How does CAS works? How does it work with garbage collector? Where is the problem and how does it work without garbage collector?
I was reading a presentation about CAS and using it on "write rarely, read many" problem and there was said, that use of CAS is convenient while you can use garbage collector, but there is problem (not specified) while you can not use garbage collector.
Can you tell me something about this? If you can sum up principle of CAS at first, it would be appreciated.
Ok, so CAS is an atomic instruction, that is there is special hardware support for it.
Its main use is to not use locks at all when implementing your data structures and other operations, since using locks, if a thread takes a page fault, a cache miss or is being descheduled by the OS for instance the thread takes the lock with it and all the rest of the threads are blocked. This obviously yields serious performance issues.
CAS is the core of lock-free programming and here and here.
CAS basically is the following:
CAS(CURRENT_VALUE, OLD_VALUE, NEW_VALUE) <=>
if CURRENT_VALUE==OLD_VALUE then CURRENT_VALUE = NEW_VALUE
You have a variable (e.g. class variable) and you have no clue if it was modified or not by other threads in the time you read from it and you want to write to it.
CAS helps you here on the write part since this CAS is done atomically (in hardware) and no lock is being implemented there, thus even if your thread goes to sleep the rest of the threads can operate on your data structure.
The issue with CAS on non-GC systems is the ABA problem and an example is the following:
You have a single linked list: HEAD->A->X->Y->Z
Thread 1: let's read A: localA = A; localA_Value = A.Value (let's say 5)
Thread 2: let's delete A: HEAD->A->X->Y->Z
Thread 3: let's add a new node at start (the malloc will find the right spot right were old A was): HEAD->A'->X->Y->Z (A'.Value = 10)
Thread 1 resumes and wants to swap A with B: CAS(localA, A', B) => but this thread expects that if CAS passes the value of A to be 5; wrong: since CAS passes given that localA and A' have the same memory location but localA.Value!=A'.Value => thus the operation shouldn't be performed.
The thing is that in GC enabled systems this will never happen since localA holds a reference to that memory location and thus A' will never get allocated to that memory location.
Related
I'm reading C++ Concurrency in Action by Anthony Williams, and don't understand its push implementation of the lock_free_stack class. Listing 7.12 to be precise
void push(T const& data)
{
counted_node_ptr new_node;
new_node.ptr=new node(data);
new_node.external_count=1;
new_node.ptr->next=head.load(std::memory_order_relaxed)
while(!head.compare_exchange_weak(new_node.ptr->next,new_node, std::memory_order_release, std::memory_order_relaxed));
}
So imagine 2 threads (A, B) calling push function. Both of them reach while loop but not start it. So they both read the same value from head.load(std::memory_order_relaxed).
Then we have the following things going on:
B thread gets swiped out for any reason
A thread starts the loop and obviously successfully adds a new node to the stack.
B thread gets back on track and also starts the loop.
And this is where it gets interesting as it seems to me.
Because there was a load operation with std::memory_order_relaxed and compare_exchange_weak(..., std::memory_order_release, ...) in case of success it looks like there is no synchronization between threads whatsoever.
I mean it's like std::memory_order_relaxed - std::memory_order_release and not std::memory_order_acquire - std::memory_order_release.
So B thread will simply add a new node to the stack but to its initial state when we had no nodes in the stack and reset head to this new node.
I was doing my research all around this subject and the best i could find was in this post Does exchange or compare_and_exchange reads last value in modification order?
So the question is, is it true? and all RMW functions see the last value in modification order? No matter what std::memory_order we used, if we use RMW operation it will synchronize with all threads (CPU and etc) and find the last value to be written to the atomic operation upon it is called?
So after some research and asking a bunch of people I believe I found the proper answer to this question, I hope it'll be a help to someone.
So the question is, is it true? and all RMW functions see the last
value in modification order?
Yes, it is true.
No matter what std::memory_order we used, if we use RMW operation it
will synchronize with all threads (CPU and etc) and find the last
value to be written to the atomic operation upon it is called?
Yes, it is also true, however there is something that needs to be highlighted.
RMW operation will synchronize only the atomic variable it works with. In our case, it is head.load
Perhaps you would like to ask why we need release - acquire semantics at all if RMW does the synchronization even with the relaxed memory order.
The answer is because RMW works only with the variable it synchronizes, but other operations which occurred before RMW might not be seen in the other thread.
lets look at the push function again:
void push(T const& data)
{
counted_node_ptr new_node;
new_node.ptr=new node(data);
new_node.external_count=1;
new_node.ptr->next=head.load(std::memory_order_relaxed)
while(!head.compare_exchange_weak(new_node.ptr->next,new_node, std::memory_order_release, std::memory_order_relaxed));
}
In this example, in case of using two push threads they won't be synchronized with each other to some extent, but it could be allowed here.
Both threads will always see the newest head because compare_exchange_weak
provides this. And a new node will be always added to the top of the stack.
However if we tried to get the value like this *(new_node.ptr->next) after this line new_node.ptr->next=head.load(std::memory_order_relaxed) things could easily turn ugly: empty pointer might be dereferenced.
This might happen because a processor can change the order of instructions and because there is no synchronization between threads the second thread could see the pointer to a top node even before that was initialized!
And this is exactly where release-acquire semantic comes to help. It ensures that all operations which happened before release operation will be seen in acquire part!
Check out and compare listings 5.5 and 5.8 in the book.
I also recommend you to read this article about how processors work, it might provide some essential information for better understanding.
memory barriers
Start with x = 0. Note there are no memory barriers in any of the code below.
volatile int x = 0
Thread 1:
while (x == 0) {}
print "Saw non-zer0"
while (x != 0) {}
print "Saw zero again!"
Thread 2:
x = 1
Is it ever possible to see the second message, "Saw zero again!", on any (real) CPU? What about on x86_64?
Similarly, in this code:
volatile int x = 0.
Thread 1:
while (x == 0) {}
x = 2
Thread 2:
x = 1
Is the final value of x guaranteed to be 2, or could the CPU caches update main memory in some arbitrary order, so that although x = 1 gets into a CPU's cache where thread 1 can see it, then thread 1 gets moved to a different cpu where it writes x = 2 to that cpu's cache, and the x = 2 gets written back to main memory before x = 1.
Yes, it's entirely possible. The compiler could, for example, have just written x to memory but still have the value in a register. One while loop could check memory while the other checks the register.
It doesn't happen due to CPU caches because cache coherency hardware logic makes the caches invisible on all CPUs you are likely to actually use.
Theoretically, the write race you talk about could happen due to posted write buffering and read prefetching. Miraculous tricks were used to make this impossible on x86 CPUs to avoid breaking legacy code. But you shouldn't expect future processors to do this.
Leaving aside for a second tricks done by the compiler (even ones allowed by language standards), I believe you're asking how the micro-architecture could behave in such scenario. Keep in mind that the code would most likely expand into a busy wait loop of cmp [x] + jz or something similar, which hides a load inside it. This means that [x] is likely to live in the cache of the core running thread 1.
At some point, thread 2 would come and perform the store. If it resides on a different core, the line would first be invalidated completely from the first core. If these are 2 threads running on the same physical core - the store would immediately affect all chronologically younger loads.
Now, the most likely thing to happen on a modern out-of-order machine is that all the loads in the pipeline at this point would be different iterations of the same first loop (since any branch predictor facing so many repetitive "taken" resolution is likely to assume the branch will continue being taken, until proven wrong), so what would happen is that the first load to encounter the new value modified by the other thread will cause the matching branch to simply flush the entire pipe from all younger operations, without the 2nd loop ever having a chance to execute.
However, it's possible that for some reason you did get to the 2nd loop (let's say the predictor issue a not-taken prediction just at the right moment when the loop condition check saw the new value) - in this case, the question boils down to this scenario:
Time -->
----------------------------------------------------------------
thread 1
cmp [x],0 execute
je ... execute (not taken)
...
cmp [x],0 execute
jne ... execute (not taken)
Can_We_Get_Here:
...
thread2
store [x],1 execute
In other words, given that most modern CPUs may execute instructions out of order, can a younger load be evaluated before an older one to the same address, allowing the store (from another thread) to change the value so it may be observed inconsistently by the loads.
My guess is that the above timeline is quite possible given the nature of out-of-order execution engines today, as they simply arbitrate and perform whatever operation is ready. However, on most x86 implementations there are safeguards to protect against such a scenario, since the memory ordering rules strictly say -
8.2.3.2 Neither Loads Nor Stores Are Reordered with Like Operations
Such mechanisms may detect this scenario and flush the machine to prevent the stale/wrong values becoming visible. So The answer is - no, it should not be possible, unless of course the software or the compiler change the nature of the code to prevent the hardware from noticing the relation. Then again, memory ordering rules are sometimes flaky, and i'm not sure all x86 manufacturers adhere to the exact same wording, but this is a pretty fundamental example of consistency, so i'd be very surprised if one of them missed it.
The answer seems to be, "this is exactly the job of the CPU cache coherency." x86 processors implement the MESI protocol, which guarantee that the second thread can't see the new value then the old.
I'm having trouble figuring out a key point in wait-free algorithm design. Suppose a data structure has a pointer to another data structure (e.g. linked list, tree, etc), how can the right time for releasing a data structure?
The problem is this, there are separate operations that can't be executed atomically without a lock. For example one thread reads the pointer to some memory, and increments the use count for that memory to prevent free while this thread is using the data, which might take long, and even if it doesn't, it's a race condition. What prevents another thread from reading the pointer, decrementing the use count and determining that it's no longer used and freeing it before the first thread incremented the use count?
The main issue is that current CPUs only have a single word CAS (compare & swap). Alternatively the problem is that I'm clueless about waitfree algorithms and data structures and after reading some papers I'm still not seeing the light.
IMHO Garbage collection can't be the answer, because it would either GC would have to be prevented from running if any single thread is inside an atomic block (which would mean it can't be guaranteed that the GC will ever run again) or the problem is simply pushed to the GC, in which case, please explain how the GC would figure out if the data is in the silly state (a pointer is read [e.g. stored in a local variable] but the the use count didn't increment yet).
PS, references to advanced tutorials on wait-free algorithms for morons are welcome.
Edit: You should assume that the problem is being solved in a non-managed language, like C or C++. After all if it were Java, we'd have no need to worry about releasing memory. Further assume that the compiler may generate code that will store temporary references to objects in registers (invisible to other threads) right before the usage counter increment, and that a thread can be interrupted between loading the object address and incrementing the counter. This of course doesn't mean that the solution must be limited to C or C++, rather that the solution should give a set of primitives that allowing the implementation of wait-free algorithms on linked data structures. I'm interested in the primitives and how they solve the problem of designing wait-free algorithms. With such primitives a wait-free algorithm can be implemented equally well in C++ and Java.
After some research I learned this.
The problem is not trivial to solve and there are several solutions each with advantages and disadvantages. The reason for the complexity comes from inter CPU synchronization issues. If not done right it might appear to work correctly 99.9% of the time, which isn't enough, or it might fail under load.
Three solutions that I found are 1) hazard pointers, 2) quiescence period based reclamation (used by the Linux kernel in the RCU implementation) 3) reference counting techniques. 4) Other 5) Combinations
Hazard pointers work by saving the currently active references in a well-known per thread location, so any thread deciding to free memory (when the counter appears to be zero) can check if the memory is still in use by anyone. An interesting improvement is to buffer request to release memory in a small array and free them up in a batch when the array is full. The advantage of using hazard pointers is that it can actually guarantee an upper bound on unreclaimed memory. The disadvantage is that it places extra burden on the reader.
Quiescence period based reclamation works by delaying the actual release of the memory until it's known that each thread has had a chance to finish working on any data that may need to be released. The way to know that this condition is satisfied is to check if each thread passed through a quiescent period (not in a critical section) after the object was removed. In the Linux kernel this means something like each task making a voluntary task switch. In a user space application it would be the end of a critical section. This can be achieved by a simple counter, each time the counter is even the thread is not in a critical section (reading shared data), each time the counter is odd the thread is inside a critical section, to move from a critical section or back all the thread needs to do is to atomically increment the number. Based on this the "garbage collector" can determine if each thread has had a chance to finish. There are several approaches, one simple one would be to queue up the requests to free memory (e.g. in a linked list or an array), each with the current generation (managed by the GC), when the GC runs it checks the state of the threads (their state counters) to see if each passed to the next generation (their counter is higher than the last time or is the same and even), any memory can be reclaimed one generation after it was freed. The advantage of this approach is that is places the least burden on the reading threads. The disadvantage is that it can't guarantee an upper bound for the memory waiting to be released (e.g. one thread spending 5 minutes in a critical section, while the data keeps changing and memory isn't released), but in practice it works out all right.
There is a number of reference counting solutions, many of them require double compare and swap, which some CPUs don't support, so can't be relied upon. The key problem remains though, taking a reference before updating the counter. I didn't find enough information to explain how this can be done simply and reliably though. So .....
There are of course a number of "Other" solutions, it's a very important topic of research with tons of papers out there. I didn't examine all of them. I only need one.
And of course the various approaches can be combined, for example hazard pointers can solve the problems of reference counting. But there's a nearly infinite number of combinations, and in some cases a spin lock might theoretically break wait-freedom, but doesn't hurt performance in practice. Somewhat like another tidbit I found in my research, it's theoretically not possible to implement wait-free algorithms using compare-and-swap, that's because in theory (purely in theory) a CAS based update might keep failing for non-deterministic excessive times (imagine a million threads on a million cores each trying to increment and decrement the same counter using CAS). In reality however it rarely fails more than a few times (I suspect it's because the CPUs spend more clocks away from CAS than there are CPUs, but I think if the algorithm returned to the same CAS on the same location every 50 clocks and there were 64 cores there could be a chance of a major problem, then again, who knows, I don't have a hundred core machine to try this). Another results of my research is that designing and implementing wait-free algorithms and data-structures is VERY challenging (even if some of the heavy lifting is outsourced, e.g. to a garbage collector [e.g. Java]), and might perform less well than a similar algorithm with carefully placed locks.
So, yeah, it's possible to free memory even without delays. It's just tricky. And if you forget to make the right operations atomic, or to place the right memory barrier, oh, well, you're toast. :-) Thanks everyone for participating.
I think atomic operations for increment/decrement and compare-and-swap would solve this problem.
Idea:
All resources have a counter which is modified with atomic operations. The counter is initially zero.
Before using a resource: "Acquire" it by atomically incrementing its counter. The resource can be used if and only if the incremented value is greater than zero.
After using a resource: "Release" it by atomically decrementing its counter. The resource should be disposed/freed if and only if the decremented value is equal to zero.
Before disposing: Atomically compare-and-swap the counter value with the minimum (negative) value. Dispose will not happen if a concurrent thread "Acquired" the resource in between.
You haven't specified a language for your question. Here goes an example in c#:
class MyResource
{
// Counter is initially zero. Resource will not be disposed until it has
// been acquired and released.
private int _counter;
public bool Acquire()
{
// Atomically increment counter.
int c = Interlocked.Increment(ref _counter);
// Resource is available if the resulting value is greater than zero.
return c > 0;
}
public bool Release()
{
// Atomically decrement counter.
int c = Interlocked.Decrement(ref _counter);
// We should never reach a negative value
Debug.Assert(c >= 0, "Resource was released without being acquired");
// Dispose when we reach zero
if (c == 0)
{
// Mark as disposed by setting counter its minimum value.
// Only do this if the counter remain at zero. Atomic compare-and-swap operation.
if (Interlocked.CompareExchange(ref _counter, int.MinValue, c) == c)
{
// TODO: Run dispose code (free stuff)
return true; // tell caller that resource is disposed
}
}
return false; // released but still in use
}
}
Usage:
// "r" is an instance of MyResource
bool acquired = false;
try
{
if (acquired = r.Acquire())
{
// TODO: Use resource
}
}
finally
{
if (acquired)
{
if (r.Release())
{
// Resource was disposed.
// TODO: Nullify variable or similar to let GC collect it.
}
}
}
I know this is not the best way but it works for me:
for shared dynamic data-structure lists I use usage counter per item
for example:
struct _data
{
DWORD usage;
bool delete;
// here add your data
_data() { usage=0; deleted=true; }
};
const int MAX = 1024;
_data data[MAX];
now when item is started to be used somwhere then
// start use of data[i]
data[i].cnt++;
after is no longer used then
// stop use of data[i]
data[i].cnt--;
if you want to add new item to list then
// add item
for (i=0;i<MAX;i++) // find first deleted item
if (data[i].deleted)
{
data[i].deleted=false;
data[i].cnt=0;
// copy/set your data
break;
}
and now in the background once in a while (on timer or whatever)
scann data[] an all undeleted items with cnt == 0 set as deleted (+ free its dynamic memory if it has any)
[Note]
to avoid multi-thread access problems implement single global lock per data list
and program it so you cannot scann data while any data[i].cnt is changing
one bool and one DWORD suffice for this if you do not want to use OS locks
// globals
bool data_cnt_locked=false;
DWORD data_cnt=0;
now any change of data[i].cnt modify like this:
// start use of data[i]
while (data_cnt_locked) Sleep(1);
data_cnt++;
data[i].cnt++;
data_cnt--;
and modify delete scan like this
while (data_cnt) Sleep(1);
data_cnt_locked=true;
Sleep(1);
if (data_cnt==0) // just to be sure
for (i=0;i<MAX;i++) // here scan for items to delete ...
if (!data[i].cnt)
if (!data[i].deleted)
{
data[i].deleted=true;
data[i].cnt=0;
// release your dynamic data ...
}
data_cnt_locked=false;
PS.
do not forget to play with the sleep times a little to suite your needs
lock free algorithm sleep times are sometimes dependent on OS task/scheduler
this is not really an lock free implementation
because while GC is at work then all is locked
but if ather than that multi access is not blocking to each other
so if you do not run GC too often you are fine
I've always been told to puts locks around variables that multiple threads will access, I've always assumed that this was because you want to make sure that the value you are working with doesn't change before you write it back
i.e.
mutex.lock()
int a = sharedVar
a = someComplexOperation(a)
sharedVar = a
mutex.unlock()
And that makes sense that you would lock that. But in other cases I don't understand why I can't get away with not using Mutexes.
Thread A:
sharedVar = someFunction()
Thread B:
localVar = sharedVar
What could possibly go wrong in this instance? Especially if I don't care that Thread B reads any particular value that Thread A assigns.
It depends a lot on the type of sharedVar, the language you're using, any framework, and the platform. In many cases, it's possible that assigning a single value to sharedVar may take more than one instruction, in which case you may read a "half-set" copy of the value.
Even when that's not the case, and the assignment is atomic, you may not see the latest value without a memory barrier in place.
MSDN Magazine has a good explanation of different problems you may encounter in multithreaded code:
Forgotten Synchronization
Incorrect Granularity
Read and Write Tearing
Lock-Free Reordering
Lock Convoys
Two-Step Dance
Priority Inversion
The code in your question is particularly vulnerable to Read/Write Tearing. But your code, having neither locks nor memory barriers, is also subject to Lock-Free Reordering (which may include speculative writes in which thread B reads a value that thread A never stored) in which side-effects become visible to a second thread in a different order from how they appeared in your source code.
It goes on to describe some known design patterns which avoid these problems:
Immutability
Purity
Isolation
The article is available here
The main problem is that the assignment operator (operator= in C++) is not always guaranteed to be atomic (not even for primitive, built in types). In plain English, that means that assignment can take more than a single clock cycle to complete. If, in the middle of that, the thread gets interrupted, then the current value of the variable might be corrupted.
Let me build off of your example:
Lets say sharedVar is some object with operator= defined as this:
object& operator=(const object& other) {
ready = false;
doStuff(other);
if (other.value == true) {
value = true;
doOtherStuff();
} else {
value = false;
}
ready = true;
return *this;
}
If thread A from your example is interrupted in the middle of this function, ready will still be false when thread B starts to run. This could mean that the object is only partially copied over, or is in some intermediate, invalid state when thread B attempts to copy it into a local variable.
For a particularly nasty example of this, think of a data structure with a removed node being deleted, then interrupted before it could be set to NULL.
(For some more information regarding structures that don't need a lock (aka, are atomic), here is another question that talks a bit more about that.)
This could go wrong, because threads can be suspended and resumed by the thread scheduler, so you can't be sure about the order these instructions are executed. It might just as well be in this order:
Thread B:
localVar = sharedVar
Thread A:
sharedVar = someFunction()
In which case localvar will be null or 0 (or some completeley unexpected value in an unsafe language), probably not what you intended.
Mutexes actually won't fix this particular issue by the way. The example you supply does not lend itself well for parallelization.
I understand about race conditions and how with multiple threads accessing the same variable, updates made by one can be ignored and overwritten by others, but what if each thread is writing the same value (not different values) to the same variable; can even this cause problems? Could this code:
GlobalVar.property = 11;
(assuming that property will never be assigned anything other than 11), cause problems if multiple threads execute it at the same time?
The problem comes when you read that state back, and do something about it. Writing is a red herring - it is true that as long as this is a single word most environments guarantee the write will be atomic, but that doesn't mean that a larger piece of code that includes this fragment is thread-safe. Firstly, presumably your global variable contained a different value to begin with - otherwise if you know it's always the same, why is it a variable? Second, presumably you eventually read this value back again?
The issue is that presumably, you are writing to this bit of shared state for a reason - to signal that something has occurred? This is where it falls down: when you have no locking constructs, there is no implied order of memory accesses at all. It's hard to point to what's wrong here because your example doesn't actually contain the use of the variable, so here's a trivialish example in neutral C-like syntax:
int x = 0, y = 0;
//thread A does:
x = 1;
y = 2;
if (y == 2)
print(x);
//thread B does, at the same time:
if (y == 2)
print(x);
Thread A will always print 1, but it's completely valid for thread B to print 0. The order of operations in thread A is only required to be observable from code executing in thread A - thread B is allowed to see any combination of the state. The writes to x and y may not actually happen in order.
This can happen even on single-processor systems, where most people do not expect this kind of reordering - your compiler may reorder it for you. On SMP even if the compiler doesn't reorder things, the memory writes may be reordered between the caches of the separate processors.
If that doesn't seem to answer it for you, include more detail of your example in the question. Without the use of the variable it's impossible to definitively say whether such a usage is safe or not.
It depends on the work actually done by that statement. There can still be some cases where Something Bad happens - for example, if a C++ class has overloaded the = operator, and does anything nontrivial within that statement.
I have accidentally written code that did something like this with POD types (builtin primitive types), and it worked fine -- however, it's definitely not good practice, and I'm not confident that it's dependable.
Why not just lock the memory around this variable when you use it? In fact, if you somehow "know" this is the only write statement that can occur at some point in your code, why not just use the value 11 directly, instead of writing it to a shared variable?
(edit: I guess it's better to use a constant name instead of the magic number 11 directly in the code, btw.)
If you're using this to figure out when at least one thread has reached this statement, you could use a semaphore that starts at 1, and is decremented by the first thread that hits it.
I would expect the result to be undetermined. As in it would vary from compiler to complier, langauge to language and OS to OS etc. So no, it is not safe
WHy would you want to do this though - adding in a line to obtain a mutex lock is only one or two lines of code (in most languages), and would remove any possibility of problem. If this is going to be two expensive then you need to find an alternate way of solving the problem
In General, this is not considered a safe thing to do unless your system provides for atomic operation (operations that are guaranteed to be executed in a single cycle).
The reason is that while the "C" statement looks simple, often there are a number of underlying assembly operations taking place.
Depending on your OS, there are a few things you could do:
Take a mutual exclusion semaphore (mutex) to protect access
in some OS, you can temporarily disable preemption, which guarantees your thread will not swap out.
Some OS provide a writer or reader semaphore which is more performant than a plain old mutex.
Here's my take on the question.
You have two or more threads running that write to a variable...like a status flag or something, where you only want to know if one or more of them was true. Then in another part of the code (after the threads complete) you want to check and see if at least on thread set that status... for example
bool flag = false
threadContainer tc
threadInputs inputs
check(input)
{
...do stuff to input
if(success)
flag = true
}
start multiple threads
foreach(i in inputs)
t = startthread(check, i)
tc.add(t) // Keep track of all the threads started
foreach(t in tc)
t.join( ) // Wait until each thread is done
if(flag)
print "One of the threads were successful"
else
print "None of the threads were successful"
I believe the above code would be OK, assuming you're fine with not knowing which thread set the status to true, and you can wait for all the multi-threaded stuff to finish before reading that flag. I could be wrong though.
If the operation is atomic, you should be able to get by just fine. But I wouldn't do that in practice. It is better just to acquire a lock on the object and write the value.
Assuming that property will never be assigned anything other than 11, then I don't see a reason for assigment in the first place. Just make it a constant then.
Assigment only makes sense when you intend to change the value unless the act of assigment itself has other side effects - like volatile writes have memory visibility side-effects in Java. And if you change state shared between multiple threads, then you need to synchronize or otherwise "handle" the problem of concurrency.
When you assign a value, without proper synchronization, to some state shared between multiple threads, then there's no guarantees for when the other threads will see that change. And no visibility guarantees means that it it possible that the other threads will never see the assignt.
Compilers, JITs, CPU caches. They're all trying to make your code run as fast as possible, and if you don't make any explicit requirements for memory visibility, then they will take advantage of that. If not on your machine, then somebody elses.