I've got few questions about threads in Java. Here is the code:
TestingThread class:
public class TestingThread implements Runnable {
Thread t;
volatile boolean pause = true;
String msg;
public TestingThread() {
t = new Thread(this, "Testing thread");
}
public void run() {
while (pause) {
//wait
}
System.out.println(msg);
}
public boolean isPause() {
return pause;
}
public void initMsg() {
msg = "Thread death";
}
public void setPause(boolean pause) {
this.pause = pause;
}
public void start() {
t.start();
}
}
And main thread class:
public class Main {
public static void main(String[] args) {
TestingThread testingThread = new TestingThread();
testingThread.start();
testingThread.initMsg();
testingThread.setPause(false);
}
}
Question list:
Should t be volatile?
Should msg be volatile?
Should setPause() be synchronized?
Is this a good example of good thread structure?
You have hit quite a subtlety with your question number 2.
In your very specific case, you:
first write msg from the main thread;
then write the volatile pause from the main thread;
then read the volatile pause from the child thread;
then read msg from the child thread.
Therefore you have transitively established a happens-before relationship between the write and the read of msg. Therefore msg itself does not have to be volatile.
In real-life code, however, you should avoid depending on such subtle behavior: better overapply volatile and sleep calmly.
Here are some relevant quotes from the Java Language Specification:
If x and y are actions of the same thread and x comes before y in program order, then hb(x, y).
If an action x synchronizes-with a following action y, then we also have hb(x, y).
Note that, in my list of actions,
1 comes before 2 in program order;
same for 3 and 4;
2 synchronizes-with 3.
As for your other questions,
ad 1: t doesn't have to be volatile because it's written to prior to thread creation and never mutated later. Starting a thread induces a happens-before on its own;
ad 3: setPause does not have to be synchronized because all it does is set the volatile var.
> Should msg be volatile?
Yes. Does it have to be in this example, No. But I urge you to use it anyway as the codes correctness becomes much clearer ;) Please note that I am assuming that we are discussing Java 5 or later, before then volatile was broken anyway.
The tricky part to understand is why this example can get away without msg being declared as volatile.
Consider this order part of main().
testingThread.start(); // starts the other thread
testingThread.initMsg(); // the other thread may or may not be in the
// while loop by now msg may or may not be
// visible to the testingThread yet
// the important thing to note is that either way
// testingThread cannot leave its while loop yet
testingThread.setPause(false); // after this volatile, all data earlier
// will be visible to other threads.
// Thus even though msg was not declared
// volatile it will piggy back the pauses
// use of volatile; as described [here](http://www.cs.umd.edu/~pugh/java/memoryModel/jsr-133-faq.html#volatile)
// and now the testingThread can leave
// its while loop
So if we now consider the testingThread
while (pause) { // pause is volatile, so it will see the change as soon
// as it is made
//wait
}
System.out.println(msg); // this line cannot be reached until the value
// of pause has been set to false by the main
// method. Which under the post Java5
// semantics will guarantee that msg will
// have been updated too.
> Should t be volatile?
It does not matter, but I would suggest making it private final.
> Should setPause() be synchronized?
Before Java 5, then yes. After Java 5 reading a volatile has the same memory barrier as entering a synchronized block. And writing to a volatile has the same memory barrier as at the end of a synchronized block. Thus unless you need the scoping of a synchronized block, which in this case you do not then you are fine with volatile.
The changes to volatile in Java 5 are documented by the author of the change here.
1&2
Volatile can be treated something like as "synchronization on variable",though the manner is different, but the result is alike, to make sure it is read-consistent.
3.
I feel it does not need to, since this.pause = pause should be an atomic statement.
4.
It is a bad example to do any while (true) {do nothing}, which will result in busy waiting, if you put Thread.sleep inside, which may help just a little bit. Please refer to http://en.wikipedia.org/wiki/Busy_waiting
One of a more appropriate way to do something like "wait until being awaken" is using the monitor object(Object in java is a monitor object), or using condition object along with a lock to do so. You may need to refer to http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/locks/Condition.html
Also, I don't think it is good idea either, that you have a local filed of thread inside your custom Runnable . Please refer to Seelenvirtuose 's comment.
Related
I am writing the same code on tutorial. But in tutorial the program never exits, my in my computer it exits after 4 seconds. Why?
tutorial with exact time where this code is shown: https://youtu.be/vzBw1LPupnA?t=169
public class Main {
private static boolean stopRequested;
public static void main(String[] args) throws InterruptedException {
Thread backgroundThread = new Thread(() -> {
int i = 0;
while (!stopRequested) {
i++;
System.out.println("i = " + i);
}
});
backgroundThread.start();
TimeUnit.SECONDS.sleep(1);
stopRequested = true;
}
}
The reason that you are seeing different behavior on your machine and in the video is because the program has unspecified behavior. (Or to put it another way, it is not thread-safe.)
You have two threads accessing and updating a shared variable without taking the necessary steps that will guarantee that changes made by one thread are visible to the other. What happens in that case is not specified.
In some cases (e.g. on some platforms) the changes will be visible, either immediately or within a short time.
On others, the changes may never be visible.
In technical terms, there must be a happens-before relationship between the write by on thread and the subsequent read by the other thread. This can be provided by both threads synchronizing on the same mutex or lock, by using a volatile variable, and in other ways. But this code doesn't do any of those things, so there is no guarantee that the state change will be visible.
For more details, read about the Java Memory Model.
The above is sufficient to explain the difference, but there may be a more direct explanation.
In practice, something like a System.out.println can lead to changes in the visibility. Underneath the covers, the println call will typically result in synchronization on the output stream's buffers. That can result in a serendipitous happens-before that is sufficient to guarantee visibility. But this behavior is not specified, so you should not rely on it.
At any rate, adding trace statements can change the behavior of multi-threaded coded. And the fact that you (apparently) added them in your version is a second possible explanation for the difference.
The bottom line here is that a program with a memory visibility flaw is broken, but you may not be able to demonstrate that it is broken.
As the excellent Answer by Stephen C says, your code is not thread-safe.
Establishing an AtomicBoolean early on addresses the visibility problem explained in that other Answer. This class is a thread-safe wrapper around its payload boolean value.
The volatile keyword is another solution. But I find the Atomic… classes simpler and more obvious.
Also, in modern Java we rarely need to address the Thread class directly. Instead, use the Executors framework. Define your task as a Runnable or Callable, and submit to an executor service.
Something like this untested code.
public class Main {
private static final AtomicBoolean stopRequested = new AtomicBoolean( false ) ;
public static void main(String[] args) throws InterruptedException {
Runnable task = () -> {
int i = 0;
while ( ! stopRequested.get() ) {
i++;
System.out.println("i = " + i);
TimeUnit.MILLISECONDS.sleep(100); // Don’t spin too fast.
}
};
ExecutorService es = Executors.newSingleThreadedExecutorService() ;
es.submit( task ) ;
TimeUnit.SECONDS.sleep(1);
stopRequested.set( true ) ;
TimeUnit.SECONDS.sleep(1);
// Shut down here executor service. Boilerplate taken from Javadoc.
es.shutdown(); // Disable new tasks from being submitted
try {
// Wait a while for existing tasks to terminate
if (!es.awaitTermination(60, TimeUnit.SECONDS)) {
es.shutdownNow(); // Cancel currently executing tasks
// Wait a while for tasks to respond to being cancelled
if (!es.awaitTermination(60, TimeUnit.SECONDS))
System.err.println("Executor service did not terminate");
}
} catch (InterruptedException ex) {
// (Re-)Cancel if current thread also interrupted
es.shutdownNow();
// Preserve interrupt status
Thread.currentThread().interrupt();
}
}
}
I am trying to see how multithreading(particularly with synchronized keyword) works.In this example I want the second thread abc1 to start executing after thread abc. So I've used synchronized keyword in run function.But the output line which says:
Initial balance in this thread is 10000
Initial balance in this thread is 10000
is what concerns me.Because the initial balance should be "-243000" as indicated in output line
Final balance after intial -243000 is 59049000
because the abc1 thread should wait for abc due to synchronized keyword.
Primarily , I want the threads to behave as if I write
abc.start
abc.join()
abc1.start()
abc1.join()
Here is my source code:
class parallel extends Thread{
account a;
public parallel(account a) {
this.a=a;
}
public synchronized void run() {
synchronized(this) {
System.out.println("Initial balance in this thread is "+a.amount);
long duplicate=a.amount;
boolean flag=true;
//System.out.println("Transaction inititated");
for(int i=0;i<10;i++) {
if(flag==true) {
//System.out.println("Deducting "+amount+"Rs from your account");
a.amount-=a.amount*2;
}
else {
//System.out.println("Depositing "+amount+"Rs from your account");
a.amount+=a.amount*2;
}
flag=!flag;
}
System.out.println("Final balance after intial "+duplicate+" is "+a.amount);
syncro.amount=a.amount;
}
}
}
class account{
public account(long rupe) {
amount=rupe;
}
long amount;
}
public class syncro {
static long amount;
public static void main(String[] args) throws InterruptedException{
//for(int i=0;i<10;i++) {
account ramesh=new account(1000);
parallel abc=new parallel(ramesh);
parallel abc1=new parallel(ramesh);
abc.start();
//abc.join();
abc1.start();
//abc1.join();
//}
//awaitTermination();
//Thread.sleep(4000);
boolean ab=true;
long cd=1000;
for(int i=0;i<10;i++) {
if(ab==true) {
//System.out.println("Deducting "+ab+"Rs from your account");
cd-=cd*2;
}
else {
//System.out.println("Depositing "+a+"Rs from your account");
cd+=cd*2;
}
ab=!ab;
}
//System.out.println("Final amount by multithreading is "+);
System.out.println("Final amount after serial order is "+cd);
}
}
You are mixing the creating of your own threads with the use of synchronized. Also, using synchronized(this) within a synchronized method is doing the same thing twice.
Synchronized is NOT about starting threads. It is about allowing only one thread to enter a certain block of code at a time.
Every object you create has a hidden field that you cannot read, but it does exist. It is of type Thread and it is called owner.
The synchronized keyword interacts with this hidden field.
synchronized (object) {
code();
}
means the following:
If object.owner == Thread.currentThread(), then just keep going and increment a counter.
If object.owner == null, then run object.owner = Thread.currentThread(), set that counter to 1, and keep going.
Otherwise (So, object.owner is some other thread), stop, freeze the thread, and wait around until the owner is set to null, and then we can go to option #2 instead.
Once we're in, run code(). When we get to the closing brace, decrement the counter. If it is 0, run object.owner = null.
Furthermore, all the above is done atomically - it is not possible for 2 threads to get into a race condition doing all this stuff. For example, if 2 threads are waiting for owner to become unset again, only one will 'get it', and the other will continue waiting. (Which one gets it? A VM impl is free to choose whatever it wants; you should assume it is arbitrary but unfair. Don't write code that depends on a certain choice, in other words).
A method that is keyworded with synchronized is just syntax sugar for wrapping ALL the code inside it in synchronized(this) for instance methods and synchronized(MyClass.this) for static methods.
Note that synchronized therefore only interacts with other synchronized blocks, and only those blocks for which the object in the parentheses is the exact same obj reference, otherwise none of this does anything. It certainly doesn't start threads! All synchronized does is potentially pause threads.
In your code, you've put ALL the run code in one gigantic synchronized block, synchronizing on your thread instance. As a general rule, when you synchronize on anything, it's public API - other code can synchronize on the same thing and affect you. Just like we don't generally write public fields in java, you should not lock on public things, and this is usually public (as in, code you don't control can hold a reference to you). So don't do that unless you're willing to spec out in your docs how your locking behaviours are set up. Instead, make an internal private final field, call it lock, and use that (private final Object lock = new Object();).
I was trying to write an example on how to use wait() and notify(), but seems that the wait() can't be notified
public class Transfer {
private int[] data;
private volatile int ptr;
private final Object lock = new Object();
public Transfer(int[] data) {
this.data = data;
this.ptr = 0;
}
public void send() {
while (ptr < data.length) {
synchronized (lock) {
try {
System.out.println("-----wait");
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
ptr++;
}
}
}
public void receive() {
while (ptr < data.length) {
synchronized (lock) {
System.out.println("current is " + data[ptr]);
System.out.println("-----notify");
lock.notifyAll();
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
////in main()
int[] data = new int[] { 111, 222, 333, 444, 555, 666, 777, 888, 999, 000 };
Transfer tf = new Transfer(data);
Thread t1 = new Thread(() -> {
tf.receive();
});
Thread t2 = new Thread(() -> {
tf.send();
});
t2.start();
t1.start();
but the result is :
-----wait
current is 111
-----notify
current is 111
-----notify
[endless repeat]
this is not what I expected, it should be :
current is 111
current is 222...
The problem with your code specifically is that you are keeping your locks way too long.
I'll first explain how wait/notify works, which is intricately connected with the concept of the monitor (synchronized), then how to do it right, and then as an encore, that you probably don't want to use this at all, it's too low level.
How does 'synchronized' work
When you write synchronized(x) you acquire the monitor - this operation can do one of three things. In all cases, x is a reference, so the reference is followed, it's about the object you find by following it.
If the reference is null, this immediately throws NPE.
If the object x points at has no current monitor, this thread becomes the monitor, the monitor count becomes 1, and code continues.
If the object x points at has a monitor but it is this thread, then the monitor count is incremented and code continues.
If the object x points at has a monitor but it is another thread, the thread will block until the monitor becomes available. Once it is available, some unfair dice show up, are rolled, and determine which of all threads 'fighting' to acquire the monitor will acquire it. Unfair in the sense that there are no guarantees made and the JVM is free to use any algorithm it wants to decide who 'wins'. If your code depends on fairness or some set order, your code is broken.
Upon reaching the } of the synchronized block, the monitor count is decremented. If it hits 0, the monitor is released (and the fight as per #4 starts, if other threads are waiting). In other words, locks are 're-entrant' in java. A thread can write synchronized(a){synchronized(a){}} and won't deadlock with itself.
Yes, this establishes comes-before stuff as per the Java Memory Model: Any fights arbitrated by a synchronized block will also ensure any writes by things that clearly came before (as established by who wins the fight) are observable by anything that clearly came after.
A method marked as 'synchronized' is effectively equivalent to wrapping the code in synchronized(this) for instance methods, and synchronized(MyClass.class) for static methods.
Monitors are not released and cannot be changed in java code* except via that } mechanism; (there is no public Thread getMonitor() {..} in j.l.Object or anywhere else) - in particular if the thread blocks for any other reason, including Thread.sleep, the monitor status does not change - your thread continues to hold on to it and thus stops all other threads from acquiring it. With one exception:
So how does wait/notify factor into this?
to wait/notify on x you MUST hold the monitor. this: x.notify();, unless it is wrapped in a synchronized(x) block, does not work.
When you wait(), the monitor is released, and the monitor count is remembered. a call to wait() requires 2 things to happen before it can continue: The 'wait' needs to be cancelled, either via a timeout, or an interrupt, or via a notify(All), and the thread needs to acquire that monitor again. If done normally (via a notify), by definition this is a fight, as whomever called notify neccessarily is still holding that monitor.
This then explains why your code does not work - your 'receiver' snippet holds on to the monitor while it sleeps. Take the sleep outside of the synchronized.
How do you use this, generally
The best way to use wait/notifyAll is not to make too many assumptions about the 'flow' of locking and unlocking. Only after acquiring the monitor, check some status. If the status is such that you need to wait for something to happen, then and only then start the wait() cycle. The thread that will cause that event to happen will first have to acquire the monitor and only then set steps to start the event. If this is not possible, that's okay - put in a failsafe, make the code that wait()s use a timeout (wait(500L) for example), so that if things fail, the while loop will fix the problem. Furthermore, there really is no good reason to ever use notify so forget that exists. notify makes no guarantees about what it'll unlock, and given that all threads that use wait ought to be checking the condition they were waiting for regardless of the behaviour of wait, notifyAll is always the right call to make.
So, it looks like this... let's say we're waiting for some file to exist.
// waiting side:
Path target = Paths.get("/file-i-am-waiting-for.txt");
synchronized (lock) {
while (!Files.isRegularFile(target)) {
try {
lock.wait(1000L);
} catch (InterruptedException e) {
// this exception occurs ONLY
// if some code explicitly called Thread.interrupt()
// on this thread. You therefore know what it means.
// usually, logging interruptedex is wrong!
// let's say here you intended it to mean: just exit
// and do nothing.
// to be clear: Interrupted does not mean:
// 'someone pressed CTRL+C' or 'the system is about to shutdown'.
return;
}
}
performOperation(target);
}
And on the 'file creation' side:
Path tgt = Paths.get("/file-i-am-waiting-for.txt");
Path create = tgt.getParent().resolve(tgt.getFileName() + ".create");
fillWithContent(create);
synchronized (lock) {
Files.move(create, tgt, StandardOpenOption.ATOMIC_MOVE);
lock.notifyAll();
}
The 'sending' (notifying) side is very simple, and note how we're using the file system to ensure that if the tgt file exists at all, it's fully formed and not a half-baked product. The receiving side uses a while loop: the notifying is itself NOT the signal to continue; it is merely the signal to re-check for the existence of this file. This is almost always how to do this stuff. Note also how all code involved with that file is always only doing things when they hold the lock, thus ensuring no clashes on that part.
But.. this is fairly low level stuff
The java.util.concurrent package has superior tooling for this stuff; for example, you may want a latch here, or a ReadWriteLock. They tend to outperform you, too.
But even juc is low level. Generally threading works best if the comm channel used between threads is inherently designed around concurrency. DBs (with a proper transaction level, such as SERIALIZABLE), or message buses like rabbitmq are such things. Why do you think script kiddies fresh off of an 8 hour course on PHP can manage to smash a website together that actually does at least hold up, thread-wise, even if it's littered with security issues? Because PHP enforces a model where all comms run through a DB because PHP is incapable of anything else in its basic deployment. As silly as these handcuffs may sound, the principle is solid, and can be applied just as easily from java.
*) sun.misc.Unsafe can do it, but it's called Unsafe for a reason.
Some closing best practices
Locks should be private; this is a rule broken by most examples and a lot of java code. You've done it right: if you're going to use synchronized, it should probably be on lock, which is private final Object lock = new Object();. Make it new Object[0] if you need it to be serializable, which arrays are, and Objects aren't.
if ever there is code in your system that does: synchronized(a) { synchronized (b) { ... }} and also code that odes: synchronized(b) { synchronized (a) { ... }} you're going to run into a deadlock at some point (each have acquired the first lock and are waiting for the second. They will be waiting forever. Be REAL careful when acquiring more than one monitor, and if you must, put in a ton of effort to ensure that you always acquire them in the same order to avoid deadlocks. Fortunately, jstack and such (tools to introspect running VMs) can tell you about deadlocks. The JVM itself, unfortunately, will just freeze in its tracks, dead as a doornail, if you deadlock it.
class Transfer {
private int[] data;
private volatile int ptr;
private final Object lock = new Object();
public Transfer(int[] data) {
this.data = data;
this.ptr = 0;
}
public void send() {
while (ptr < data.length) {
synchronized (lock) {
try {
System.out.println("-----wait");
lock.notifyAll();
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
ptr++;
}
}
}
public void receive() {
while (ptr < data.length) {
synchronized (lock) {
System.out.println("current is " + data[ptr]);
System.out.println("-----notify");
try {
lock.notifyAll();
lock.wait();
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
"Thread.sleep" does not release the lock. So you need "lock.wait" to release the lock and let other thread proceed. Then after "send" increment the pointer, it should also notify so that other thread who is stuck at receive can now proceed.
Is is okay to synchronize all methods which mutate the state of an object, but not synchronize anything which is atomic? In this case, just returning a field?
Consider:
public class A
{
private int a = 5;
private static final Object lock = new Object();
public void incrementA()
{
synchronized(lock)
{
a += 1;
}
}
public int getA()
{
return a;
}
}
I've heard people argue that it's possible for getA() and incrementA() to be called at roughly the same time and have getA() return to wrong thing. However it seems like, in the case that they're called at the same time, even if the getter is synchronized you can get the wrong thing. In fact the "right thing" doesn't even seem defined if these are called concurrently. The big thing for me is that the state remains consistent.
I've also heard talk about JIT optimizations. Given an instance of the above class and the following code(the code would be depending on a to be set in another thread):
while(myA.getA() < 10)
{
//incrementA is not called here
}
it is apparently a legal JIT optimization to change this to:
int temp = myA.getA();
while(temp < 10)
{
//incrementA is not called here
}
which can obviously result in an infinite loop.
Why is this a legal optimization? Would this be illegal if a was volatile?
Update
I did a little bit of testing into this.
public class Test
{
private int a = 5;
private static final Object lock = new Object();
public void incrementA()
{
synchronized(lock)
{
a += 1;
}
}
public int getA()
{
return a;
}
public static void main(String[] args)
{
final Test myA = new Test();
Thread t = new Thread(new Runnable(){
public void run() {
while(true)
{
try {
Thread.sleep(100);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
myA.incrementA();
}
}});
t.start();
while(myA.getA() < 15)
{
System.out.println(myA.getA());
}
}
}
Using several different sleep times, this worked even when a is not volatile. This of course isn't conclusive, it still may be legal. Does anyone have some examples that could trigger such JIT behaviour?
Is is okay to synchronize all methods which mutate the state of an object, but not synchronize anything which is atomic? In this case, just returning a field?
Depends on the particulars. It is important to realize that synchronization does two important things. It is not just about atomicity but it is also required because of memory synchronization. If one thread updates the a field, then other threads may not see the update because of memory caching on the local processor. Making the int a field be volatile solves this problem. Making both the get and the set method be synchronized will as well but it is more expensive.
If you want to be able to change and read a from multiple threads, the best mechanism is to use an AtomicInteger.
private AtomicInteger a = new AtomicInteger(5);
public void setA(int a) {
// no need to synchronize because of the magic of the `AtomicInteger` code
this.a.set(a);
}
public int getA() {
// AtomicInteger also takes care of the memory synchronization
return a.get();
}
I've heard people argue that it's possible for getA() and setA() to be called at roughly the same time and have getA() return to wrong thing.
This is true but you can get the wrong value if getA() is called after setA() as well. A bad cache value can stick forever.
which can obviously result in an infinite loop. Why is this a legal optimization?
It is a legal optimization because threads running with their own memory cache asynchronously is one of the important reasons why you see performance improvements with them. If all memory accesses where synchronized with main memory then the per-CPU memory caches would not be used and threaded programs would run a lot slower.
Would this be illegal if a was volatile?
It is not legal if there is some way for a to be altered – by another thread possibly. If a was final then the JIT could make that optimization. If a was volatile or the get method marked as synchronized then it would certainly not be a legal optimization.
It's not thread safe because that getter does not ensure that a thread will see the latest value, as the value may be stale. Having the getter be synchronized ensures that any thread calling the getter will see the latest value instead of a possible stale one.
You basically have two options:
1) Make your int volatile
2) Use an atomic type like AtomicInt
using a normal int without synchronization is not thread safe at all.
Your best solution is to use an AtomicInteger, they were basically designed for exactly this use case.
If this is more of a theoretical "could this be done question", I think something like the following would be safe (but still not perform as well as an AtomicInteger):
public class A
{
private volatile int a = 5;
private static final Object lock = new Object();
public void incrementA()
{
synchronized(lock)
{
final int tmp = a + 1;
a = tmp;
}
}
public int getA()
{
return a;
}
}
The short answer is your example will be thread-safe, if
the variable is declared as volatile, or
the getter is declared as synchronized.
The reason that your example class A is not thread-safe is that one can create a program using it that doesn't have a "well-formed execution" (see JLS 17.4.7).
For instance, consider
// in thread #1
int a1 = A.getA();
Thread.sleep(...);
int a2 = A.getA();
if (a1 == a2) {
System.out.println("no increment");
// in thread #2
A.incrementA();
in the scenario that the increment happens during the sleep.
For this execution to be well-formed, there must be a "happens before" (HB) chain between the assignment to a in incrementA called by thread #2, and the subsequent read of a in getA called by thread #1.
If the two threads synchronize using the same lock object, then there is a HB between one thread releasing the lock and a second thread acquiring the lock. So we get this:
thread #2 acquires lock --HB-->
thread #2 reads a --HB-->
thread #2 writes a --HB-->
thread #2 releases lock --HB-->
thread #1 acquires lock --HB-->
thread #1 reads a
If two threads share a a volatile variable, there is a HB between any write and any subsequent read (without an intervening write). So we typically get this:
thread #2 acquires lock --HB-->
thread #2 reads a --HB-->
thread #2 writes a --HB-->
thread #1 reads a
Note that incrementA needs to be synchronized to avoid race conditions with other threads calling incrementA.
If neither of the above is true, we get this:
thread #2 acquires lock --HB-->
thread #2 reads a --HB-->
thread #2 writes a // No HB!!
thread #1 reads a
Since there is no HB between the write by thread #2 and the subsequent read by thread #1, the JLS does not guarantee that the latter will see the value written by the former.
Note that this is a simplified version of the rules. For the complete version, you need to read all of JLS Chapter 17.
A warning is showing every time I synchronize on a non-final class field. Here is the code:
public class X
{
private Object o;
public void setO(Object o)
{
this.o = o;
}
public void x()
{
synchronized (o) // synchronization on a non-final field
{
}
}
}
so I changed the coding in the following way:
public class X
{
private final Object o;
public X()
{
o = new Object();
}
public void x()
{
synchronized (o)
{
}
}
}
I am not sure the above code is the proper way to synchronize on a non-final class field. How can I synchronize a non final field?
First of all, I encourage you to really try hard to deal with concurrency issues on a higher level of abstraction, i.e. solving it using classes from java.util.concurrent such as ExecutorServices, Callables, Futures etc.
That being said, there's nothing wrong with synchronizing on a non-final field per se. You just need to keep in mind that if the object reference changes, the same section of code may be run in parallel. I.e., if one thread runs the code in the synchronized block and someone calls setO(...), another thread can run the same synchronized block on the same instance concurrently.
Synchronize on the object which you need exclusive access to (or, better yet, an object dedicated to guarding it).
It's really not a good idea - because your synchronized blocks are no longer really synchronized in a consistent way.
Assuming the synchronized blocks are meant to be ensuring that only one thread accesses some shared data at a time, consider:
Thread 1 enters the synchronized block. Yay - it has exclusive access to the shared data...
Thread 2 calls setO()
Thread 3 (or still 2...) enters the synchronized block. Eek! It think it has exclusive access to the shared data, but thread 1 is still furtling with it...
Why would you want this to happen? Maybe there are some very specialized situations where it makes sense... but you'd have to present me with a specific use case (along with ways of mitigating the sort of scenario I've given above) before I'd be happy with it.
I agree with one of John's comment: You must always use a final lock dummy while accessing a non-final variable to prevent inconsistencies in case of the variable's reference changes. So in any cases and as a first rule of thumb:
Rule#1: If a field is non-final, always use a (private) final lock dummy.
Reason #1: You hold the lock and change the variable's reference by yourself. Another thread waiting outside the synchronized lock will be able to enter the guarded block.
Reason #2: You hold the lock and another thread changes the variable's reference. The result is the same: Another thread can enter the guarded block.
But when using a final lock dummy, there is another problem: You might get wrong data, because your non-final object will only be synchronized with RAM when calling synchronize(object). So, as a second rule of thumb:
Rule#2: When locking a non-final object you always need to do both: Using a final lock dummy and the lock of the non-final object for the sake of RAM synchronisation. (The only alternative will be declaring all fields of the object as volatile!)
These locks are also called "nested locks". Note that you must call them always in the same order, otherwise you will get a dead lock:
public class X {
private final LOCK;
private Object o;
public void setO(Object o){
this.o = o;
}
public void x() {
synchronized (LOCK) {
synchronized(o){
//do something with o...
}
}
}
}
As you can see I write the two locks directly on the same line, because they always belong together. Like this, you could even do 10 nesting locks:
synchronized (LOCK1) {
synchronized (LOCK2) {
synchronized (LOCK3) {
synchronized (LOCK4) {
//entering the locked space
}
}
}
}
Note that this code won't break if you just acquire an inner lock like synchronized (LOCK3) by another threads. But it will break if you call in another thread something like this:
synchronized (LOCK4) {
synchronized (LOCK1) { //dead lock!
synchronized (LOCK3) {
synchronized (LOCK2) {
//will never enter here...
}
}
}
}
There is only one workaround around such nested locks while handling non-final fields:
Rule #2 - Alternative: Declare all fields of the object as volatile. (I won't talk here about the disadvantages of doing this, e.g. preventing any storage in x-level caches even for reads, aso.)
So therefore aioobe is quite right: Just use java.util.concurrent. Or begin to understand everything about synchronisation and do it by yourself with nested locks. ;)
For more details why synchronisation on non-final fields breaks, have a look into my test case: https://stackoverflow.com/a/21460055/2012947
And for more details why you need synchronized at all due to RAM and caches have a look here: https://stackoverflow.com/a/21409975/2012947
I'm not really seeing the correct answer here, that is, It's perfectly alright to do it.
I'm not even sure why it's a warning, there is nothing wrong with it. The JVM makes sure that you get some valid object back (or null) when you read a value, and you can synchronize on any object.
If you plan on actually changing the lock while it's in use (as opposed to e.g. changing it from an init method, before you start using it), you have to make the variable that you plan to change volatile. Then all you need to do is to synchronize on both the old and the new object, and you can safely change the value
public volatile Object lock;
...
synchronized (lock) {
synchronized (newObject) {
lock = newObject;
}
}
There. It's not complicated, writing code with locks (mutexes) is actally quite easy. Writing code without them (lock free code) is what's hard.
EDIT: So this solution (as suggested by Jon Skeet) might have an issue with atomicity of implementation of "synchronized(object){}" while object reference is changing. I asked separately and according to Mr. erickson it is not thread safe - see: Is entering synchronized block atomic?. So take it as example how to NOT do it - with links why ;)
See the code how it would work if synchronised() would be atomic:
public class Main {
static class Config{
char a='0';
char b='0';
public void log(){
synchronized(this){
System.out.println(""+a+","+b);
}
}
}
static Config cfg = new Config();
static class Doer extends Thread {
char id;
Doer(char id) {
this.id = id;
}
public void mySleep(long ms){
try{Thread.sleep(ms);}catch(Exception ex){ex.printStackTrace();}
}
public void run() {
System.out.println("Doer "+id+" beg");
if(id == 'X'){
synchronized (cfg){
cfg.a=id;
mySleep(1000);
// do not forget to put synchronize(cfg) over setting new cfg - otherwise following will happend
// here it would be modifying different cfg (cos Y will change it).
// Another problem would be that new cfg would be in parallel modified by Z cos synchronized is applied on new object
cfg.b=id;
}
}
if(id == 'Y'){
mySleep(333);
synchronized(cfg) // comment this and you will see inconsistency in log - if you keep it I think all is ok
{
cfg = new Config(); // introduce new configuration
// be aware - don't expect here to be synchronized on new cfg!
// Z might already get a lock
}
}
if(id == 'Z'){
mySleep(666);
synchronized (cfg){
cfg.a=id;
mySleep(100);
cfg.b=id;
}
}
System.out.println("Doer "+id+" end");
cfg.log();
}
}
public static void main(String[] args) throws InterruptedException {
Doer X = new Doer('X');
Doer Y = new Doer('Y');
Doer Z = new Doer('Z');
X.start();
Y.start();
Z.start();
}
}
AtomicReference suits for your requirement.
From java documentation about atomic package:
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);
Sample code:
String initialReference = "value 1";
AtomicReference<String> someRef =
new AtomicReference<String>(initialReference);
String newReference = "value 2";
boolean exchanged = someRef.compareAndSet(initialReference, newReference);
System.out.println("exchanged: " + exchanged);
In above example, you replace String with your own Object
Related SE question:
When to use AtomicReference in Java?
If o never changes for the lifetime of an instance of X, the second version is better style irrespective of whether synchronization is involved.
Now, whether there's anything wrong with the first version is impossible to answer without knowing what else is going on in that class. I would tend to agree with the compiler that it does look error-prone (I won't repeat what the others have said).
Just adding my two cents: I had this warning when I used component that is instantiated through designer, so it's field cannot really be final, because constructor cannot takes parameters. In other words, I had quasi-final field without the final keyword.
I think that's why it is just warning: you are probably doing something wrong, but it might be right as well.