When to use Volatile modifier? [duplicate] - java

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Closed 10 years ago.
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When exactly do you use the volatile keyword in Java?
When and why volatile modifier is required in java?
I am interested in seeing a real world usage of a volatile modified primitive or object reference.

volatile modifier will tell the JVM to be cautious of the threads which runs concurrently. Essentially, volatile is used to indicate that a variable's value will be modified by different threads.
Declaring a volatile Java variable means:
The value of this variable will never be cached thread-locally: all reads and writes will go straight to "main memory"
Access to the variable acts as though it is enclosed in a synchronized block, synchronized on itself.
We say "acts as though" in the second point, because to the programmer at least (and probably in most JVM implementations) there is no actual lock object involved.
The volatile modifier tells the compiler that the variable modified by volatile can be changed unexpectedly by other parts of your program. One of these situations involves multithreaded programs.
In a multithreaded program, sometimes, two or more threads share the same instance variable. For efficiency considerations, each thread can keep its own, private copy of such a shared variable.
The real (or master) copy of the variable is updated at various times, such as when a synchronized method is entered. While this approach works fine, it may be inefficient at times. In some cases, all that really matters is that the master copy of a variable always reflects its current state.
To ensure this, simply specify the variable as volatile, which tells the compiler that it must always use the master copy of a volatile variable (or, at least, always keep any private copies up to date with the master copy, and vice versa). Also, accesses to the master variable must be executed in the precise order in which they are executed on any private copy.

If you are working with the multi-threaded programming, the volatile keyword will be more useful. When multiple
threads using the same variable, each thread will have its own copy of the local cache for that variable. So, when it’s
updating the value, it is actually updated in the local cache not in the main variable memory. The other thread which
is using the same variable doesn’t know anything about the values changed by the another thread. To avoid this problem,
if you declare a variable as volatile, then it will not be stored in the local cache. Whenever thread are updating the
values, it is updated to the main memory. So, other threads can access the updated value.
Declaring a variable volatile means
There will be no cache maintained means all the changes made in main memory.
Access to this variable acts as synchronized block, even though it is in synchronized unit.
Example -
public class Snippet implements Runnable{
volatile int num =0;
public void run(){
Thread t = Thread.currentThread();
String name = t.getName();
if(name.equals("Thread1")){
num=10;
}
else{
System.out.println("value of num is :"+num);
}
}
public static void main(String args[]) throws InterruptedException{
Runnable r = new Snippet();
Thread t1 = new Thread(r);
t1.setName("Thread1");
t1.start();
Thread.sleep(1000);
Thread t2 = new Thread(r);
t2.setName("Thread2");
t2.start();
}
}

(This answer assumes Java 5+ -- before that, volatile had weaker guarantees.)
It's useful when you want to ensure a memory barrier, aka a formal "happens-before" relationship, between a write to a field and a subsequent read to that field by a separate thread. Synchronization can also give you that relationship, as well as other multithreading guarantees, but it's a tad slower and can create synchronization bottlenecks.
One use case is in concurrent collection classes (like ConcurrentHashMap, or LinkedBlockingQueue) where, in conjunction with things like atomic compare-and-set (CAS) operations, you can write correct thread-safe code without having to use synchronized.

You got good answers for the first question. The second one:
Can any one give me real time scenario of it
IMO, you should never ever you volatile. There are better tools for multithreaded apps. It's a bit bizarre that such a high level language has this keyword. Here is a good read (It's about C#, but Java is similar in this matter).

Related

Safe multithreading in java

I am new to multi threading in java.
I have gone through some online references but can't get clarity regarding how to properly implement thread concurrency and addressing resource access conflicts.
(like where to use synchronized and volatile and how to design code that dont even need them).
Can somebody suggest some guidelines or provide any valuable online references you have come across for implementing a safer multi threading project?
Thanks in advance.
Didn't go through your code, but here's something important to begin using synchronize and volatile keywords.
Essentially, volatile is used to indicate that a variable's value will be modified by different threads.
Declaring a volatile Java variable means:
The value of this variable will never be cached thread-locally: all reads and writes will go straight to "main memory"; This means that threads are making changes directly to a (volatile)variable where other threads also have a hold on. Everyone(every thread) has control and they can make changes which are reflected globally.
Here is an excellent example to understand more about volatile variables
If a variable is not declared volatile : The problem with threads not seeing the latest value of a variable because it has not yet been written back to main memory by another thread, is called a "visibility" problem. The updates of one thread are not visible to other threads
Declaring a synchronized Java variable means:
Synchronized blocks in Java are marked with the synchronized keyword and is synchronized on some object. All synchronized blocks synchronized on the same object can only have one thread executing inside them at the same time. All other threads attempting to enter the synchronized block are blocked until the thread inside the synchronized block exits the block.
Usage :
If you want a count variable to be incremented by some threads then make it volatile.
public class SharedObject {
public volatile int counter = 0;
}
However if you need your counter increment to be atomic( one thread at a time) make it synchronized too.
public synchronized void add(int value){
this.counter += value;
}

Java variable shared between two process [duplicate]

My teacher in an upper level Java class on threading said something that I wasn't sure of.
He stated that the following code would not necessarily update the ready variable. According to him, the two threads don't necessarily share the static variable, specifically in the case when each thread (main thread versus ReaderThread) is running on its own processor and therefore doesn't share the same registers/cache/etc and one CPU won't update the other.
Essentially, he said it is possible that ready is updated in the main thread, but NOT in the ReaderThread, so that ReaderThread will loop infinitely.
He also claimed it was possible for the program to print 0 or 42. I understand how 42 could be printed, but not 0. He mentioned this would be the case when the number variable is set to the default value.
I thought perhaps it is not guaranteed that the static variable is updated between the threads, but this strikes me as very odd for Java. Does making ready volatile correct this problem?
He showed this code:
public class NoVisibility {
private static boolean ready;
private static int number;
private static class ReaderThread extends Thread {
public void run() {
while (!ready) Thread.yield();
System.out.println(number);
}
}
public static void main(String[] args) {
new ReaderThread().start();
number = 42;
ready = true;
}
}
There isn't anything special about static variables when it comes to visibility. If they are accessible any thread can get at them, so you're more likely to see concurrency problems because they're more exposed.
There is a visibility issue imposed by the JVM's memory model. Here's an article talking about the memory model and how writes become visible to threads. You can't count on changes one thread makes becoming visible to other threads in a timely manner (actually the JVM has no obligation to make those changes visible to you at all, in any time frame), unless you establish a happens-before relationship.
Here's a quote from that link (supplied in the comment by Jed Wesley-Smith):
Chapter 17 of the Java Language Specification defines the happens-before relation on memory operations such as reads and writes of shared variables. The results of a write by one thread are guaranteed to be visible to a read by another thread only if the write operation happens-before the read operation. The synchronized and volatile constructs, as well as the Thread.start() and Thread.join() methods, can form happens-before relationships. In particular:
Each action in a thread happens-before every action in that thread that comes later in the program's order.
An unlock (synchronized block or method exit) of a monitor happens-before every subsequent lock (synchronized block or method entry) of that same monitor. And because the happens-before relation is transitive, all actions of a thread prior to unlocking happen-before all actions subsequent to any thread locking that monitor.
A write to a volatile field happens-before every subsequent read of that same field. Writes and reads of volatile fields have similar memory consistency effects as entering and exiting monitors, but do not entail mutual exclusion locking.
A call to start on a thread happens-before any action in the started thread.
All actions in a thread happen-before any other thread successfully returns from a join on that thread.
He was talking about visibility and not to be taken too literally.
Static variables are indeed shared between threads, but the changes made in one thread may not be visible to another thread immediately, making it seem like there are two copies of the variable.
This article presents a view that is consistent with how he presented the info:
http://jeremymanson.blogspot.com/2008/11/what-volatile-means-in-java.html
First, you have to understand a little something about the Java memory model. I've struggled a bit over the years to explain it briefly and well. As of today, the best way I can think of to describe it is if you imagine it this way:
Each thread in Java takes place in a separate memory space (this is clearly untrue, so bear with me on this one).
You need to use special mechanisms to guarantee that communication happens between these threads, as you would on a message passing system.
Memory writes that happen in one thread can "leak through" and be seen by another thread, but this is by no means guaranteed. Without explicit communication, you can't guarantee which writes get seen by other threads, or even the order in which they get seen.
...
But again, this is simply a mental model to think about threading and volatile, not literally how the JVM works.
Basically it's true, but actually the problem is more complex. Visibility of shared data can be affected not only by CPU caches, but also by out-of-order execution of instructions.
Therefore Java defines a Memory Model, that states under which circumstances threads can see consistent state of the shared data.
In your particular case, adding volatile guarantees visibility.
They are "shared" of course in the sense that they both refer to the same variable, but they don't necessarily see each other's updates. This is true for any variable, not just static.
And in theory, writes made by another thread can appear to be in a different order, unless the variables are declared volatile or the writes are explicitly synchronized.
Within a single classloader, static fields are always shared. To explicitly scope data to threads, you'd want to use a facility like ThreadLocal.
When you initialize static primitive type variable java default assigns a value for static variables
public static int i ;
when you define the variable like this the default value of i = 0;
thats why there is a possibility to get you 0.
then the main thread updates the value of boolean ready to true. since ready is a static variable , main thread and the other thread reference to the same memory address so the ready variable change. so the secondary thread get out from while loop and print value.
when printing the value initialized value of number is 0. if the thread process has passed while loop before main thread update number variable. then there is a possibility to print 0
#dontocsata
you can go back to your teacher and school him a little :)
few notes from the real world and regardless what you see or be told.
Please NOTE, the words below are regarding this particular case in the exact order shown.
The following 2 variable will reside on the same cache line under virtually any know architecture.
private static boolean ready;
private static int number;
Thread.exit (main thread) is guaranteed to exit and exit is guaranteed to cause a memory fence, due to the thread group thread removal (and many other issues). (it's a synchronized call, and I see no single way to be implemented w/o the sync part since the ThreadGroup must terminate as well if no daemon threads are left, etc).
The started thread ReaderThread is going to keep the process alive since it is not a daemon one!
Thus ready and number will be flushed together (or the number before if a context switch occurs) and there is no real reason for reordering in this case at least I can't even think of one.
You will need something truly weird to see anything but 42. Again I do presume both static variables will be in the same cache line. I just can't imagine a cache line 4 bytes long OR a JVM that will not assign them in a continuous area (cache line).

Is a static variable shared between two threads? [duplicate]

This question already has answers here:
Are static variables shared between threads?
(7 answers)
Closed 7 years ago.
In a multithread program if we have a static variable then does each thread have its own copy of the variable and is the change made by one thread visible to the other in case of static variable ?? Please explain
All variables1 are shared between all threads.
Only code visibility is defined by the variable type (eg. static/member, public/private) - but not thread visibility. (The volatile modifier affects thread access but is often not sufficient by itself.)
Using correct synchronization (and/or volatile variables) is required for "thread-safe" access for variables, and all data that can be reached from such, that can be accessed by multiple threads. Without correct synchronization there is no guarantee that another thread will 'see' a change to a [static] variable.
1 It is possible to create ThreadLocal 'variables' (via an indirect object), but this should be a rare case.
In a multithread program if we have a static variable then does each thread have its own copy of the variable
No. If you need that, you can use a ThreadLocal
and is the change made by one thread visible to the other in case of static variable
Not immediately. Another thread might still have the old value cached. To guarantee that the change is visible to other threads, you have multiple options. You can use a lock, a synchronized block/method, make the variable volatile or use some existing thread-safe utility class like AtomicReference.
If multiple threads are accessing the same static variable, all threads can see the changes.
As a result, if multiple threads are accessing the same variable, you will need to use locks for the variable to be thread-safe.
Yes, Static variables are shared between threads unlike local variables.
For more reference : http://java67.blogspot.com/2012/11/what-is-static-class-variable-method.html
There is a classic example ----Producers and Consumers. This inspires us that ,if you want to find the variable's change in other thread ,you must consider each Thread's possession time in CPU. EX: The variable may be just changed in one thread,and the other thread don't hava the use of CPU.So it can't find the change.
Of course,you can use
synchronized
to limit the thread's behavior(when multithreading) that changes the static variable.But you must know cross use of CPU , that means you don't know multiple threads' running way if you don't define their priority clearly.

is a volatile variable synchronized? (java)

Say that I have a private variable and I have a setVariable() method for it which is synchronized, isn't it exactly the same as using volatile modifier?
No. Volatile means the variable isn't cached in any per-thread cache, and its value is always retrieved from main memory when needed. Synchronization means that those per-thread caches will be kept in sync at certain points. In theory, using a volatile variable can come with a great speed penalty if many threads need to read the value of the variable, but it is changed only rarely.
No, calling a synchronized getXXX/setXXX method is not the same as reading/writing to a volatile variable.
Multiple threads can concurrently read from or write to a volatile variable. But only one thread at a time can read from or write to a variable that is guarded by a synchronized block.
volatile variables are not synchronized (at least, not in the way synchronized stuff is synchronized). What volatile does is ensure that a variable is retrieved each time it's used (ie: it prevents certain kinds of optimization), and IIRC that it's read and written in the correct order. This could conceivably emulate some kinds of synchronization, but it can't work the same if your setter has to set more than one thing. (If you set two volatile variables, for example, there will be a point where one is set and the other isn't.)
Actually No.
volatile is actually weaker form of synchronization, when field is declared as a volatile the compiler and runtime understands that this variable is shared and operations on it shouldn't be reordered with other memory operations. Volatile variable aren't cached in registers or in caches where they are hidden from other processors, so a read of a volatile variable always return a recent write by any thread.
just an example :
First thread run :
while(stopped){
... do something
}
Second thread run :
stopped = true;
it's useful to declare stopped as a volatile boolean for the first thread to have a fresh value of it.
There is no any relation.
Basically
Volatile => it always retrieves parameter's latest value
Synchronized => it serves only 1 thread at the same time

java threads synchronization

In the class below, is the method getIt() thread safe and why?
public class X {
private long myVar;
public void setIt(long var){
myVar = var;
}
public long getIt() {
return myVar;
}
}
It is not thread-safe. Variables of type long and double in Java are treated as two separate 32-bit variables. One thread could be writing and have written half the value when another thread reads both halves. In this situation, the reader would see a value that was never supposed to exist.
To make this thread-safe you can either declare myVar as volatile (Java 1.5 or later) or make both setIt and getIt synchronized.
Note that even if myVar was a 32-bit int you could still run into threading issues where one thread could be reading an out of date value that another thread has changed. This could occur because the value has been cached by the CPU. To resolve this, you again need to declare myVar as volatile (Java 1.5 or later) or make both setIt and getIt synchronized.
It's also worth noting that if you are using the result of getIt in a subsequent setIt call, e.g. x.setIt(x.getIt() * 2), then you probably want to synchronize across both calls:
synchronized(x)
{
x.setIt(x.getIt() * 2);
}
Without the extra synchronization, another thread could change the value in between the getIt and setIt calls causing the other thread's value to be lost.
This is not thread-safe. Even if your platform guarantees atomic writes of long, the lack of synchronized makes it possible that one thread calls setIt() and even after this call has finished it is possible that another thread can call getIt() and this call could return the old value of myVar.
The synchronized keyword does more than an exclusive access of one thread to a block or a method. It also guarantees that the second thread is informed about a change of a variable.
So you either have to mark both methods as synchronized or mark the member myVar as volatile.
There's a very good explanation about synchronization here:
Atomic actions cannot be interleaved, so they can be used without fear of thread interference. However, this does not eliminate all need to synchronize atomic actions, because memory consistency errors are still possible. Using volatile variables reduces the risk of memory consistency errors, because any write to a volatile variable establishes a happens-before relationship with subsequent reads of that same variable. This means that changes to a volatile variable are always visible to other threads. What's more, it also means that when a thread reads a volatile variable, it sees not just the latest change to the volatile, but also the side effects of the code that led up the change.
No, it's not. At least, not on platforms that lack atomic 64-bit memory accesses.
Suppose that Thread A calls setIt, copies 32 bits into memory where the backing value is, and is then pre-empted before it can copy the other 32 bits.
Then Thread B calls getIt.
No it is not, because longs are not atomic in java, so one thread could have written 32 bits of the long in the setIt method, and then the getIt could read the value, and then setIt could set the other 32 bits.
So the end result is that getIt returns a value that was never valid.
It ought to be, and generally is, but is not guaranteed to be thread safe. There could be issues with different cores having different versions in CPU cache, or the store/retrieve not being atomic for all architectures. Use the AtomicLong class.
The getter is not thread safe because it’s not guarded by any mechanism that guarantees the most up-to-date visibility. Your choices are:
making myVar final (but then you can’t mutate it)
making myVar volatile
use synchronized to accessing myVar
AFAIK, Modern JVMs no longer split long and double operations. I don't know of any reference which states this is still a problem. For example, see AtomicLong which doesn't use synchronization in Sun's JVM.
Assuming you want to be sure it is not a problem then you can use synchronize both get() and set(). However, if you are performing an operation like add, i.e. set(get()+1) then this synchronization doesn't buy you much, you still have to synchronize the object for the whole operation. (A better way around this is to use a single operation for add(n) which is synchronized)
However, a better solution is to use an AtomicLong. This supports atomic operations like get, set and add and DOESN'T use synchronization.
Since it is a read only method. You should synchronize the set method.
EDIT : I see why the get method needs to be synchronized as well. Good job explaining Phil Ross.

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