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Java variable shared between two process [duplicate]

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).

Volatile Vs Atomic [duplicate]

This question already has answers here:
What is the difference between atomic / volatile / synchronized?
(7 answers)
Closed 9 years ago.
I read somewhere below line.
Java volatile keyword doesn't means atomic, its common misconception
that after declaring volatile, ++ operation will be atomic, to make
the operation atomic you still need to ensure exclusive access using
synchronized method or block in Java.
So what will happen if two threads attack a volatile primitive variable at same time?
Does this mean that whosoever takes lock on it, that will be setting its value first. And if in meantime, some other thread comes up and read old value while first thread was changing its value, then doesn't new thread will read its old value?
What is the difference between Atomic and volatile keyword?

The effect of the volatile keyword is approximately that each individual read or write operation on that variable is made atomically visible to all threads.
Notably, however, an operation that requires more than one read/write -- such as i++, which is equivalent to i = i + 1, which does one read and one write -- is not atomic, since another thread may write to i between the read and the write.
The Atomic classes, like AtomicInteger and AtomicReference, provide a wider variety of operations atomically, specifically including increment for AtomicInteger.

Volatile and Atomic are two different concepts. Volatile ensures, that a certain, expected (memory) state is true across different threads, while Atomics ensure that operation on variables are performed atomically.
Take the following example of two threads in Java:
Thread A:
value = 1;
done = true;
Thread B:
if (done)
System.out.println(value);
Starting with value = 0 and done = false the rule of threading tells us, that it is undefined whether or not Thread B will print value. Furthermore value is undefined at that point as well! To explain this you need to know a bit about Java memory management (which can be complex), in short: Threads may create local copies of variables, and the JVM can reorder code to optimize it, therefore there is no guarantee that the above code is run in exactly that order. Setting done to true and then setting value to 1 could be a possible outcome of the JIT optimizations.
volatile only ensures, that at the moment of access of such a variable, the new value will be immediately visible to all other threads and the order of execution ensures, that the code is at the state you would expect it to be. So in case of the code above, defining done as volatile will ensure that whenever Thread B checks the variable, it is either false, or true, and if it is true, then value has been set to 1 as well.
As a side-effect of volatile, the value of such a variable is set thread-wide atomically (at a very minor cost of execution speed). This is however only important on 32-bit systems that i.E. use long (64-bit) variables (or similar), in most other cases setting/reading a variable is atomic anyways. But there is an important difference between an atomic access and an atomic operation. Volatile only ensures that the access is atomically, while Atomics ensure that the operation is atomically.
Take the following example:
i = i + 1;
No matter how you define i, a different Thread reading the value just when the above line is executed might get i, or i + 1, because the operation is not atomically. If the other thread sets i to a different value, in worst case i could be set back to whatever it was before by thread A, because it was just in the middle of calculating i + 1 based on the old value, and then set i again to that old value + 1. Explanation:
Assume i = 0
Thread A reads i, calculates i+1, which is 1
Thread B sets i to 1000 and returns
Thread A now sets i to the result of the operation, which is i = 1
Atomics like AtomicInteger ensure, that such operations happen atomically. So the above issue cannot happen, i would either be 1000 or 1001 once both threads are finished.

There are two important concepts in multithreading environment:
atomicity
visibility
The volatile keyword eradicates visibility problems, but it does not deal with atomicity. volatile will prevent the compiler from reordering instructions which involve a write and a subsequent read of a volatile variable; e.g. k++.
Here, k++ is not a single machine instruction, but three:
copy the value to a register;
increment the value;
place it back.
So, even if you declare a variable as volatile, this will not make this operation atomic; this means another thread can see a intermediate result which is a stale or unwanted value for the other thread.
On the other hand, AtomicInteger, AtomicReference are based on the Compare and swap instruction. CAS has three operands: a memory location V on which to operate, the expected old value A, and the new value B. CAS atomically updates V to the new value B, but only if the value in V matches the expected old value A; otherwise, it does nothing. In either case, it returns the value currently in V. The compareAndSet() methods of AtomicInteger and AtomicReference take advantage of this functionality, if it is supported by the underlying processor; if it is not, then the JVM implements it via spin lock.

As Trying as indicated, volatile deals only with visibility.
Consider this snippet in a concurrent environment:
boolean isStopped = false;
:
:
while (!isStopped) {
// do some kind of work
}
The idea here is that some thread could change the value of isStopped from false to true in order to indicate to the subsequent loop that it is time to stop looping.
Intuitively, there is no problem. Logically if another thread makes isStopped equal to true, then the loop must terminate. The reality is that the loop will likely never terminate even if another thread makes isStopped equal to true.
The reason for this is not intuitive, but consider that modern processors have multiple cores and that each core has multiple registers and multiple levels of cache memory that are not accessible to other processors. In other words, values that are cached in one processor's local memory are not visisble to threads executing on a different processor. Herein lies one of the central problems with concurrency: visibility.
The Java Memory Model makes no guarantees whatsoever about when changes that are made to a variable in one thread may become visible to other threads. In order to guarantee that updates are visisble as soon as they are made, you must synchronize.
The volatile keyword is a weak form of synchronization. While it does nothing for mutual exclusion or atomicity, it does provide a guarantee that changes made to a variable in one thread will become visible to other threads as soon as it is made. Because individual reads and writes to variables that are not 8-bytes are atomic in Java, declaring variables volatile provides an easy mechanism for providing visibility in situations where there are no other atomicity or mutual exclusion requirements.

The volatile keyword is used:
to make non atomic 64-bit operations atomic: long and double. (all other, primitive accesses are already guaranteed to be atomic!)
to make variable updates guaranteed to be seen by other threads + visibility effects: after writing to a volatile variable, all the variables that where visible before writing that variable become visible to another thread after reading the same volatile variable (happen-before ordering).
The java.util.concurrent.atomic.* classes are, according to the java docs:
A small toolkit of classes that support lock-free thread-safe
programming on single variables. In essence, the classes in this
package extend the notion of volatile values, fields, and array
elements to those that also provide an atomic conditional update
operation of the form:
boolean compareAndSet(expectedValue, updateValue);
The atomic classes are built around the atomic compareAndSet(...) function that maps to an atomic CPU instruction. The atomic classes introduce the happen-before ordering as the volatile variables do. (with one exception: weakCompareAndSet(...)).
From the java docs:
When a thread sees an update to an atomic variable caused by a
weakCompareAndSet, it does not necessarily see updates to any other
variables that occurred before the weakCompareAndSet.
To your question:
Does this mean that whosoever takes lock on it, that will be setting
its value first. And in if meantime, some other thread comes up and
read old value while first thread was changing its value, then doesn't
new thread will read its old value?
You don't lock anything, what you are describing is a typical race condition that will happen eventually if threads access shared data without proper synchronization. As already mentioned declaring a variable volatile in this case will only ensure that other threads will see the change of the variable (the value will not be cached in a register of some cache that is only seen by one thread).
What is the difference between AtomicInteger and volatile int?
AtomicInteger provides atomic operations on an int with proper synchronization (eg. incrementAndGet(), getAndAdd(...), ...), volatile int will just ensure the visibility of the int to other threads.

So what will happen if two threads attack a volatile primitive variable at same time?
Usually each one can increment the value. However sometime, both will update the value at the same time and instead of incrementing by 2 total, both thread increment by 1 and only 1 is added.
Does this mean that whosoever takes lock on it, that will be setting its value first.
There is no lock. That is what synchronized is for.
And in if meantime, some other thread comes up and read old value while first thread was changing its value, then doesn't new thread will read its old value?
Yes,
What is the difference between Atomic and volatile keyword?
AtomicXxxx wraps a volatile so they are basically same, the difference is that it provides higher level operations such as CompareAndSwap which is used to implement increment.
AtomicXxxx also supports lazySet. This is like a volatile set, but doesn't stall the pipeline waiting for the write to complete. It can mean that if you read a value you just write you might see the old value, but you shouldn't be doing that anyway. The difference is that setting a volatile takes about 5 ns, bit lazySet takes about 0.5 ns.

Is unsynchronized read of integer threadsafe in java?

I see this code quite frequently in some OSS unit tests, but is it thread safe ? Is the while loop guaranteed to see the correct value of invoc ?
If no; nerd points to whoever also knows which CPU architecture this may fail on.
private int invoc = 0;
private synchronized void increment() {
invoc++;
}
public void isItThreadSafe() throws InterruptedException {
for (int i = 0; i < TOTAL_THREADS; i++) {
new Thread(new Runnable() {
public void run() {
// do some stuff
increment();
}
}).start();
}
while (invoc != TOTAL_THREADS) {
Thread.sleep(250);
}
}

No, it's not threadsafe. invoc needs to be declared volatile, or accessed while synchronizing on the same lock, or changed to use AtomicInteger. Just using the synchronized method to increment invoc, but not synchronizing to read it, isn't good enough.
The JVM does a lot of optimizations, including CPU-specific caching and instruction reordering. It uses the volatile keyword and locking to decide when it can optimize freely and when it has to have an up-to-date value available for other threads to read. So when the reader doesn't use the lock the JVM can't know not to give it a stale value.
This quote from Java Concurrency in Practice (section 3.1.3) discusses how both writes and reads need to be synchronized:
Intrinsic locking can be used to guarantee that one thread sees the effects of another in a predictable manner, as illustrated by Figure 3.1. When thread A executes a synchronized block, and subsequently thread B enters a synchronized block guarded by the same lock, the values of variables that were visible to A prior to releasing the lock are guaranteed to be visible to B upon acquiring the lock. In other words, everything A did in or prior to a synchronized block is visible to B when it executes a synchronized block guarded by the same lock. Without synchronization, there is no such guarantee.
The next section (3.1.4) covers using volatile:
The Java language also provides an alternative, weaker form of synchronization, volatile variables, to ensure that updates to a variable are propagated predictably to other threads. When a field is declared volatile, the compiler and runtime are put on notice that this variable is shared and that operations on it should not be reordered with other memory operations. Volatile variables are not cached in registers or in caches where they are hidden from other processors, so a read of a volatile variable always returns the most recent write by any thread.
Back when we all had single-CPU machines on our desktops we'd write code and never have a problem until it ran on a multiprocessor box, usually in production. Some of the factors that give rise to the visiblity problems, things like CPU-local caches and instruction reordering, are things you would expect from any multiprocessor machine. Elimination of apparently unneeded instructions could happen for any machine, though. There's nothing forcing the JVM to ever make the reader see the up-to-date value of the variable, you're at the mercy of the JVM implementors. So it seems to me this code would not be a good bet for any CPU architecture.

Well!
private volatile int invoc = 0;
Will do the trick.
And see Are java primitive ints atomic by design or by accident? which sites some of the relevant java definitions. Apparently int is fine, but double & long might not be.
edit, add-on. The question asks, "see the correct value of invoc ?". What is "the correct value"? As in the timespace continuum, simultaneity doesn't really exist between threads. One of the above posts notes that the value will eventually get flushed, and the other thread will get it. Is the code "thread safe"? I would say "yes", because it won't "misbehave" based on the vagaries of sequencing, in this case.

Theoretically, it is possible that the read is cached. Nothing in Java memory model prevents that.
Practically, that is extremely unlikely to happen (in your particular example). The question is, whether JVM can optimize across a method call.
read #1
method();
read #2
For JVM to reason that read#2 can reuse the result of read#1 (which can be stored in a CPU register), it must know for sure that method() contains no synchronization actions. This is generally impossible - unless, method() is inlined, and JVM can see from the flatted code that there's no sync/volatile or other synchronization actions between read#1 and read#2; then it can safely eliminate read#2.
Now in your example, the method is Thread.sleep(). One way to implement it is to busy loop for certain times, depending on CPU frequency. Then JVM may inline it, and then eliminate read#2.
But of course such implementation of sleep() is unrealistic. It is usually implemented as a native method that calls OS kernel. The question is, can JVM optimize across such a native method.
Even if JVM has knowledge of internal workings of some native methods, therefore can optimize across them, it's improbable that sleep() is treated that way. sleep(1ms) takes millions of CPU cycles to return, there is really no point optimizing around it to save a few reads.
--
This discussion reveals the biggest problem of data races - it takes too much effort to reason about it. A program is not necessarily wrong, if it is not "correctly synchronized", however to prove it's not wrong is not an easy task. Life is much simpler, if a program is correctly synchronized and contains no data race.

As far as I understand the code it should be safe. The bytecode can be reordered, yes. But eventually invoc should be in sync with the main thread again. Synchronize guarantees that invoc is incremented correctly so there is a consistent representation of invoc in some register. At some time this value will be flushed and the little test succeeds.
It is certainly not nice and I would go with the answer I voted for and would fix code like this because it smells. But thinking about it I would consider it safe.

If you're not required to use "int", I would suggest AtomicInteger as an thread-safe alternative.

is this class thread safe?

consider this class,with no instance variables and only methods which are non-synchronous can we infer from this info that this class in Thread-safe?
public class test{
public void test1{
// do something
}
public void test2{
// do something
}
public void test3{
// do something
}
}

It depends entirely on what state the methods mutate. If they mutate no shared state, they're thread safe. If they mutate only local state, they're thread-safe. If they only call methods that are thread-safe, they're thread-safe.

Not being thread safe means that if multiple threads try to access the object at the same time, something might change from one access to the next, and cause issues. Consider the following:
int incrementCount() {
this.count++;
// ... Do some other stuff
return this.count;
}
would not be thread safe. Why is it not? Imagine thread 1 accesses it, count is increased, then some processing occurs. While going through the function, another thread accesses it, increasing count again. The first thread, which had it go from, say, 1 to 2, would now have it go from 1 to 3 when it returns. Thread 2 would see it go from 1 to 3 as well, so what happened to 2?
In this case, you would want something like this (keeping in mind that this isn't any language-specific code, but closest to Java, one of only 2 I've done threading in)
int incrementCount() synchronized {
this.count++;
// ... Do some other stuff
return this.count;
}
The synchronized keyword here would make sure that as long as one thread is accessing it, no other threads could. This would mean that thread 1 hits it, count goes from 1 to 2, as expected. Thread 2 hits it while 1 is processing, it has to wait until thread 1 is done. When it's done, thread 1 gets a return of 2, then thread 2 goes throguh, and gets the expected 3.
Now, an example, similar to what you have there, that would be entirely thread-safe, no matter what:
int incrementCount(int count) {
count++;
// ... Do some other stuff
return this.count;
}
As the only variables being touched here are fully local to the function, there is no case where two threads accessing it at the same time could try working with data changed from the other. This would make it thread safe.
So, to answer the question, assuming that the functions don't modify anything outside of the specific called function, then yes, the class could be deemed to be thread-safe.

Consider the following quote from an article about thread safety ("Java theory and practice: Characterizing thread safety"):
In reality, any definition of thread safety is going to have a certain degree of circularity, as it must appeal to the class's specification -- which is an informal, prose description of what the class does, its side effects, which states are valid or invalid, invariants, preconditions, postconditions, and so on. (Constraints on an object's state imposed by the specification apply only to the externally visible state -- that which can be observed by calling its public methods and accessing its public fields -- rather than its internal state, which is what is actually represented in its private fields.)
Thread safety
For a class to be thread-safe, it first must behave correctly in a single-threaded environment. If a class is correctly implemented, which is another way of saying that it conforms to its specification, no sequence of operations (reads or writes of public fields and calls to public methods) on objects of that class should be able to put the object into an invalid state, observe the object to be in an invalid state, or violate any of the class's invariants, preconditions, or postconditions.
Furthermore, for a class to be thread-safe, it must continue to behave correctly, in the sense described above, when accessed from multiple threads, regardless of the scheduling or interleaving of the execution of those threads by the runtime environment, without any additional synchronization on the part of the calling code. The effect is that operations on a thread-safe object will appear to all threads to occur in a fixed, globally consistent order.
So your class itself is thread-safe, as long as it doesn't have any side effects. As soon as the methods mutate any external objects (e.g. some singletons, as already mentioned by others) it's not any longer thread-safe.

Depends on what happens inside those methods. If they manipulate / call any method parameters or global variables / singletons which are not themselves thread safe, the class is not thread safe either.
(yes I see that the methods as shown here here have no parameters, but no brackets either, so this is obviously not full working code - it wouldn't even compile as is.)

yes, as long as there are no instance variables. method calls using only input parameters and local variables are inherently thread-safe. you might consider making the methods static too, to reflect this.

If it has no mutable state - it's thread safe. If you have no state - you're thread safe by association.

No, I don't think so.
For example, one of the methods could obtain a (non-thread-safe) singleton object from another class and mutate that object.

Yes - this class is thread safe but this does not mean that your application is.
An application is thread safe if the threads in it cannot concurrently access heap state. All objects in Java (and therefore all of their fields) are created on the heap. So, if there are no fields in an object then it is thread safe.
In any practical application, objects will have state. If you can guarantee that these objects are not accessed concurrently then you have a thread safe application.
There are ways of optimizing access to shared state e.g. Atomic variables or with carful use of the volatile keyword, but I think this is going beyond what you've asked.
I hope this helps.

java threads synchronization

In the class below, is the method getIt() thread safe and why?
public class X {
private long myVar;
public void setIt(long var){
myVar = var;
}
public long getIt() {
return myVar;
}
}

It is not thread-safe. Variables of type long and double in Java are treated as two separate 32-bit variables. One thread could be writing and have written half the value when another thread reads both halves. In this situation, the reader would see a value that was never supposed to exist.
To make this thread-safe you can either declare myVar as volatile (Java 1.5 or later) or make both setIt and getIt synchronized.
Note that even if myVar was a 32-bit int you could still run into threading issues where one thread could be reading an out of date value that another thread has changed. This could occur because the value has been cached by the CPU. To resolve this, you again need to declare myVar as volatile (Java 1.5 or later) or make both setIt and getIt synchronized.
It's also worth noting that if you are using the result of getIt in a subsequent setIt call, e.g. x.setIt(x.getIt() * 2), then you probably want to synchronize across both calls:
synchronized(x)
{
x.setIt(x.getIt() * 2);
}
Without the extra synchronization, another thread could change the value in between the getIt and setIt calls causing the other thread's value to be lost.

This is not thread-safe. Even if your platform guarantees atomic writes of long, the lack of synchronized makes it possible that one thread calls setIt() and even after this call has finished it is possible that another thread can call getIt() and this call could return the old value of myVar.
The synchronized keyword does more than an exclusive access of one thread to a block or a method. It also guarantees that the second thread is informed about a change of a variable.
So you either have to mark both methods as synchronized or mark the member myVar as volatile.
There's a very good explanation about synchronization here:
Atomic actions cannot be interleaved, so they can be used without fear of thread interference. However, this does not eliminate all need to synchronize atomic actions, because memory consistency errors are still possible. Using volatile variables reduces the risk of memory consistency errors, because any write to a volatile variable establishes a happens-before relationship with subsequent reads of that same variable. This means that changes to a volatile variable are always visible to other threads. What's more, it also means that when a thread reads a volatile variable, it sees not just the latest change to the volatile, but also the side effects of the code that led up the change.

No, it's not. At least, not on platforms that lack atomic 64-bit memory accesses.
Suppose that Thread A calls setIt, copies 32 bits into memory where the backing value is, and is then pre-empted before it can copy the other 32 bits.
Then Thread B calls getIt.

No it is not, because longs are not atomic in java, so one thread could have written 32 bits of the long in the setIt method, and then the getIt could read the value, and then setIt could set the other 32 bits.
So the end result is that getIt returns a value that was never valid.

It ought to be, and generally is, but is not guaranteed to be thread safe. There could be issues with different cores having different versions in CPU cache, or the store/retrieve not being atomic for all architectures. Use the AtomicLong class.

The getter is not thread safe because it’s not guarded by any mechanism that guarantees the most up-to-date visibility. Your choices are:
making myVar final (but then you can’t mutate it)
making myVar volatile
use synchronized to accessing myVar

AFAIK, Modern JVMs no longer split long and double operations. I don't know of any reference which states this is still a problem. For example, see AtomicLong which doesn't use synchronization in Sun's JVM.
Assuming you want to be sure it is not a problem then you can use synchronize both get() and set(). However, if you are performing an operation like add, i.e. set(get()+1) then this synchronization doesn't buy you much, you still have to synchronize the object for the whole operation. (A better way around this is to use a single operation for add(n) which is synchronized)
However, a better solution is to use an AtomicLong. This supports atomic operations like get, set and add and DOESN'T use synchronization.

Since it is a read only method. You should synchronize the set method.
EDIT : I see why the get method needs to be synchronized as well. Good job explaining Phil Ross.