads/writes and execution sequence re-orderings. It is important for expecting some particular behaviour in multi-threaded environment.

You have two big problems:
1) As many pointed out, you are not using ThreadLocal properly so you don't actually have any "thread local" variables.
2) Your code is equivalent to:
MyClass someInstance = new Class();
for (...)
... new Thread(someInstance);
so you should expect to see 1 on and 4 off. However your code is badly synchronized, so you get random results. The problem is that although you declare flag as volatile, this is not enough for good synchronization since you do the check on flag in printNameTillFlagSet and then change the flag value just after that method call in run. There is a gap here where many threads can see the flag as true. You should check the flag value and change it within a synchronized block.

You main thread where you have a ThreadLocal object is being passed to all the threads. So the same instance is being passed.
So its as good as
new Thread(new MyClass());
What you could try is have an object being called by different threads with a thread local variable. This will be a proper test for ThreadLocal where each thread will get its own instance of the variable.