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How does thread synchronization work in Kotlin?

I have been experimenting with Kotlin synchronization and I do not understand from the docs on how the locking mechanism works on thread synchronization over common resources and thus attempted to write this piece of code which further complicates my understanding.
fun main() {
val myList = mutableListOf(1)
thread {
myList.forEach {
while (true) {
println("T1 : $it")
}
}
}
thread {
synchronized(myList) {
while (true) {
myList[0] = 9999
println("**********\n**********\n**********\n")
}
}
}
}
myList is the common resource in question.
The first thread is a simple read operation that I intend to keep the resource utilized in read mode. The second is another thread which requests a lock in order to modify the common resource.
Though the first thread does not contain any synchronization, I would expect it to internally handle this so that a while a function like map or forEach is in progress over a resource, another thread should not be able to lock it otherwise the elements being iterated over may change while the map/forEach is in progress (even though that operation may be paused for a bit while another thread has a lock over it).
The output I see instead shows that both the threads are executing in parallel. Both of them are printing the first element in the list and the stars respectively. But in the second thread, even though the stars are being printed, myList[0] is never set to 9999 because the first thread continues to print 1.

Threading and synchronisation are JVM features, not specific to Kotlin. If you can follow Java, there are many resources out there which can explain them fully. But the short answer is: they're quite low-level, and tricky to get right, so please exercise due caution. And if a higher-level construction (work queues/executors, map/reduce, actors...) or immutable objects can do what you need, life will be easier if you use that instead!
But here're the basics. First, in the JVM, every object has a lock, which can be used to control access to something. (That something is usually the object the lock belongs to, but need not be...) The lock can be taken by the code in a particular thread; while it's holding that lock, any other thread which tries to take the lock will block until the first thread releases it.
And that's pretty much all there is! The synchronised keyword (actually a function) is used to claim a lock; either that belonging to a given object or (if none's given) 'this' object.
Note that holding a lock prevents other threads holding the lock; it doesn't prevent anything else. So I'm afraid your expectation is wrong. That's why you're seeing the threads happily running simultaneously.
Ideally, every class would be written with some consideration for how it interacts with multithreading; it could document itself as 'immutable' (no mutable state to worry about), 'thread-safe' (safe to call from multiple threads simultaneously), 'conditionally thread-safe' (safe to call from multiple threads if certain patterns are adhered to), 'thread-compatible' (taking no special precautions but callers can do their own synchronisation to make it safe), or 'thread-hostile' (impossible to use from multiple threads). But in practice, most don't.
In fact, most turn out to be thread-compatible; and that applies to much of the Java and Kotlin collection classes. So you can do your own synchronisation (as per your synchronized block); but you have to take care to synchronise every possible access to the list -- otherwise, a race condition could leave your list in an inconsistent state.
(And that can mean more than just a dodgy value somewhere. I had a server app with a thread that got stuck in a busy-loop -- chewing up 100% of a CPU but never continuing with the rest of the code -- because I had one thread update a HashMap while another thread was reading it, and I'd missed the synchronisation on one of those. Most embarrassing.)
So, as I said, if you can use a higher-level construction instead, your life will be easier!

Second thread is not changing the value of the first list element, as == means compare, not assign. You need to use = tio change the value e.g. myList[0] = 9999. However in your code it's not guaranteed that the change from the second thread will become visible in the first thread as thread one is not synchronising on myList.
If you are targeting JVM you should read about JVM memory model e.g. what is #Volatile. You current approach does not guarantee that first thread will ever see changes from the second one. You can simplify your code to below broken example:
var counter = 1
fun main() {
thread {
while (counter++ < 1000) {
println("T1: $counter")
}
}
thread {
while (counter++ < 1000) {
println("T2: $counter")
}
}
}
Which can print strange results like:
T2: 999
T1: 983
T2: 1000
This can be fixed in few ways e.g. by using synchronisations.

CPU usage is 100% during Thread.onSpinWait()

I'm writing a backtesting raw data collector for my crypto trading bot and I've run into a weird optimization issue.
I constantly have 30 runnables in an Executors.newCachedThreadPool() running get requests from an API. Since the API has a request limit of 1200 per minute I have this bit of code in my runnable:
while (minuteRequests.get() >= 1170) {
Thread.onSpinWait();
}
Yes, minuteRequests is an AtomicInteger, so I'm not running into any issues there.
Everything works, the issue is that even though I'm using the recommended busy-waiting onSpinWait method, I shoot from 24% CPU usage or so to 100% when the waiting is initiated. For reference I'm running this on a 3900X (24 thread).
Any recommendations on how to better handle this situation?

My recommendation would be to not do busy waiting at all.
The javadocs for Thread.onSpinWait say this:
Indicates that the caller is momentarily unable to progress, until the occurrence of one or more actions on the part of other activities. By invoking this method within each iteration of a spin-wait loop construct, the calling thread indicates to the runtime that it is busy-waiting. The runtime may take action to improve the performance of invoking spin-wait loop constructions.
Note the highlighted section uses the word may rather than will. That means that it also may not do anything. Also "improve the performance" does not mean that your code will be objectively efficient.
The javadoc also implies that the improvements may be hardware dependent.
In short, this is the right way to use onSpinwait ... but you are expecting too much of it. It won't make your busy-wait code efficient.
So what would I recommend you actually do?
I would recommend that you replace the AtomicInteger with a Semaphore (javadoc). This particular loop would be replaced by the following:
semaphore.acquire();
This blocks1 until 1 "permit" is available and acquires it. Refer to the class javadocs for an explanation of how semaphores work.
Note: since you haven't show us the complete implementation of your rate limiting, it is not clear how your current approach actually works. Therefore, I can't tell you exactly how to replace AtomicInteger with Semaphore throughout.
1 - The blocked thread is "parked" until some other thread releases a permit. While it is parked, the thread does not run and is not associated with a CPU core. The core is either left idle (typically in a low power state) or it is assigned to some other thread. This is typically handled by the operating system's thread scheduler. When another thread releases a permit, the Semaphore.release method will tell the OS to unpark one of the threads that is blocked in acquire.

difference between wait() and yield()

So far what I have understood about wait() and yield () methods is that yield() is called when the thread is not carrying out any task and lets the CPU execute some other thread. wait() is used when some thread is put on hold and usually used in the concept of synchronization. However, I fail to understand the difference in their functionality and i'm not sure if what I have understood is right or wrong. Can someone please explain the difference between them(apart from the package they are present in).

aren't they both doing the same task - waiting so that other threads can execute?
Not even close, because yield() does not wait for anything.
Every thread can be in one of a number of different states: Running means that the thread is actually running on a CPU, Runnable means that nothing is preventing the thread from running except, maybe the availability of a CPU for it to run on. All of the other states can be lumped into a category called blocked. A blocked thread is a thread that is waiting for something to happen before it can become runnable.
The operating system preempts running threads on a regular basis: Every so often (between 10 times per second and 100 times per second on most operating systems) the OS tags each running thread and says, "your turn is up, go to the back of the run queue' (i.e., change state from running to runnable). Then it lets whatever thread is at the head of the run queue use that CPU (i.e., become running again).
When your program calls Thread.yield(), it's saying to the operating system, "I still have work to do, but it might not be as important as the work that some other thread is doing. Please send me to the back of the run queue right now." If there is an available CPU for the thread to run on though, then it effectively will just keep running (i.e., the yield() call will immediately return).
When your program calls foobar.wait() on the other hand, it's saying to the operating system, "Block me until some other thread calls foobar.notify().
Yielding was first implemented on non-preemptive operating systems and, in non-preemptive threading libraries. On a computer with only one CPU, the only way that more than one thread ever got to run was when the threads explicitly yielded to one another.
Yielding also was useful for busy waiting. That's where a thread waits for something to happen by sitting in a tight loop, testing the same condition over and over again. If the condition depended on some other thread to do some work, the waiting thread would yield() each time around the loop in order to let the other thread do its work.
Now that we have preemption and multiprocessor systems and libraries that provide us with higher-level synchronization objects, there is basically no reason why an application programs would need to call yield() anymore.

wait is for waiting on a condition. This might not jump into the eye when looking at the method as it is entirely up to you to define what kind of condition it is. But the API tries to force you to use it correctly by requiring that you own the monitor of the object on which you are waiting, which is necessary for a correct condition check in a multi-threaded environment.
So a correct use of wait looks like:
synchronized(object) {
while( ! /* your defined condition */)
object.wait();
/* execute other critical actions if needed */
}
And it must be paired with another thread executing code like:
synchronized(object) {
/* make your defined condition true */)
object.notify();
}
In contrast Thread.yield() is just a hint that your thread might release the CPU at this point of time. It’s not specified whether it actually does anything and, regardless of whether the CPU has been released or not, it has no impact on the semantics in respect to the memory model. In other words, it does not create any relationship to other threads which would be required for accessing shared variables correctly.
For example the following loop accessing sharedVariable (which is not declared volatile) might run forever without ever noticing updates made by other threads:
while(sharedVariable != expectedValue) Thread.yield();
While Thread.yield might help other threads to run (they will run anyway on most systems), it does not enforce re-reading the value of sharedVariable from the shared memory. Thus, without other constructs enforcing memory visibility, e.g. decaring sharedVariable as volatile, this loop is broken.

The first difference is that yield() is a Thread method , wait() is at the origins Object method inheritid in thread as for all classes , that in the shape, in the background (using java doc)
wait()
Causes the current thread to wait until another thread invokes the notify() method or the notifyAll() method for this object. In other words, this method behaves exactly as if it simply performs the call wait(0).
yield()
A hint to the scheduler that the current thread is willing to yield its current use of a processor. The scheduler is free to ignore this hint.
and here you can see the difference between yield() and wait()

Yield(): When a running thread is stopped to give its space to another thread with a high priority, this is called Yield.Here the running thread changes to runnable thread.
Wait(): A thread is waiting to get resources from a thread to continue its execution.

I've been taught that conditions in concurrency do not necessarily need to be written in a loop, against what the oracle doc says

So basically I am learning a bit more serious concurrency (studying how things actually work, instead of just using random stuff if needed).
And my proffesor, when I asked him about this, said me that he and his colleagues hadn't been able to reproduce a spurious wake up, and believes that line is an old line not deleted (like, it was there, java got "better", it's not longer needed, the line is still there), and that is not the case.
Link:
http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/locks/Condition.html
It's right below the point called:
Implementation Considerations
In his opinion, a condition that looked kind of like this:
lock.lock()
if (p>q) {
lock.newCondition().await
}
Would be perfectly fine, since he says a spurious wake up can not happen, it wouldn't be needed a loop:
lock.lock()
while (p>q) {
lock.newCondition().await
}
I am MORE than likely mixing things and understanding both the doc and my teacher the wrong way, but I do have spent some time trying to understand why each thing, and can't come with an "answer" of my own, I either believe one or the other (not like it matters, it's pure I-want-to-learn).
My teacher does spend time telling us how explaining concurrency in java it's pretty silly, but I didn't choose it either, so there's that.

Would be perfectly fine, since he says a spurious wake up can not happen, it wouldnt be needed a loop:
Your teacher is wrong for two reasons:
Spurious wakeups do happen. It may not happen on the architecture that they tested on but if you don't take it into account, when you move your application to a different piece of hardware or a different OS revision, you will see problems. It may also be that the spurious interrupts happen occasionally during an exceptional kernel event such as a signal getting delivered at precisely the wrong time. Again, your application may run fine in testing but when you move it into production with a lot higher load, the frequency of the exceptional event may increase...
The underlying problem is that certain native thread implementations may choose to wakeup all conditions associated with an application instead of the specific one that was notified. This is well documented in the javadocs for Object.wait():
As in the one argument version, interrupts and spurious wakeups are possible, and this method should always be used in a loop:
Here's one example of an architecture that has this limitation. I'll quote from this interesting blog entry:
Internally, wait is implemented as a call to the 'futex' system call. Each blocking system call on Linux returns abruptly when the process receives a signal -- because calling signal handler from kernel call is tricky. What if the signal handler calls some other system function? And a new signal arrives? It's easy to run out of kernel stack for a process. Exactly because each system call can be interrupted, when glibc calls any blocking function, like 'read', it does it in a loop, and if 'read' returns EINTR, calls 'read' again.
The while loop is also very important to protect against race conditions -- especially in multiple thread producer/consumer models. If you have multiple threads that are consuming from a queue (for example), a notification that there are items in the queue may wakeup a thread but by the time it is able to get the lock, another thread has already dequeued the item.
This is well documented on my page here with a sample program that demonstrates the race condition without the use of while.
Producer Consumer Thread Race Conditions
In your example, thread A may be waiting in await() while another thread B may be waiting to get the lock(). Thread C has the lock and is adding to the queue.
// B is here waiting for the lock
lock.lock()
while (p > q) {
// A is here waiting for the signal
lock.newCondition().await();
}
// dequeue
lock.unlock();
Then if the producer adds something to the queue and calls signal() the thread A moves from the WAIT state to the BLOCKED state to get the lock itself. But it may be behind thread B which is already waiting. Once the lock is released, thread B dequeues the element, not thread A. When thread A then gets a chance to dequeue, the queue is empty. Without the while loop, you can get out-of-bounds exceptions or other problems by trying to dequeue from an empty queue.
See my link for more explicit details of the race.

It is still necessary. Your professor is not necessarily incorrect, but has created a strawman argument to knock down.
There are two reasons why you must protect your conditions in a loop.
The first is spurious wake-up. Your professor seems to have been unable to reproduce this, and it may likely not be a problem on the platforms he tests on. This does not mean it is unreproduceable on all platforms.
The second is that between the times that you wake up and actually go to do the logic, the condition may no longer be true. You must guard against this potential race condition. This is also notoriously difficult to reproduce in the lab, and will probably only happen in bizarre circumstances in production.

Are static variables shared between threads?

My teacher in an upper level Java class on threading said something that I wasn't sure of.
He stated that the following code would not necessarily update the ready variable. According to him, the two threads don't necessarily share the static variable, specifically in the case when each thread (main thread versus ReaderThread) is running on its own processor and therefore doesn't share the same registers/cache/etc and one CPU won't update the other.
Essentially, he said it is possible that ready is updated in the main thread, but NOT in the ReaderThread, so that ReaderThread will loop infinitely.
He also claimed it was possible for the program to print 0 or 42. I understand how 42 could be printed, but not 0. He mentioned this would be the case when the number variable is set to the default value.
I thought perhaps it is not guaranteed that the static variable is updated between the threads, but this strikes me as very odd for Java. Does making ready volatile correct this problem?
He showed this code:
public class NoVisibility {
private static boolean ready;
private static int number;
private static class ReaderThread extends Thread {
public void run() {
while (!ready) Thread.yield();
System.out.println(number);
}
}
public static void main(String[] args) {
new ReaderThread().start();
number = 42;
ready = true;
}
}

There isn't anything special about static variables when it comes to visibility. If they are accessible any thread can get at them, so you're more likely to see concurrency problems because they're more exposed.
There is a visibility issue imposed by the JVM's memory model. Here's an article talking about the memory model and how writes become visible to threads. You can't count on changes one thread makes becoming visible to other threads in a timely manner (actually the JVM has no obligation to make those changes visible to you at all, in any time frame), unless you establish a happens-before relationship.
Here's a quote from that link (supplied in the comment by Jed Wesley-Smith):
Chapter 17 of the Java Language Specification defines the happens-before relation on memory operations such as reads and writes of shared variables. The results of a write by one thread are guaranteed to be visible to a read by another thread only if the write operation happens-before the read operation. The synchronized and volatile constructs, as well as the Thread.start() and Thread.join() methods, can form happens-before relationships. In particular:
Each action in a thread happens-before every action in that thread that comes later in the program's order.
An unlock (synchronized block or method exit) of a monitor happens-before every subsequent lock (synchronized block or method entry) of that same monitor. And because the happens-before relation is transitive, all actions of a thread prior to unlocking happen-before all actions subsequent to any thread locking that monitor.
A write to a volatile field happens-before every subsequent read of that same field. Writes and reads of volatile fields have similar memory consistency effects as entering and exiting monitors, but do not entail mutual exclusion locking.
A call to start on a thread happens-before any action in the started thread.
All actions in a thread happen-before any other thread successfully returns from a join on that thread.

He was talking about visibility and not to be taken too literally.
Static variables are indeed shared between threads, but the changes made in one thread may not be visible to another thread immediately, making it seem like there are two copies of the variable.
This article presents a view that is consistent with how he presented the info:
http://jeremymanson.blogspot.com/2008/11/what-volatile-means-in-java.html
First, you have to understand a little something about the Java memory model. I've struggled a bit over the years to explain it briefly and well. As of today, the best way I can think of to describe it is if you imagine it this way:
Each thread in Java takes place in a separate memory space (this is clearly untrue, so bear with me on this one).
You need to use special mechanisms to guarantee that communication happens between these threads, as you would on a message passing system.
Memory writes that happen in one thread can "leak through" and be seen by another thread, but this is by no means guaranteed. Without explicit communication, you can't guarantee which writes get seen by other threads, or even the order in which they get seen.
...
But again, this is simply a mental model to think about threading and volatile, not literally how the JVM works.

Basically it's true, but actually the problem is more complex. Visibility of shared data can be affected not only by CPU caches, but also by out-of-order execution of instructions.
Therefore Java defines a Memory Model, that states under which circumstances threads can see consistent state of the shared data.
In your particular case, adding volatile guarantees visibility.

They are "shared" of course in the sense that they both refer to the same variable, but they don't necessarily see each other's updates. This is true for any variable, not just static.
And in theory, writes made by another thread can appear to be in a different order, unless the variables are declared volatile or the writes are explicitly synchronized.

Within a single classloader, static fields are always shared. To explicitly scope data to threads, you'd want to use a facility like ThreadLocal.

When you initialize static primitive type variable java default assigns a value for static variables
public static int i ;
when you define the variable like this the default value of i = 0;
thats why there is a possibility to get you 0.
then the main thread updates the value of boolean ready to true. since ready is a static variable , main thread and the other thread reference to the same memory address so the ready variable change. so the secondary thread get out from while loop and print value.
when printing the value initialized value of number is 0. if the thread process has passed while loop before main thread update number variable. then there is a possibility to print 0

#dontocsata
you can go back to your teacher and school him a little :)
few notes from the real world and regardless what you see or be told.
Please NOTE, the words below are regarding this particular case in the exact order shown.
The following 2 variable will reside on the same cache line under virtually any know architecture.
private static boolean ready;
private static int number;
Thread.exit (main thread) is guaranteed to exit and exit is guaranteed to cause a memory fence, due to the thread group thread removal (and many other issues). (it's a synchronized call, and I see no single way to be implemented w/o the sync part since the ThreadGroup must terminate as well if no daemon threads are left, etc).
The started thread ReaderThread is going to keep the process alive since it is not a daemon one!
Thus ready and number will be flushed together (or the number before if a context switch occurs) and there is no real reason for reordering in this case at least I can't even think of one.
You will need something truly weird to see anything but 42. Again I do presume both static variables will be in the same cache line. I just can't imagine a cache line 4 bytes long OR a JVM that will not assign them in a continuous area (cache line).