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What does "subsequent read" mean in the context of volatile variables?

Java memory visibility documentation says that:
A write to a volatile field happens-before every subsequent read of that same field.
I'm confused what does subsequent means in context of multithreading. Does this sentence implies some global clock for all processors and cores. So for example I assign value to variable in cycle c1 in some thread and then second thread is able to see this value in subsequent cycle c1 + 1?

It sounds to me like it's saying that it provides lockless acquire/release memory-ordering semantics between threads. See Jeff Preshing's article explaining the concept (mostly for C++, but the main point of the article is language neutral, about the concept of lock-free acquire/release synchronization.)
In fact Java volatile provides sequential consistency, not just acq/rel. There's no actual locking, though. See Jeff Preshing's article for an explanation of why the naming matches what you'd do with a lock.)
If a reader sees the value you wrote, then it knows that everything in the producer thread before that write has also already happened.
This ordering guarantee is only useful in combination with other guarantees about ordering within a single thread.
e.g.
int data[100];
volatile bool data_ready = false;
Producer:
data[0..99] = stuff;
// release store keeps previous ops above this line
data_ready = true;
Consumer:
while(!data_ready){} // spin until we see the write
// acquire-load keeps later ops below this line
int tmp = data[99]; // gets the value from the producer
If data_ready was not volatile, reading it wouldn't establish a happens-before relationship between two threads.
You don't have to have a spinloop, you could be reading a sequence number, or an array index from a volatile int, and then reading data[i].
I don't know Java well. I think volatile actually gives you sequential-consistency, not just release/acquire. A sequential-release store isn't allowed to reorder with later loads, so on typical hardware it needs an expensive memory barrier to make sure the local core's store buffer is flushed before any later loads are allowed to execute.
Volatile Vs Atomic explains more about the ordering volatile gives you.
Java volatile is just an ordering keyword; it's not equivalent to C11 _Atomic or C++11 std::atomic<T> which also give you atomic RMW operations. In Java, volatile_var++ is not an atomic increment, it a separate load and store, like volatile_var = volatile_var + 1. In Java, you need a class like AtomicInteger to get an atomic RMW.
And note that C/C++ volatile doesn't imply atomicity or ordering at all; it only tells the compiler to assume that the value can be modified asynchronously. This is only a small part of what you need to write lockless for anything except the simplest cases.

It means that once a certain Thread writes to a volatile field, all other Thread(s) will observe (on the next read) that written value; but this does not protect you against races though.
Threads have their caches, and those caches will be invalidated and updated with that newly written value via cache coherency protocol.
EDIT
Subsequent means whenever that happens after the write itself. Since you don't know the exact cycle/timing when that will happen, you usually say when some other thread observes the write, it will observer all the actions done before that write; thus a volatile establishes the happens-before guarantees.
Sort of like in an example:
// Actions done in Thread A
int a = 2;
volatile int b = 3;
// Actions done in Thread B
if(b == 3) { // observer the volatile write
// Thread B is guaranteed to see a = 2 here
}
You could also loop (spin wait) until you see 3 for example.

Peter's answer gives the rationale behind the design of the Java memory model.
In this answer I'm attempting to give an explanation using only the concepts defined in the JLS.
In Java every thread is composed by a set of actions.
Some of these actions have the potential to be observable by other threads (e.g. writing a shared variable), these
are called synchronization actions.
The order in which the actions of a thread are written in the source code is called the program order.
An order defines what is before and what is after (or better, not before).
Within a thread, each action has a happens-before relationship (denoted by <) with the next (in program order) action.
This relationship is important, yet hard to understand, because it's very fundamental: it guarantees that if A < B then
the "effects" of A are visible to B.
This is indeed what we expect when writing the code of a function.
Consider
Thread 1 Thread 2
A0 A'0
A1 A'1
A2 A'2
A3 A'3
Then by the program order we know A0 < A1 < A2 < A3 and that A'0 < A'1 < A'2 < A'3.
We don't know how to order all the actions.
It could be A0 < A'0 < A'1 < A'2 < A1 < A2 < A3 < A'3 or the sequence with the primes swapped.
However, every such sequence must have that the single actions of each thread are ordered according to the thread's program order.
The two program orders are not sufficient to order every action, they are partial orders, in opposition of the
total order we are looking for.
The total order that put the actions in a row according to a measurable time (like a clock) they happened is called the execution order.
It is the order in which the actions actually happened (it is only requested that the actions appear to be happened in
this order, but that's just an optimization detail).
Up until now, the actions are not ordered inter-thread (between two different threads).
The synchronization actions serve this purpose.
Each synchronization action synchronizes-with at least another synchronization action (they usually comes in pairs, like
a write and a read of a volatile variable, a lock and the unlock of a mutex).
The synchronize-with relationship is the happens-before between thread (the former implies the latter), it is exposed as
a different concept because 1) it slightly is 2) happens-before are enforced naturally by the hardware while synchronize-with
may require software intervention.
happens-before is derived from the program order, synchronize-with from the synchronization order (denoted by <<).
The synchronization order is defined in terms of two properties: 1) it is a total order 2) it is consistent with each thread's
program order.
Let's add some synchronization action to our threads:
Thread 1 Thread 2
A0 A'0
S1 A'1
A1 S'1
A2 S'2
S2 A'3
The program orders are trivial.
What is the synchronization order?
We are looking for something that by 1) includes all of S1, S2, S'1 and S'2 and by 2) must have S1 < S2 and S'1 < S'2.
Possible outcomes:
S1 < S2 < S'1 < S'2
S1 < S'1 < S'2 < S2
S'1 < S1 < S'2 < S'2
All are synchronization orders, there is not one synchronization order but many, the question of above is wrong, it
should be "What are the synchronization orders?".
If S1 and S'1 are so that S1 << S'1 than we are restricting the possible outcomes to the ones where S1 < S'2 so the
outcome S'1 < S1 < S'2 < S'2 of above is now forbidden.
If S2 << S'1 then the only possible outcome is S1 < S2 < S'1 < S'2, when there is only a single outcome I believe we have
sequential consistency (the converse is not true).
Note that if A << B these doesn't mean that there is a mechanism in the code to force an execution order where A < B.
Synchronization actions are affected by the synchronization order they do not impose any materialization of it.
Some synchronization actions (e.g. locks) impose a particular execution order (and thereby a synchronization order) but some don't (e.g. reads/writes of volatiles).
It is the execution order that create the synchronization order, this is completely orthogonal to the synchronize-with relationship.
Long story short, the "subsequent" adjective refers to any synchronization order, that is any valid (according to each thread
program order) order that encompasses all the synchronization actions.
The JLS then continues defining when a data race happens (when two conflicting accesses are not ordered by happens-before)
and what it means to be happens-before consistent.
Those are out of scope.

I'm confused what does subsequent means in context of multithreading. Does this sentence implies some global clock for all processors and cores...?
Subsequent means (according to the dictionary) coming after in time. There certainly is a global clock across all CPUs in a computer (think X Ghz) and the document is trying to say that if thread-1 did something at clock tick 1 then thread-2 does something on another CPU at clock tick 2, it's actions are considered subsequent.
A write to a volatile field happens-before every subsequent read of that same field.
The key phrase that could be added to this sentence to make it more clear is "in another thread". It might make more sense to understand it as:
A write to a volatile field happens-before every subsequent read of that same field in another thread.
What this is saying that if a read of a volatile field happens in Thread-2 after (in time) the write in Thread-1, then Thread-2 will be guaranteed to see the updated value. Further up in the documentation you point to is the section (emphasis mine):
... 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.
Notice the highlighted phrase. The Java compiler is free to reorder instructions in any one thread's execution for optimization purposes as long as the reordering doesn't violate the definition of the language – this is called execution order and is critically different than program order.
Let's look at the following example with variables a and b that are non-volatile ints initialized to 0 with no synchronized clauses. What is shown is program order and the time in which the threads are encountering the lines of code.
Time Thread-1 Thread-2
1 a = 1;
2 b = 2;
3 x = a;
4 y = b;
5 c = a + b; z = x + y;
If Thread-1 adds a + b at Time 5, it is guaranteed to be 3. However, if Thread-2 adds x + y at Time 5, it might get 0, 1, 2, or 3 depends on race conditions. Why? Because the compiler might have reordered the instructions in Thread-1 to set a after b because of efficiency reasons. Also, Thread-1 may not have appropriately published the values of a and b so that Thread-2 might get out of date values. Even if Thread-1 gets context-switched out or crosses a write memory barrier and a and b are published, Thread-2 needs to cross a read barrier to update any cached values of a and b.
If a and b were marked as volatile then the write to a must happen-before (in terms of visibility guarantees) the subsequent read of a on line 3 and the write to b must happen-before the subsequent read of b on line 4. Both threads would get 3.
We use volatile and synchronized keywords in java to ensure happens-before guarantees. A write memory barrier is crossed when assigning a volatile or exiting a synchronized block and a read barrier is crossed when reading a volatile or entering a synchronized block. The Java compiler cannot reorder write instructions past these memory barriers so the order of updates is assured. These keywords control instruction reordering and insure proper memory synchronization.
NOTE: volatile is unnecessary in a single-threaded application because program order assures the reads and writes will be consistent. A single-threaded application might see any value of (non-volatile) a and b at times 3 and 4 but it always sees 3 at Time 5 because of language guarantees. So although use of volatile changes the reordering behavior in a single-threaded application, it is only required when you share data between threads.

This is more a definition of what will not happen rather than what will happen.
Essentially it is saying that once a write to an atomic variable has happened there cannot be any other thread that, on reading the variable, will read a stale value.
Consider the following situation.
Thread A is continuously incrementing an atomic value a.
Thread B occasionally reads A.a and exposes that value as a
non-atomic b variable.
Thread C occasionally reads both A.a and B.b.
Given that a is atomic it is possible to reason that from the point of view of C, b may occasionally be less than a but will never be greater than a.
If a was not atomic no such guarantee could be given. Under certain caching situations it would be quite possible for C to see b progress beyond a at any time.
This is a simplistic demonstration of how the Java memory model allows you to reason about what can and cannot happen in a multi-threaded environment. In real life the potential race conditions between reading and writing to data structures can be much more complex but the reasoning process is the same.

how synchronized keyword works internally

I read the below program and answer in a blog.
int x = 0;
boolean bExit = false;
Thread 1 (not synchronized)
x = 1;
bExit = true;
Thread 2 (not synchronized)
if (bExit == true)
System.out.println("x=" + x);
is it possible for Thread 2 to print “x=0”?
Ans : Yes ( reason : Every thread has their own copy of variables. )
how do you fix it?
Ans: By using make both threads synchronized on a common mutex or make both variable volatile.
My doubt is : If we are making the 2 variable as volatile then the 2 threads will share the variables from the main memory. This make a sense, but in case of synchronization how it will be resolved as both the thread have their own copy of variables.
Please help me.

This is actually more complicated than it seems. There are several arcane things at work.
Caching
Saying "Every thread has their own copy of variables" is not exactly correct. Every thread may have their own copy of variables, and they may or may not flush these variables into the shared memory and/or read them from there, so the whole thing is non-deterministic. Moreover, the very term flushing is really implementation-dependent. There are strict terms such as memory consistency, happens-before order, and synchronization order.
Reordering
This one is even more arcane. This
x = 1;
bExit = true;
does not even guarantee that Thread 1 will first write 1 to x and then true to bExit. In fact, it does not even guarantee that any of these will happen at all. The compiler may optimize away some values if they are not used later. The compiler and CPU are also allowed to reorder instructions any way they want, provided that the outcome is indistinguishable from what would happen if everything was really in program order. That is, indistinguishable for the current thread! Nobody cares about other threads until...
Synchronization comes in
Synchronization does not only mean exclusive access to resources. It is also not just about preventing threads from interfering with each other. It's also about memory barriers. It can be roughly described as each synchronization block having invisible instructions at the entry and exit, the first one saying "read everything from the shared memory to be as up-to-date as possible" and the last one saying "now flush whatever you've been doing there to the shared memory". I say "roughly" because, again, the whole thing is an implementation detail. Memory barriers also restrict reordering: actions may still be reordered, but the results that appear in the shared memory after exiting the synchronized block must be identical to what would happen if everything was indeed in program order.
All that only works, of course, only if both blocks use the same locking object.
The whole thing is described in details in Chapter 17 of the JLS. In particular, what's important is the so-called "happens-before order". If you ever see in the documentation that "this happens-before that", it means that everything the first thread does before "this" will be visible to whoever does "that". This may even not require any locking. Concurrent collections are a good example: one thread puts there something, another one reads that, and that magically guarantees that the second thread will see everything the first thread did before putting that object into the collection, even if those actions had nothing to do with the collection itself!
Volatile variables
One last warning: you better give up on the idea that making variables volatile will solve things. In this case maybe making bExit volatile will suffice, but there are so many troubles that using volatiles can lead to that I'm not even willing to go into that. But one thing is for sure: using synchronized has much stronger effect than using volatile, and that goes for memory effects too. What's worse, volatile semantics changed in some Java version so there may exist some versions that still use the old semantics which was even more obscure and confusing, whereas synchronized always worked well provided you understand what it is and how to use it.
Pretty much the only reason to use volatile is performance because synchronized may cause lock contention and other troubles. Read Java Concurrency in Practice to figure all that out.
Q & A
1) You wrote "now flush whatever you've been doing there to the shared
memory" about synchronized blocks. But we will see only the variables
that we access in the synchronize block or all the changes that the
thread call synchronize made (even on the variables not accessed in the
synchronized block)?
Short answer: it will "flush" all variables that were updated during the synchronized block or before entering the synchronized block. And again, because flushing is an implementation detail, you don't even know whether it will actually flush something or do something entirely different (or doesn't do anything at all because the implementation and the specific situation already somehow guarantee that it will work).
Variables that wasn't accessed inside the synchronized block obviously won't change during the execution of the block. However, if you change some of those variables before entering the synchronized block, for example, then you have a happens-before relationship between those changes and whatever happens in the synchronized block (the first bullet in 17.4.5). If some other thread enters another synchronized block using the same lock object then it synchronizes-with the first thread exiting the synchronized block, which means that you have another happens-before relationship here. So in this case the second thread will see the variables that the first thread updated prior to entering the synchronized block.
If the second thread tries to read those variables without synchronizing on the same lock, then it is not guaranteed to see the updates. But then again, it isn't guaranteed to see the updates made inside the synchronized block as well. But this is because of the lack of the memory-read barrier in the second thread, not because the first one didn't "flush" its variables (memory-write barrier).
2) In this chapter you post (of JLS) it is written that: "A write to a
volatile field (§8.3.1.4) happens-before every subsequent read of that
field." Doesn't this mean that when the variable is volatile you will
see only changes of it (because it is written write happens-before
read, not happens-before every operation between them!). I mean
doesn't this mean that in the example, given in the description of the
problem, we can see bExit = true, but x = 0 in the second thread if
only bExit is volatile? I ask, because I find this question here: http://java67.blogspot.bg/2012/09/top-10-tricky-java-interview-questions-answers.html
and it is written that if bExit is volatile the program is OK. So the
registers will flush only bExits value only or bExits and x values?
By the same reasoning as in Q1, if you do bExit = true after x = 1, then there is an in-thread happens-before relationship because of the program order. Now since volatile writes happen-before volatile reads, it is guaranteed that the second thread will see whatever the first thread updated prior to writing true to bExit. Note that this behavior is only since Java 1.5 or so, so older or buggy implementations may or may not support this. I have seen bits in the standard Oracle implementation that use this feature (java.concurrent collections), so you can at least assume that it works there.
3) Why monitor matters when using synchronized blocks about memory
visibility? I mean when try to exit synchronized block aren't all
variables (which we accessed in this block or all variables in the
thread - this is related to the first question) flushed from registers
to main memory or broadcasted to all CPU caches? Why object of
synchronization matters? I just cannot imagine what are relations and
how they are made (between object of synchronization and memory).
I know that we should use the same monitor to see this changes, but I
don't understand how memory that should be visible is mapped to
objects. Sorry, for the long questions, but these are really
interesting questions for me and it is related to the question (I
would post questions exactly for this primer).
Ha, this one is really interesting. I don't know. Probably it flushes anyway, but Java specification is written with high abstraction in mind, so maybe it allows for some really weird hardware where partial flushes or other kinds of memory barriers are possible. Suppose you have a two-CPU machine with 2 cores on each CPU. Each CPU has some local cache for every core and also a common cache. A really smart VM may want to schedule two threads on one CPU and two threads on another one. Each pair of the threads uses its own monitor, and VM detects that variables modified by these two threads are not used in any other threads, so it only flushes them as far as the CPU-local cache.
See also this question about the same issue.
4) I thought that everything before writing a volatile will be up to
date when we read it (moreover when we use volatile a read that in
Java it is memory barrier), but the documentation don't say this.
It does:
17.4.5.
If x and y are actions of the same thread and x comes before y in program order, then hb(x, y).
If hb(x, y) and hb(y, z), then hb(x, z).
A write to a volatile field (§8.3.1.4) happens-before every subsequent
read of that field.
If x = 1 comes before bExit = true in program order, then we have happens-before between them. If some other thread reads bExit after that, then we have happens-before between write and read. And because of the transitivity, we also have happens-before between x = 1 and read of bExit by the second thread.
5) Also, if we have volatile Person p does we have some dependency
when we use p.age = 20 and print(p.age) or have we memory barrier in
this case(assume age is not volatile) ? - I think - No
You are correct. Since age is not volatile, then there is no memory barrier, and that's one of the trickiest things. Here is a fragment from CopyOnWriteArrayList, for example:
Object[] elements = getArray();
E oldValue = get(elements, index);
if (oldValue != element) {
int len = elements.length;
Object[] newElements = Arrays.copyOf(elements, len);
newElements[index] = element;
setArray(newElements);
} else {
// Not quite a no-op; ensures volatile write semantics
setArray(elements);
Here, getArray and setArray are trivial setter and getter for the array field. But since the code changes elements of the array, it is necessary to write the reference to the array back to where it came from in order for the changes to the elements of the array to become visible. Note that it is done even if the element being replaced is the same element that was there in the first place! It is precisely because some fields of that element may have changed by the calling thread, and it's necessary to propagate these changes to future readers.
6) And is there any happens before 2 subsequent reads of volatile
field? I mean does the second read will see all changes from thread
which reads this field before it(of course we will have changes only
if volatile influence visibility of all changes before it - which I am
a little confused whether it is true or not)?
No, there is no relationship between volatile reads. Of course, if one thread performs a volatile write and then two other thread perform volatile reads, they are guaranteed to see everything at least up to date as it was before the volatile write, but there is no guarantee of whether one thread will see more up-to-date values than the other. Moreover, there is not even strict definition of one volatile read happening before another! It is wrong to think of everything happening on a single global timeline. It is more like parallel universes with independent timelines that sometimes sync their clocks by performing synchronization and exchanging data with memory barriers.

It depends on the implementation which decides if threads will keep a copy of the variables in their own memory. In case of class level variables threads have a shared access and in case of local variables threads will keep a copy of it. I will provide two examples which shows this fact , please have a look at it.
And in your example if I understood it correctly your code should look something like this--
package com.practice.multithreading;
public class LocalStaticVariableInThread {
static int x=0;
static boolean bExit = false;
public static void main(String[] args) {
Thread t1=new Thread(run1);
Thread t2=new Thread(run2);
t1.start();
t2.start();
}
static Runnable run1=()->{
x = 1;
bExit = true;
};
static Runnable run2=()->{
if (bExit == true)
System.out.println("x=" + x);
};
}
Output
x=1
I am getting this output always. It is because the threads share the variable and the when it is changed by one thread other thread can see it. But in real life scenarios we can never say which thread will start first, since here the threads are not doing anything we can see the expected result.
Now take this example--
Here if you make the i variable inside the for-loop` as static variable then threads won t keep a copy of it and you won t see desired outputs, i.e. the count value will not be 2000 every time even if u have synchronized the count increment.
package com.practice.multithreading;
public class RaceCondition2Fixed {
private int count;
int i;
/*making it synchronized forces the thread to acquire an intrinsic lock on the method, and another thread
cannot access it until this lock is released after the method is completed. */
public synchronized void increment() {
count++;
}
public static void main(String[] args) {
RaceCondition2Fixed rc= new RaceCondition2Fixed();
rc.doWork();
}
private void doWork() {
Thread t1 = new Thread(new Runnable() {
#Override
public void run() {
for ( i = 0; i < 1000; i++) {
increment();
}
}
});
Thread t2 = new Thread(new Runnable() {
#Override
public void run() {
for ( i = 0; i < 1000; i++) {
increment();
}
}
});
t1.start();
t2.start();
try {
t1.join();
t2.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
/*if we don t use join then count will be 0. Because when we call t1.start() and t2.start()
the threads will start updating count in the spearate threads, meanwhile the main thread will
print the value as 0. So. we need to wait for the threads to complete. */
System.out.println(Thread.currentThread().getName()+" Count is : "+count);
}
}

Does synchronized keyword prevent reordering in Java?

Suppose I have the following code in Java
a = 5;
synchronized(lock){
b = 5;
}
c = 5;
Does synchronized prevent reordering? There is no dependency between a, b and c. Would assignment to a first happen then to b and then to c? If I did not have synchronized, the statements can be reordered in any way the JVM chooses right?

Locking the assignment to b will, at the very least, introduce an acquire-fence before the assignment, and a release-fence after the assignment.
This prevents instructions after the acquire-fence to be moved above the fence, and instructions before the release-fence to be moved below the fence.
Using the ↓↑ notation:
a = 5;
↓
b = 5;
↑
c = 5;
The ↓ prevents instructions from being moved above it.
The ↑ prevents instructions from being moved below it.

Does synchronized prevent reordering?
It prevents some re-ordering. You can still have re-ordering outside the synchronized block and inside the synchronized block, but not from inside a synchronized block, to outside it.
There is no dependency between a, b and c.
That makes no difference.
Would assignment to a first happen then to b and then to c?
Yes. But as has been noted, this is not guaranteed for all JVMs. (See below)
If I did not have synchronized, the statements can be reordered in any way the JVM chooses right?
Yes, by the JVM and/or the CPU instruction optimiser and/or CPU cache, but it is unlikely given there is no obvious reason to suspect that changing the order of a = 5; and b = 5; will improve performance.
What you could see is a change of visibility for the cache. i.e. another thread reading these values could see the b = 5; before a = 5; e.g. they are on different cache lines, if it is not also synchronized.

Does synchronized prevent reordering?
Partially, see below.
Would assignment to a first happen then to b and then to c?
No. As dcastro pointed out, actions can be moved into synchronized blocks. So the compiler would be allowed to generate code, that corresponds to the following statements:
synchronized (lock){
a = 5;
b = 5;
c = 5;
}
And the compiler is also allowed to reorder statements within a synchronized block, that have no dependency to each other. So the compiler can also generate code that corresponds to the following statements:
synchronized (lock){
c = 5;
b = 5;
a = 5;
}
If I did not have synchronized, the statements can be reordered in any way the JVM chooses right?
Well, I think that's the wrong question, and it's also the wrong way to think about the Java Memory Model. The Java Memory Model is not defined in terms of reorderings. Actually, it's much easier than most people think. Basically, there is only one important rule, that you can find in §17.4.5 in the Java Language Specification:
A program is correctly synchronized if and only if all sequentially consistent executions are free of data races. If a program is correctly synchronized, then all executions of the program will appear to be sequentially consistent.
In other words: If you completely ignore reorderings, and for all possible executions of the program, there is no way that two threads access the same memory location, which is neither volatile nor atomic, and at least one of the actions is a write operation, then all executions will appear, as if there are no reorderings.
In short: Avoid data races, and you will never observe a reordering.

Double check locking and code reordering in Java [duplicate]

This question already has answers here:
Why is volatile used in double checked locking
(8 answers)
Closed 4 years ago.
In some article I read that double check locking is broken. As the compiler can reorder the sequence of constructors.
Ss allocate memory for an object
Then return the address to a reference variable
Then initialize the state of the object
While typically one would expect:
It should be as allocated memory for the object
then initialize the state of object
then return the address to the reference variable.
Again, when using the synchronized keyword, the code reorder never happens as per JMM specification.
Why is compiler reordering the sequence of constructor events when it is inside the synchronized() block?
I saw a lot of posts here about DCL, but I am expecting the description based on JMM and compiler reordering.

The compiler is free to reorder instructions within a synchronized block. And the compiler is free to reorder instructions before (as long as they stay before) or after (as long as they stay after) the synchronized block. However, the compiler is not free to reorder instructions across the synchronized block boundaries (block start or block end).
Thus, the construction and assignment which are wholly within the synchronized block can be reordered, and an outside viewer which has not correctly synchronized can see the assignment before the construction.

First of all:
Again when using the synchronized keyword, the code reorder never happens as per the JMM specification.
The above statement is not fully accurate. The JMM defined the happens-before relationship.
The JLS only defines the program order and happens-before order. See 17.4.5. Happens-before Order.
It has effects on the reordering of instructions. For example,
int x = 1;
synch(obj) {
y = 2;
}
int z = 3;
Now for the above piece of code, the below types of reordering are possible.
synch(obj) {
int x = 1;
y = 2;
int z = 3;
}
The above is a valid reordering.
See Roach Motels and The Java Memory Model.
synch(obj) {
int z = 3;
y = 2;
int x = 1;
}
The above is also a valid reordering.
What is not possible is that y=2 will only be executed after the lock has been acquired and before the lock is released this is what guaranteed given by JMM. Also to see the correct effects from another thread, we need to access y inside the synchronized block only.
Now I come to DCL.
See the code of DCL.
if (singleton == null)
synch(obj) {
if(singleton == null) {
singleton == new Singleton()
}
}
return singleton;
Now the problem in the above approach is:
singleton = new Singleton() is not a single instruction. But a set of instructions. It is quite possible that a singleton reference is assigned an object reference first, before fully initializing the constructor.
So if 1 happens then it's quite possible the other thread reads a singleton reference as a non null and thus is seeing a partially constructed object.
The above effects can be controlled by making a singleton as volatile which also establishes happens-before guarantees and visibility.

Why is compiler reordering the sequence of constructor events when it is inside the synchronized() block?
It would typically do this to make the code run faster.
The Java Language Specification (JLS) says that the implementation (for example, the compiler) is allowed to reorder instructions and sequences of instructions subject to certain constraints.
The problem is that the broken variants of DCL make assumptions that fall outside of what the JLS says can be made. The result is an execution that the JLS says is not well-formed. Whether this manifests itself as an actual bug / unexpected behaviour depends on the compiler version, the hardware and various other things.
But the point is that the compiler isn't doing anything wrong. The fault is in the DCL code.
I just want to add that the JIT compiler is often not reordering the events per se. what it is often doing is removing constraints on hardware-level memory read/write actions. For example, by removing the constraint that a particular memory write is flushed to main memory, you allow the hardware to defer (or even skip entirely) a slow write-to-memory, and just write to the L1 cache. By contrast, the end of a synchronized block will force the cached writes to main memory, incurring extra memory traffic and (probably) a pipeline stalls.

Interpretation of "program order rule" in Java concurrency

Program order rule states "Each action in a thread happens-before every action in that thread that comes later in the program order"
1.I read in another thread that an action is
reads and writes to variables
locks and unlocks of monitors
starting and joining with threads
Does this mean that reads and writes can be changed in order, but reads and writes cannot change order with actions specified in 2nd or 3rd lines?
2.What does "program order" mean?
Explanation with an examples would be really helpful.
Additional related question
Suppose I have the following code:
long tick = System.nanoTime(); //Line1: Note the time
//Block1: some code whose time I wish to measure goes here
long tock = System.nanoTime(); //Line2: Note the time
Firstly, it's a single threaded application to keep things simple. Compiler notices that it needs to check the time twice and also notices a block of code that has no dependency with surrounding time-noting lines, so it sees a potential to reorganize the code, which could result in Block1 not being surrounded by the timing calls during actual execution (for instance, consider this order Line1->Line2->Block1). But, I as a programmer can see the dependency between Line1,2 and Block1. Line1 should immediately precede Block1, Block1 takes a finite amount of time to complete, and immediately succeeded by Line2.
So my question is: Am I measuring the block correctly?
If yes, what is preventing the compiler from rearranging the order.
If no, (which is think is correct after going through Enno's answer) what can I do to prevent it.
P.S.: I stole this code from another question I asked in SO recently.

It probably helps to explain why such rule exist in the first place.
Java is a procedural language. I.e. you tell Java how to do something for you. If Java executes your instructions not in the order you wrote, it would obviously not work. E.g. in the below example, if Java would do 2 -> 1 -> 3 then the stew would be ruined.
1. Take lid off
2. Pour salt in
3. Cook for 3 hours
So, why does the rule not simply say "Java executes what you wrote in the order you wrote"? In a nutshell, because Java is clever. Take the following example:
1. Take eggs out of the freezer
2. Take lid off
3. Take milk out of the freezer
4. Pour egg and milk in
5. Cook for 3 hours
If Java was like me, it'll just execute it in order. However Java is clever enough to understand that it's more efficient AND that the end result would be the same should it do 1 -> 3 -> 2 -> 4 -> 5 (you don't have to walk to the freezer again, and that doesn't change the recipe).
So what the rule "Each action in a thread happens-before every action in that thread that comes later in the program order" is trying to say is, "In a single thread, your program will run as if it was executed in the exact order you wrote it. We might change the ordering behind the scene but we make sure that none of that would change the output.
So far so good. Why does it not do the same across multiple threads? In multi-thread programming, Java isn't clever enough to do it automatically. It will for some operations (e.g. joining threads, starting threads, when a lock (monitor) is used etc.) but for other stuff you need to explicitly tell it to not do reordering that would change the program output (e.g. volatile marker on fields, use of locks etc.).
Note:
Quick addendum about "happens-before relationship". This is a fancy way of saying no matter what reordering Java might do, stuff A will happen before stuff B. In our weird later stew example, "Step 1 & 3 happens-before step 4 "Pour egg and milk in" ". Also for example, "Step 1 & 3 do not need a happens-before relationship because they don't depend on each other in any way"
On the additional question & response to the comment
First, let us establish what "time" means in the programming world. In programming, we have the notion of "absolute time" (what's the time in the world now?) and the notion of "relative time" (how much time has passed since x?). In an ideal world, time is time but unless we have an atomic clock built in, the absolute time would have to be corrected time to time. On the other hand, for relative time we don't want corrections as we are only interested in the differences between events.
In Java, System.currentTime() deals with absolute time and System.nanoTime() deals with relative time. This is why the Javadoc of nanoTime states, "This method can only be used to measure elapsed time and is not related to any other notion of system or wall-clock time".
In practice, both currentTimeMillis and nanoTime are native calls and thus the compiler can't practically prove if a reordering won't affect the correctness, which means it will not reorder the execution.
But let us imagine we want to write a compiler implementation that actually looks into native code and reorders everything as long as it's legal. When we look at the JLS, all that it tells us is that "You can reorder anything as long as it cannot be detected". Now as the compiler writer, we have to decide if the reordering would violate the semantics. For relative time (nanoTime), it would clearly be useless (i.e. violates the semantics) if we'd reorder the execution. Now, would it violate the semantics if we'd reorder for absolute time (currentTimeMillis)? As long as we can limit the difference from the source of the world's time (let's say the system clock) to whatever we decide (like "50ms")*, I say no. For the below example:
long tick = System.currentTimeMillis();
result = compute();
long tock = System.currentTimeMillis();
print(result + ":" + tick - tock);
If the compiler can prove that compute() takes less than whatever maximum divergence from the system clock we can permit, then it would be legal to reorder this as follows:
long tick = System.currentTimeMillis();
long tock = System.currentTimeMillis();
result = compute();
print(result + ":" + tick - tock);
Since doing that won't violate the spec we defined, and thus won't violate the semantics.
You also asked why this is not included in the JLS. I think the answer would be "to keep the JLS short". But I don't know much about this realm so you might want to ask a separate question for that.
*: In actual implementations, this difference is platform dependent.

The program order rule guarantees that, within individual threads, reordering optimizations introduced by the compiler cannot produce different results from what would have happened if the program had been executed in serial fashion. It makes no guarantees about what order the thread's actions may appear to occur in to any other threads if its state is observed by those threads without synchronization.
Note that this rule speaks only to the ultimate results of the program, and not to the order of individual executions within that program. For instance, if we have a method which makes the following changes to some local variables:
x = 1;
z = z + 1;
y = 1;
The compiler remains free to reorder these operations however it sees best fit to improve performance. One way to think of this is: if you could reorder these ops in your source code and still obtain the same results, the compiler is free to do the same. (And in fact, it can go even further and completely discard operations which are shown to have no results, such as invocations of empty methods.)
With your second bullet point the monitor lock rule comes into play: "An unlock on a monitor happens-before every subsequent lock on that main monitor lock." (Java Concurrency in Practice p. 341) This means that a thread acquiring a given lock will have a consistent view of the actions which occurred in other threads before releasing that lock. However, note that this guarantee only applies when two different threads release or acquire the same lock. If Thread A does a bunch of stuff before releasing Lock X, and then Thread B acquires Lock Y, Thread B is not assured to have a consistent view of A's pre-X actions.
It is possible for reads and writes to variables to be reordered with start and join if a.) doing so doesn't break within-thread program order, and b.) the variables have not had other "happens-before" thread synchronization semantics applied to them, say by storing them in volatile fields.
A simple example:
class ThreadStarter {
Object a = null;
Object b = null;
Thread thread;
ThreadStarter(Thread threadToStart) {
this.thread = threadToStart;
}
public void aMethod() {
a = new BeforeStartObject();
b = new BeforeStartObject();
thread.start();
a = new AfterStartObject();
b = new AfterStartObject();
a.doSomeStuff();
b.doSomeStuff();
}
}
Since the fields a and b and the method aMethod() are not synchronized in any way, and the action of starting thread does not change the results of the writes to the fields (or the doing of stuff with those fields), the compiler is free to reorder thread.start() to anywhere in the method. The only thing it could not do with the order of aMethod() would be to move the order of writing one of the BeforeStartObjects to a field after writing an AfterStartObject to that field, or to move one of the doSomeStuff() invocations on a field before the AfterStartObject is written to it. (That is, assuming that such reordering would change the results of the doSomeStuff() invocation in some way.)
The critical thing to bear in mind here is that, in the absence of synchronization, the thread started in aMethod() could theoretically observe either or both of the fields a and b in any of the states which they take on during the execution of aMethod() (including null).
Additional question answer
The assignments to tick and tock cannot be reordered with respect to the code in Block1 if they are to be actually used in any measurements, for example by calculating the difference between them and printing the result as output. Such reordering would clearly break Java's within-thread as-if-serial semantics. It changes the results from what would have been obtained by executing instructions in the specified program order. If the assignments aren't used for any measurements and have no side-effects of any kind on the program result, they'll likely be optimized away as no-ops by the compiler rather than being reordered.

Before I answer the question,
reads and writes to variables
Should be
volatile reads and volatile writes (of the same field)
Program order doesn't guarantee this happens before relationship, rather the happens-before relationship guarantees program order
To your questions:
Does this mean that reads and writes can be changed in order, but reads and writes cannot change order with actions specified in 2nd or 3rd lines?
The answer actually depends on what action happens first and what action happens second. Take a look at the JSR 133 Cookbook for Compiler Writers. There is a Can Reorder grid that lists the allowed compiler reordering that can occur.
For instance a Volatile Store can be re-ordered above or below a Normal Store but a Volatile Store cannot be be reordered above or below a Volatile Load. This is all assuming intrathread semantics still hold.
What does "program order" mean?
This is from the JLS
Among all the inter-thread actions performed by each thread t, the
program order of t is a total order that reflects the order in which
these actions would be performed according to the intra-thread
semantics of t.
In other words, if you can change the writes and loads of a variable in such a way that it will preform exactly the same way as you wrote it then it maintains program order.
For instance
public static Object getInstance(){
if(instance == null){
instance = new Object();
}
return instance;
}
Can be reordered to
public static Object getInstance(){
Object temp = instance;
if(instance == null){
temp = instance = new Object();
}
return temp;
}

it simply mean though the thread may be multiplxed, but the internal order of the thread's action/operation/instruction would remain constant (relatively)
thread1: T1op1, T1op2, T1op3...
thread2: T2op1, T2op2, T2op3...
though the order of operation (Tn'op'M) among thread may vary, but operations T1op1, T1op2, T1op3 within a thread will always be in this order, and so as the T2op1, T2op2, T2op3
for ex:
T2op1, T1op1, T1op2, T2op2, T2op3, T1op3

Java tutorial http://docs.oracle.com/javase/tutorial/essential/concurrency/memconsist.html says that happens-before relationship is simply a guarantee that memory writes by one specific statement are visible to another specific statement. Here is an illustration
int x;
synchronized void x() {
x += 1;
}
synchronized void y() {
System.out.println(x);
}
synchronized creates a happens-before relationship, if we remove it there will be no guarantee that after thread A increments x thread B will print 1, it may print 0