As I understand it, volatile helps in memory visibility and synchronized helps in achieving execution control. Volatile just guarantees that the value read by the thread would have the latest value written to it.
Consider the following:
public class Singleton{
private static volatile Singleton INSTANCE = null;
private Singleton(){}
public static Singleton getInstance(){
if(INSTANCE==null){
synchronized(Integer.class){
if(INSTANCE==null){
INSTANCE = new Singleton();
}
}
}
return INSTANCE;
}
}
In the above piece of code, we use double-checked locking. This helps us create only one instance of Singleton and this is communicated to the other threads by the creating thread as soon as possible. This is what the keyword volatile does. We need the above synchronized block because the delay in the thread reading the INSTANCE variable as null and initializing the object could cause a race condition.
Now consider the following:
public class Singleton{
private static Singleton INSTANCE = null;
private Singleton(){}
public static synchronized Singleton getInstance(){
if(INSTANCE==null){
INSTANCE = new Singleton();
}
return INSTANCE;
}
}
Say we have 2 threads t1 and t2 trying to get the Singleton object. Thread t1 enters the getInstance() method first and creates the INSTANCE object. Now this newly created object should be visible to all the other threads. If the INSTANCE variable is not volatile then how do we make sure that the object is still not in t1's memory and visible to other threads. How soon is the above INSTANCE initialized by t1 visible to other threads ?
Does this mean that it is advisable to always make variables volatile with synchronized ?
In what scenarios would we not require the variable to be volatile ?
P.S I have read other questions on StackOverflow but could not find the answer to my question. Please comment before down-voting.
My question arises from the explanation given here
I think what you're missing is this from JLS 17.4.4:
An unlock action on monitor m synchronizes-with all subsequent lock actions on m (where "subsequent" is defined according to the synchronization order).
Which is very similar to the bullet about volatile variables:
A write to a volatile variable v (§8.3.1.4) synchronizes-with all subsequent reads of v by any thread (where "subsequent" is defined according to the synchronization order).
Then in 17.4.5:
If an action x synchronizes-with a following action y, then we also have hb(x, y).
... where hb is the "happens-before" relation.
Then:
If one action happens-before another, then the first is visible to and ordered before the second.
The memory model is incredibly complicated and I don't claim to be an expert, but my understanding is that the implication of the quoted parts is that the second pattern you've shown is safe without the variable being volatile - and indeed any variable which is only modified and read within synchronization blocks for the same monitor is safe without being volatile. The more interesting aspect (to me) is what happens to the variables within the object that the variable's value refers to. If Singleton isn't immutable, you've still potentially got problems there - but that's one step removed.
To put it more concretely, if two threads call getInstance() when INSTANCE is null, one of those threads will lock the monitor first. The write action of a non-null reference to INSTANCE happens-before the unlock operation, and that unlock operation happens-before the lock operation of the other thread. The lock operation happens-before the read of the INSTANCE variable, therefore the write happens-before the read... at which point, we are guaranteed that the write is visible to the reading thread.
This explanation of what is happening here is entirely wrong, as I misunderstood the Java Memory Model. See Jon Skeet's answer.
Safe lazy initialization
The action you are attempting in this case is "lazy-initialization", and that particular pattern is useful for instances, but sub-optimal for static variables. For static variables, the lazy initialization holder class idiom is preferred.
The following quote and code block are copied directly from Item 71 of Effective Java (2nd Edition), by Josh Bloch:
Because there is no locking if the field is already initialized, it
is critical that the field be declared volatile.
// Double-check idiom for lazy initialization of instance fields
private volatile FieldType field;
FieldType getField() {
FieldType result = field;
if (result == null) { // First check (no locking)
synchronized(this) {
result = field;
if (result == null) // Second check (with locking)
field = result = computeFieldValue();
}
}
return result;
}
In one of his talks, he recommended to copy this structure exactly when performing lazy initialization for instance fields, as it is optimal in such situations, and it is very easy to break it by changing it.
What is actually happening?
EDIT: This section is incorrect.
The volatile keyword means that all read and write operations for the variable are atomic; that is, they happen as one single step from the perspective of anything else. Additionally, volatile variables are always read from and written to main memory, not processor cache. The combination of these two properties guarantees that, as soon as a volatile variable variable is modified on one thread, subsequent reads on another thread will read the updated value. This guarantee is not present for non-volatile variables.
The double-check idiom does not guarantee that only one instance is created. Rather, it is so that, once the variable is initialized, future calls to getInstance() do not need to enter a synchronized block, which is expensive.
The guarantee that it is not initialized twice is made by the fact that (a) it is a volatile field, and (b) it is checked (again) inside of the synchronized block. The outer check helps efficiency; the inner check guarantees single initialization.
I highly recommend reading Item 71 of Effective Java (2nd Edition) for a more complete explanation. I also recommend the book in general as being fantastic.
UPDATE:
The local result variable used reduces the number of accesses of the volatile field needed, which improves performance. If the local variable was left out, and all reads and writes directly accessed the volatile field, it should have the same result, but take slightly longer.
Related
From Head First design patterns book, the singleton pattern with double checked locking has been implemented as below:
public class Singleton {
private volatile static Singleton instance;
private Singleton() {}
public static Singleton getInstance() {
if (instance == null) {
synchronized (Singleton.class) {
if (instance == null) {
instance = new Singleton();
}
}
}
return instance;
}
}
I don't understand why volatile is being used. Doesn't volatile usage defeat the purpose of using double checked locking i.e performance?
A good resource for understanding why volatile is needed comes from the JCIP book. Wikipedia has a decent explanation of that material as well.
The real problem is that Thread A may assign a memory space for instance before it is finished constructing instance. Thread B will see that assignment and try to use it. This results in Thread B failing because it is using a partially constructed version of instance.
As quoted by #irreputable, volatile is not expensive. Even if it is expensive, consistency should be given priority over performance.
There is one more clean elegant way for Lazy Singletons.
public final class Singleton {
private Singleton() {}
public static Singleton getInstance() {
return LazyHolder.INSTANCE;
}
private static class LazyHolder {
private static final Singleton INSTANCE = new Singleton();
}
}
Source article : Initialization-on-demand_holder_idiom from wikipedia
In software engineering, the Initialization on Demand Holder (design pattern) idiom is a lazy-loaded singleton. In all versions of Java, the idiom enables a safe, highly concurrent lazy initialization with good performance
Since the class does not have any static variables to initialize, the initialization completes trivially.
The static class definition LazyHolder within it is not initialized until the JVM determines that LazyHolder must be executed.
The static class LazyHolder is only executed when the static method getInstance is invoked on the class Singleton, and the first time this happens the JVM will load and initialize the LazyHolder class.
This solution is thread-safe without requiring special language constructs (i.e. volatile or synchronized).
Well, there's no double-checked locking for performance. It is a broken pattern.
Leaving emotions aside, volatile is here because without it by the time second thread passes instance == null, first thread might not construct new Singleton() yet: no one promises that creation of the object happens-before assignment to instance for any thread but the one actually creating the object.
volatile in turn establishes happens-before relation between reads and writes, and fixes the broken pattern.
If you are looking for performance, use holder inner static class instead.
Declaring the variable as volatile guarantees that all accesses to it actually read its current value from memory.
Without volatile, the compiler may optimize away the memory accesses to the variable (such as keeping its value in a register), so only the first use of the variable reads the actual memory location holding the variable. This is a problem if the variable is modified by another thread between the first and second access; the first thread has only a copy of the first (pre-modified) value, so the second if statement tests a stale copy of the variable's value.
If you didn't have it, a second thread could get into the synchronized block after the first set it to null, and your local cache would still think it was null.
The first one is not for correctness (if it were you are correct that it would be self defeating) but rather for optimization.
A volatile read is not really expensive in itself.
You can design a test to call getInstance() in a tight loop, to observe the impact of a volatile read; however that test is not realistic; in such situation, programmer usually would call getInstance() once and cache the instance for the duration of use.
Another impl is by using a final field (see wikipedia). This requires an additional read, which may become more expensive than the volatile version. The final version may be faster in a tight loop, however that test is moot as previously argued.
The reason why you need volatile is because volatile has 2 semantics in Java
variable visibility between threads
stop re-ordering
So the problem without volatile in the double checked lock is that statement
instance = new Singleton()
have 3 main steps in bytecode which can be viewed by command javap -c Singleton.class
17: new #3 // class Singleton
20: dup
21: invokespecial #4 // Method "<init>":()V
Allocate memory space for the object (not initialized yet)
Create a variable to point to this space memory address
Call constructor to initialize the object
These 3 steps can be re-ordered during runtime by CPU or JVM which can be a case you will get an instance not fully initialized yet.
By having volatile JVM will insert monitorenter and monitorexit to avoid re-ordering as below.
10: monitorenter
11: getstatic #2 // Field instance:LSingleton;
14: ifnonnull 27
17: new #3 // class Singleton
20: dup
21: invokespecial #4 // Method "<init>":()V
24: putstatic #2 // Field instance:LSingleton;
27: aload_0
28: monitorexit
So volative is required for singleton.
Double checked locking is a technique to prevent creating another instance of singleton when call to getInstance method is made in multithreading environment.
Pay attention
Singleton instance is checked twice before initialization.
Synchronized critical section is used only after first checking singleton instance for that reason to improve performance.
volatile keyword on the declaration of the instance member. This will tell the compiler to always read from, and write to, main memory and not the CPU cache. With volatile variable guaranteeing happens-before relationship, all the write will happen before any read of instance variable.
Disadvantages
Since it requires the volatile keyword to work properly, it's not compatible with Java 1.4 and lower versions. The problem is that an out-of-order write may allow the instance reference to be returned before the singleton constructor is executed.
Performance issue because of decline cache for volatile variable.
Singleton instance is checked two times before initialization.
It's quite verbose and it makes the code difficult to read.
There are several realization of singleton pattern each one with advantages and disadvantages.
Eager loading singleton
Double-checked locking singleton
Initialization-on-demand holder idiom
The enum based singleton
Detailed description each of them is too verbose so I just put a link to a good article - All you want to know about Singleton
For the following simplified class:
public class MutableInteger {
private int value;
public MutableInteger(int initial) {
synchronized(this) { // is this necessary for memory visibility?
this.value = initial;
}
}
public synchronized int get() {
return this.value;
}
public synchronized void increment() {
this.value++;
}
...
}
I guess the general question is for mutable variables guarded by synchronization is it necessary to synchronize when setting the initial value in the constructor?
You're right, without the synchronized block in the constructor there is no visibility guarantee for non-final fields, as can be seen in this example.
However in practice I would rather use volatile fields or the Atomic* classes in situations like this.
Update: It is also important to mention here that in order for your program to be correctly synchronized (as defined by the JLS), you will need to publish the reference to your object in a safe manner. The cited example doesn't do that, hence why you may see the wrong value in non-final fields. But if you publish the object reference correctly (i.e. by assigning it to a final field of another object, or by creating it before calling Thread.start()), it is guaranteed that your object will be seen at least as up-to-date as the time of publishing, therefore making the synchronized block in the constructor unnecessary.
Though you've accepted an answer, let me add my two cents.
Based on what I've read, synchronization or making the field volatile would not grantee the following visibility.
A thread T1 may see a not-null value for this, but unless you've made the field value final, there's a good chance of thread T1 seeing the default value of value.
The value could be a volatile or been accessed within synchronized blocks (monitor acquire and release), either way provided that the correct execution order was followed, there's happens-before edge from the write to the read of value. There's no argument on that.
But it's not the happens before edge that we have to consider here, but the correct publication of the object itself(MutableInteger).
Creating an object is twofold where the JVM first allocates a heap space and then start initializing fields. A thread may see a not-null reference of an object but an uninitialized field of that as long as the said field is not final (Assuming reference has been correctly published).
Are multiple threads guaranteed to see the same version of a shared object to which they have a reference? Here is a code sample:
public static void main(String[] args) {
final AtomicBoolean flag = new AtomicBoolean(false);
new Thread(){
public void run() { possibly read and mutate flag }
}.start();
new Thread(){
public void run() { possibly read and mutate flag }
}.start();
while (!flag.get()) {
Thread.yield();
}
}
To be clear, I am wondering whether writes by the child threads to the shared object are seen by the parent and sibling threads.
Are multiple threads guaranteed to see the same version of a shared local variable in their scope.
In general, it depends on what you mean by "the same version". It also depends on the nature of the variable (e.g. how it is declared and initialized) ... and on how the threads use it.
(In general, Java doesn't do "versions" of variables. A thread accessing a shared variable or object will either see the latest state, or it won't. If it sees a state that isn't the latest state, then there are no guarantees as to what it will see. In particular, it may see something that doesn't directly correspond to any notional version of the object ... due to word-tearing and other cache-related memory artefacts.)
In your example you are using a final local variable within an inner class (in this case you have two anonymous inner classes). When you do that, the compiler creates a corresponding synthetic variable in the inner class that is initialized with the value of the variable in the method scope. The compiled inner class then refers to the value of the synthetic variable instead of the original variable.
In your example, it is guaranteed that the inner classes (e.g. your threads) will see the same (reference) value as in the original variable. Furthermore, it is guaranteed that they will (at least initially) see a consistent snapshot of whatever object it is that it references. (And since it is an AtomicXxxx class, it will always be consistent for all threads that can access it. Guaranteed.)
OK, so what about other cases:
If flag was a static or instance field that was also final, then we wouldn't have synthetic variables, and each nested class would be referencing the same shared variable. But it would all still work.
If flag was a static or instance field and it wasn't final, but nothing changed the field (after creating of the threads) then it would still be OK. (Though you could argue that this is fragile ... because something could change the field.)
If flag was a static or instance field and it wasn't final or volatile, then the threads would initially see the same state as the parent thread. But if either the original thread or any of the other threads changed the variable (etcetera), then the others are not guaranteed to see the new state ... unless they respective threads synchronize properly.
I would like to know if changes to flag made in one thread are seen immediately by the other two threads.
As I said above, it depends ...
In your example, the answer is "yes", because you use a final reference to AtomicBoolean.
If you had declared flag as a boolean and marked it as volatile, then the answer would be "yes".
If you had declared flag as a boolean and non-volatile, then the answer would be "no".
If flag was a final reference to an ordinary object with a mutable non-volatile boolean field, then the answer would also be "no". (The threads would all see the same object, but they wouldn't consistently see the latest state. The solution would be to use synchronized getters and setters, or equivalent.)
Yes, the two threads share the same final AtomicBoolean which is a class used to set the truth value. The variable flag itself can't be recreated because it is final. But you can perform actions on it to set value. Just like a final int[] can't be assigned to different size but you can change the value of what's inside.
final AtomicBoolean flag = new AtomicBoolean(false);
new Thread(){
public void run(){
try {
Thread.sleep(100);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
flag.set(true);
}
}.start();
new Thread(){
public void run(){
flag.set(false);
}
}.start();
Thread.sleep(200);// comment this line, you see different results
System.out.println(flag);
In this case, yes. The Java Language Specification says that calling Thread.start() synchronizes-with all previous actions on the calling thread:
An action that starts a thread synchronizes-with the first action in the thread it starts.
This creates a happens-before relationship between all writes on your main thread (including any writes the constructor of the AtomicBoolean made to initialize itself) are made visible to the thread your main thread started.
A call to start() on a thread happens-before any actions in the started thread.
So basically you are good to go. Your AtomicBoolean object is visible to both threads, and they both see the same object.
This pattern is called Safe Publication, btw. You use it to safely publish an object you create (like your AtomicBoolean) so that other threads can see it. (And yes, Thread.start() isn't on the list there of ways to safely publish an object because Thread.start() isn't general enough. But it's the same idea, and works the same way.)
The local variable is not shared1, and being final means that there would be no worry of it changing even if it was the case. (A question about member variables would result in a different response although, excluding constructor leakage, a final member would provide the same guarantees.)
The same object is shared across threads; it will be the same object and will adhere to the defined AtomicBoolean contract.
A boolean value that may be updated atomically. See the java.util.concurrent.atomic package specification for description of the properties of atomic variables.
In short the package documentation specifies the following which in turn guarantees happens-before relationships.
get has the memory effects of reading a volatile variable.
set has the memory effects of writing (assigning) a volatile variable.
There are many questions relating to the thread-safey of volatile (and AtomicXYZ objects), eg. see Is a volatile int in Java thread-safe? and Is AtomicBoolean needed to create a cancellable thread?
1 Anonymous types, including Java 8 lambdas, do not create closures/lexical bindings to variables in scope and as such are not capable of sharing local variables; rather variables are synthesized with the value of the final (or effectively final) variable from the enclosing scope which is bound when the anonymous type is instantiated.
Consider the following class:
public class MyClass
{
private MyObject obj;
public MyClass()
{
obj = new MyObject();
}
public void methodCalledByOtherThreads()
{
obj.doStuff();
}
}
Since obj was created on one thread and accessed from another, could obj be null when methodCalledByOtherThread is called? If so, would declaring obj as volatile be the best way to fix this issue? Would declaring obj as final make any difference?
Edit:
For clarity, I think my main question is:
Can other threads see that obj has been initialized by some main thread or could obj be stale (null)?
For the methodCalledByOtherThreads to be called by another thread and cause problems, that thread would have to get a reference to a MyClass object whose obj field is not initialized, ie. where the constructor has not yet returned.
This would be possible if you leaked the this reference from the constructor. For example
public MyClass()
{
SomeClass.leak(this);
obj = new MyObject();
}
If the SomeClass.leak() method starts a separate thread that calls methodCalledByOtherThreads() on the this reference, then you would have problems, but this is true regardless of the volatile.
Since you don't have what I'm describing above, your code is fine.
It depends on whether the reference is published "unsafely". A reference is "published" by being written to a shared variable; another thread reads the variable to get the reference. If there is no relationship of happens-before(write, read), the publication is called unsafe. An example of unsafe publication is through a non-volatile static field.
#chrylis 's interpretation of "unsafe publication" is not accurate. Leaking this before constructor exit is orthogonal to the concept of unsafe publication.
Through unsafe publication, another thread may observe the object in an uncertain state (hence the name); in your case, field obj may appear to be null to another thread. Unless, obj is final, then it cannot appear to be null even if the host object is published unsafely.
This is all too technical and it requires further readings to understand. The good news is, you don't need to master "unsafe publication", because it is a discouraged practice anyway. The best practice is simply: never do unsafe publication; i.e. never do data race; i.e. always read/write shared data through proper synchronization, by using synchronized, volatile or java.util.concurrent.
If we always avoid unsafe publication, do we still need final fields? The answer is no. Then why are some objects (e.g. String) designed to be "thread safe immutable" by using final fields? Because it's assumed that they can be used in malicious code that tries to create uncertain state through deliberate unsafe publication. I think this is an overblown concern. It doesn't make much sense in server environments - if an application embeds malicious code, the server is compromised, period. It probably makes a bit of sense in Applet environment where JVM runs untrusted codes from unknown sources - even then, this is an improbable attack vector; there's no precedence of this kind of attack; there are a lot of other more easily exploitable security holes, apparently.
This code is fine because the reference to the instance of MyClass can't be visible to any other threads before the constructor returns.
Specifically, the happens-before relation requires that the visible effects of actions occur in the same order as they're listed in the program code, so that in the thread where the MyClass is constructed, obj must be definitely assigned before the constructor returns, and the instantiating thread goes directly from the state of not having a reference to the MyClass object to having a reference to a fully-constructed MyClass object.
That thread can then pass a reference to that object to another thread, but all of the construction will have transitively happened-before the second thread can call any methods on it. This might happen through the constructing thread's launching the second thread, a synchronized method, a volatile field, or the other concurrency mechanisms, but all of them will ensure that all of the actions that took place in the instantiating thread are finished before the memory barrier is passed.
Note that if a reference to this gets passed out of the class inside the constructor somewhere, that reference might go floating around and get used before the constructor is finished. That's what's known as unsafe publishing of the object, but code such as yours that doesn't call non-final methods from the constructor (or directly pass out references to this) is fine.
Your other thread could see a null object. A volatile object could possibly help, but an explicit lock mechanism (or a Builder) would likely be a better solution.
Have a look at Java Concurrency in Practice - Sample 14.12
This class (if taken as is) is NOT thread safe. In two words: there is reordering of instructions in java (Instruction reordering & happens-before relationship in java) and when in your code you're instantiating MyClass, under some circumstances you may get following set of instructions:
Allocate memory for new instance of MyClass;
Return link to this block of memory;
Link to this not fully initialized MyClass is available for other threads, they can call "methodCalledByOtherThreads()" and get NullPointerException;
Initialize internals of MyClass.
In order to prevent this and make your MyClass really thread safe - you either have to add "final" or "volatile" to the "obj" field. In this case Java's memory model (starting from Java 5 on) will guarantee that during initialization of MyClass, reference to alocated for it block of memory will be returned only when all internals are initialized.
For more details I would strictly recommend you to read nice book "Java Concurrency in Practice". Exactly your case is described on the pages 50-51 (section 3.5.1). I would even say - you just can write correct multithreaded code without reading that book! :)
The originally picked answer by #Sotirios Delimanolis is wrong. #ZhongYu 's answer is correct.
There is the visibility issue of the concern here. So if MyClass is published unsafely, anything could happen.
Someone in the comment asked for evidence - one can check Listing 3.15 in the book Java Concurrency in Practice:
public class Holder {
private int n;
// Initialize in thread A
public Holder(int n) { this.n = n; }
// Called in thread B
public void assertSanity() {
if (n != n) throw new AssertionError("This statement is false.");
}
}
Someone comes up an example to verify this piece of code:
coding a proof for potential concurrency issue
As to the specific example of this post:
public class MyClass{
private MyObject obj;
// Initialize in thread A
public MyClass(){
obj = new MyObject();
}
// Called in thread B
public void methodCalledByOtherThreads(){
obj.doStuff();
}
}
If MyClass is initialized in Thread A, there is no guarantee that thread B will see this initialization (because the change might stay in the cache of the CPU that Thread A runs on and has not propagated into main memory).
Just as #ZhongYu has pointed out, because the write and read happens at 2 independent threads, so there is no happens-before(write, read) relation.
To fix this, as the original author has mentioned, we can declare private MyObject obj as volatile, which will ensure that the reference itself will be visible to other threads in timely manner
(https://www.logicbig.com/tutorials/core-java-tutorial/java-multi-threading/volatile-ref-object.html) .
I was studying this to understand the behavior of final fields in the new JMM (5 onwards). This concept is clear: guaranteed visibility of initialized final fields to all threads after the object is properly constructed.
But then at the end of the section, I read this, which simply confuses me:
Now, having said all of this, if, after a thread constructs an immutable object (that is, an object that only contains final fields), you want to ensure that it is seen correctly by all of the other thread, you still typically need to use synchronization. There is no other way to ensure, for example, that the reference to the immutable object will be seen by the second thread.
Does this means that though individual final fields (that compose an immutable object) do not have synchronization(say, visibility here) issues. But the immutable object itself when first created in a thread may not be visible (as properly created) in other threads?
If so, though we can share initialized immutable objects across threads without any thread-un-safe worries, but at the time of creation, they need 'special care' for thread safety just like for other mutables?
The semantics of final fields, as defined in section 17.5 of the JLS, guarantee that:
A thread that can only see a reference to an object after that object has been completely initialized is guaranteed to see the correctly initialized values for that object's final fields.
In other words, it says that if a thread sees a completely initialized object, then it is guaranteed to see it's final fields correctly initialized.
However, there's no guarantee about the object being visible to a given thread. It is a different problem.
If you do not use some kind of synchronization to publish a reference of your object, then the other thread might never be able to see a reference to it.
Consider the following code:
final class A {
private final int x;
A(int x) { this.x = x; }
public getX() { return x; }
}
class Main {
static volatile A a1 = null;
static A a2 = null;
public static void main(String[] args) {
new Thread(new Runnable() { void run() { try {
while (a1 == null) Thread.sleep(50);
System.out.println(a1.getX()); } catch (Throwable t) {}
}}).start()
new Thread(new Runnable() { void run() { try {
while (a2 == null) Thread.sleep(50);
System.out.println(a2.getX()); } catch (Throwable t) {}
}}).start()
a1 = new A(1); a2 = new A(1);
}
}
Note that the a1 field is volatile. This ensures that, eventually, a write to this field will be made visible to all threads reading it some time later. The field a2 is not volatile (so, a write to this field by one thread might never get noticed by other threads).
In this code, we can be sure that thread 1 will finish executing (that is, it will see that a1 != null. However, it might happen that thread 2 will halt, as it will never see the write to the field a2, since it is not volatile.
you want to ensure that it is seen correctly by all of the other thread, you still typically need to use synchronization. There is no other way to ensure, for example, that the reference to the immutable object will be seen by the second thread.
I would be slightly leery of a text that turns typically into no other way in the space of a sentence. In fact, which is true depends on what exactly we mean by "use synchronization".
The relevant parts of the Java Language Specification are:
Two actions can be ordered by a happens-before relationship. If one action happens-before another, then the first is visible to and ordered before the second.
and
More specifically, if two actions share a happens-before relationship, they do not necessarily have to appear to have happened in that order to any code with which they do not share a happens-before relationship. Writes in one thread that are in a data race with reads in another thread may, for example, appear to occur out of order to those reads.
Happens-before can be established in a number of ways:
If we have two actions x and y, we write hb(x, y) to indicate that x happens-before y.
If x and y are actions of the same thread and x comes before y in program order, then hb(x, y).
There is a happens-before edge from the end of a constructor of an object to the start of a finalizer (§12.6) for that object.
If an action x synchronizes-with a following action y, then we also have hb(x, y).
If hb(x, y) and hb(y, z), then hb(x, z).
where
Synchronization actions induce the synchronized-with relation on actions, defined as follows:
An unlock action on monitor m synchronizes-with all subsequent lock actions on m (where subsequent is defined according to the synchronization order).
A write to a volatile variable (§8.3.1.4) v synchronizes-with all subsequent reads of v by any thread (where subsequent is defined according to the synchronization order).
An action that starts a thread synchronizes-with the first action in the thread it starts.
The write of the default value (zero, false or null) to each variable synchronizes-with the first action in every thread. Although it may seem a little strange to write a default value to a variable before the object containing the variable is allocated, conceptually every object is created at the start of the program with its default initialized values.
The final action in a thread T1 synchronizes-with any action in another thread T2 that detects that T1 has terminated. T2 may accomplish this by calling T1.isAlive() or T1.join().
If thread T1 interrupts thread T2, the interrupt by T1 synchronizes-with any point where any other thread (including T2) determines that T2 has been interrupted (by having an InterruptedException thrown or by invoking Thread.interrupted or Thread.isInterrupted).
By making the fields final, you ensure their assignment happens-before the completion of the constructor. What you still need to ensure is that the completion of the constructor happens-before the object is accessed. If that access occurs in a different thread, you need to establish synchronizes-with, using any of the 6 ways shown above. Typically used are:
Start the reading thread after initialization has completed. In practice, initializing the object in the main thread prior to starting other threads accomplishes this nicely.
Declare the field that other threads use to access the object volatile. For instance:
class CacheHolder {
private static volatile Cache cache;
public static Cache instance() {
if (cache == null) {
// note that several threads may get here at the same time,
// in which case several caches will be constructed.
cache = new Cache();
}
return cache;
}
}
Do both the initial assignment and the reading of the field in a synchronized block.
class CacheHolder {
private static Cache cache;
public synchronized static Cache instance() {
if (cache == null) {
cache = new Cache();
}
return cache;
}
}
Making all the fields final will ensure they are published to other threads properly. That comment is probably referring to the following scenario:
private myField;
public void createSomething()
{
myField = new MyImmutableClass();
}
In this case you still need proper synchronization around any access to myField, or other threads might never see the newly created object.
I believe author referred to the situation when immutable object is referenced by non-final field. If reference itself is final, additional synchronization is not required.
Additional consideration is that above applies only to the object fields which are initialized inside object's constructor.