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).
Related
I'm going through Goetze's Java Concurrency in practice and am stuck on the section on memory visibility of shared variables when not using the synchronized keyword.
The code is as follows
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;
}
}
The author says that this class could loop forever because the value of ready might never become visibile to the reader thread.
I don't understand this statement.
The way I see it is that first the main thread starts and sets the number and ready to true. But the other thread has its own stack and its own value of number and ready which is not synced with the main memory and these two threads have only their own copies of the variables.
Now the readerthread will supposedly keep in the loop forever. I want to know why the Thread.yield variable will not yield to the main thread and then the main thread should flush to the main memory and then the readerthread should pick up this new value and terminate the loop and print the correct value because that too should have been synched.
So I guess some questions I have are.
How often does the value in the cpu's cache get flushed/synced with main memory?
Can the value be not synced with the main memory is that also a possibility?
Why would this happen?
Does this memory visibility happen also when there is only once cpu core and one cpu cache or does it happen always?
I am having some trouble understanding the memory visibility problem though I understand race conditions and deadlocks. Is this something architecture specific?
How often does the value in the cpu's cache get flushed/synced with main memory?
Undefined. Cache flushing happens when the visibility guarantees specified in the JLS say that it needs to happen.
Can the value be not synced with the main memory is that also a possibility?
Yes.
Why would this happen?
Generally speaking, caches get flushed for a reason. The happens-before relationships indicate the places where a cache flush may be necessary.
Does this memory visibility happen also when there is only once cpu core and one cpu cache or does it happen always?
If there is only one core, then cache flushing is not an issue1.
I am having some trouble understanding the memory visibility problem though I understand race conditions and deadlocks. Is this something architecture specific?
Yes and no. The memory visibility may manifest differently depending on the hardware architecture among other things, but the way to write your code to give well-defined behavior is architecture independent.
If you really need a deep understanding of the memory visibility problem, you need to understand the Memory Model. It is described in laymans terms in Goetz et al Chapter 16, and specified in the JLS.
I want to know why the Thread.yield() call will not yield to the main thread and then the main thread should flush to the main memory
The Thread.yield() may yield to another runnable thread. However, by the time that yield() is called, it is quite likely that the main thread is no longer runnable. (Or it may still be running.)
The yield() does not create a happens-before between any statements in the main and child threads. Absent that happens-before relation, the runtime is not obliged to ensure that the result of the assignment by the main thread is visible to the child thread.
While Thread.yield() might perform a cache flush2, it would be a flush of the child thread's caches, not the parent thread's caches.
Hence, the child thread's loop may continue indefinitely.
1 - Actually, that may be an over-simplification. For example, in a system with one core and multiple hyperthreads with their own caches, cache flushing would be needed.
2 - For example, if the yield() does result in a context switch, then the context switch typically includes a cache flush as part of the thread state saving performed by the OS. However, yield() won't necessary result in a context switch. And besides, this aspect is not specified by the JLS.
Field visibility means that threads observer field values from the cache memory and can have different state from other cache in another core of the CPU. JVM does not guarantee field visibilty for different threads accessing shared resource and programmer needs to use sycchronized to prevent reading incorrect state or use volatile to guarantee the changes are flushed to other cache.
I have read below references:
http://tutorials.jenkov.com/java-concurrency/volatile.html
https://www.geeksforgeeks.org/volatile-keyword-in-java/
https://docs.oracle.com/javase/tutorial/essential/concurrency/atomic.html
But, I am still not clear on what is the expected behavior in below case:
I have a thread pool in which thread is reused.
I have a non-volatile var which is accessed by different threads form that thread pool.
Threads are run in sequential.(Yes, they can be run in one thread. But, just for this case. So, don't ask why not use one thread)
And the part I am not clear is that. Do I still need volatile to make sure the change made in previous thread is visible to the next thread execution.
Like does java flash thread local cache to memory after each execution?
And does the thread reload the local cache before each execution?
It keep complain there is code section that is not quoted. So, I have to do this. Please help fix.
Java Memory Model design can answer your question:
In the Java Memory Model a volatile field has a store barrier inserted after a write to it and a load barrier inserted before a read of it. Qualified final fields of a class have a store barrier inserted after their initialisation to ensure these fields are visible once the constructor completes when a reference to the object is available.
see https://dzone.com/articles/memory-barriersfences
In other words: when a thread tries to read from volatile var JMM forces all CPUs owning this memory area to write back to memory from the local CPU's cache. Then CPU loads its value to the local cache if necessary.
And the part I am not clear is that. Do I still need volatile to make sure the change made in previous thread is visible to the next thread execution.
Yes and no. Volatile keyword is just for making a value of variable visible to other threads. But if you need only one thread read-write at the moment = you need synchronization. If you need to provide just visibility = volatile keyword is enough.
If you do not use the volatile keyword and share the value through the threads it might be not up to date or even be corrupted. For LONG type Java do 2 writes by 32bits and they are not atomic. Let's say another thread can read the value between these two writes happened.
It doesn't matter whether the Threads are running sequentially for parallely, if you don't use volatile keyword there is no visibility guarantee. That means there is no guarantee that the other thread will see the latest copy as the thread may read the values from register.
I am wondering at the difference between declaring a variable as volatile and always accessing the variable in a synchronized(this) block in Java?
According to this article http://www.javamex.com/tutorials/synchronization_volatile.shtml there is a lot to be said and there are many differences but also some similarities.
I am particularly interested in this piece of info:
...
access to a volatile variable never has the potential to block: we're only ever doing a simple read or write, so unlike a synchronized block we will never hold on to any lock;
because accessing a volatile variable never holds a lock, it is not suitable for cases where we want to read-update-write as an atomic operation (unless we're prepared to "miss an update");
What do they mean by read-update-write? Isn't a write also an update or do they simply mean that the update is a write that depends on the read?
Most of all, when is it more suitable to declare variables volatile rather than access them through a synchronized block? Is it a good idea to use volatile for variables that depend on input? For instance, there is a variable called render that is read through the rendering loop and set by a keypress event?
It's important to understand that there are two aspects to thread safety.
execution control, and
memory visibility
The first has to do with controlling when code executes (including the order in which instructions are executed) and whether it can execute concurrently, and the second to do with when the effects in memory of what has been done are visible to other threads. Because each CPU has several levels of cache between it and main memory, threads running on different CPUs or cores can see "memory" differently at any given moment in time because threads are permitted to obtain and work on private copies of main memory.
Using synchronized prevents any other thread from obtaining the monitor (or lock) for the same object, thereby preventing all code blocks protected by synchronization on the same object from executing concurrently. Synchronization also creates a "happens-before" memory barrier, causing a memory visibility constraint such that anything done up to the point some thread releases a lock appears to another thread subsequently acquiring the same lock to have happened before it acquired the lock. In practical terms, on current hardware, this typically causes flushing of the CPU caches when a monitor is acquired and writes to main memory when it is released, both of which are (relatively) expensive.
Using volatile, on the other hand, forces all accesses (read or write) to the volatile variable to occur to main memory, effectively keeping the volatile variable out of CPU caches. This can be useful for some actions where it is simply required that visibility of the variable be correct and order of accesses is not important. Using volatile also changes treatment of long and double to require accesses to them to be atomic; on some (older) hardware this might require locks, though not on modern 64 bit hardware. Under the new (JSR-133) memory model for Java 5+, the semantics of volatile have been strengthened to be almost as strong as synchronized with respect to memory visibility and instruction ordering (see http://www.cs.umd.edu/users/pugh/java/memoryModel/jsr-133-faq.html#volatile). For the purposes of visibility, each access to a volatile field acts like half a synchronization.
Under the new memory model, it is still true that volatile variables cannot be reordered with each other. The difference is that it is now no longer so easy to reorder normal field accesses around them. Writing to a volatile field has the same memory effect as a monitor release, and reading from a volatile field has the same memory effect as a monitor acquire. In effect, because the new memory model places stricter constraints on reordering of volatile field accesses with other field accesses, volatile or not, anything that was visible to thread A when it writes to volatile field f becomes visible to thread B when it reads f.
-- JSR 133 (Java Memory Model) FAQ
So, now both forms of memory barrier (under the current JMM) cause an instruction re-ordering barrier which prevents the compiler or run-time from re-ordering instructions across the barrier. In the old JMM, volatile did not prevent re-ordering. This can be important, because apart from memory barriers the only limitation imposed is that, for any particular thread, the net effect of the code is the same as it would be if the instructions were executed in precisely the order in which they appear in the source.
One use of volatile is for a shared but immutable object is recreated on the fly, with many other threads taking a reference to the object at a particular point in their execution cycle. One needs the other threads to begin using the recreated object once it is published, but does not need the additional overhead of full synchronization and it's attendant contention and cache flushing.
// Declaration
public class SharedLocation {
static public volatile SomeObject someObject=new SomeObject(); // default object
}
// Publishing code
SharedLocation.someObject=new SomeObject(...); // new object is published
// Using code
// Note: do not simply use SharedLocation.someObject.xxx(), since although
// someObject will be internally consistent for xxx(), a subsequent
// call to yyy() might be inconsistent with xxx() if the object was
// replaced in between calls.
private String getError() {
SomeObject myCopy=SharedLocation.someObject; // gets current copy
...
int cod=myCopy.getErrorCode();
String txt=myCopy.getErrorText();
return (cod+" - "+txt);
}
// And so on, with myCopy always in a consistent state within and across calls
// Eventually we will return to the code that gets the current SomeObject.
Speaking to your read-update-write question, specifically. Consider the following unsafe code:
public void updateCounter() {
if(counter==1000) { counter=0; }
else { counter++; }
}
Now, with the updateCounter() method unsynchronized, two threads may enter it at the same time. Among the many permutations of what could happen, one is that thread-1 does the test for counter==1000 and finds it true and is then suspended. Then thread-2 does the same test and also sees it true and is suspended. Then thread-1 resumes and sets counter to 0. Then thread-2 resumes and again sets counter to 0 because it missed the update from thread-1. This can also happen even if thread switching does not occur as I have described, but simply because two different cached copies of counter were present in two different CPU cores and the threads each ran on a separate core. For that matter, one thread could have counter at one value and the other could have counter at some entirely different value just because of caching.
What's important in this example is that the variable counter was read from main memory into cache, updated in cache and only written back to main memory at some indeterminate point later when a memory barrier occurred or when the cache memory was needed for something else. Making the counter volatile is insufficient for thread-safety of this code, because the test for the maximum and the assignments are discrete operations, including the increment which is a set of non-atomic read+increment+write machine instructions, something like:
MOV EAX,counter
INC EAX
MOV counter,EAX
Volatile variables are useful only when all operations performed on them are "atomic", such as my example where a reference to a fully formed object is only read or written (and, indeed, typically it's only written from a single point). Another example would be a volatile array reference backing a copy-on-write list, provided the array was only read by first taking a local copy of the reference to it.
volatile is a field modifier, while synchronized modifies code blocks and methods. So we can specify three variations of a simple accessor using those two keywords:
int i1;
int geti1() {return i1;}
volatile int i2;
int geti2() {return i2;}
int i3;
synchronized int geti3() {return i3;}
geti1() accesses the value currently stored in i1 in the current thread.
Threads can have local copies of variables, and the data does not have to be the same as the data held in other threads.In particular, another thread may have updated i1 in it's thread, but the value in the current thread could be different from that updated value. In fact Java has the idea of a "main" memory, and this is the memory that holds the current "correct" value for variables. Threads can have their own copy of data for variables, and the thread copy can be different from the "main" memory. So in fact, it is possible for the "main" memory to have a value of 1 for i1, for thread1 to have a value of 2 for i1 and for thread2 to have a value of 3 for i1 if thread1 and thread2 have both updated i1 but those updated value has not yet been propagated to "main" memory or other threads.
On the other hand, geti2() effectively accesses the value of i2 from "main" memory. A volatile variable is not allowed to have a local copy of a variable that is different from the value currently held in "main" memory. Effectively, a variable declared volatile must have it's data synchronized across all threads, so that whenever you access or update the variable in any thread, all other threads immediately see the same value. Generally volatile variables have a higher access and update overhead than "plain" variables. Generally threads are allowed to have their own copy of data is for better efficiency.
There are two differences between volitile and synchronized.
Firstly synchronized obtains and releases locks on monitors which can force only one thread at a time to execute a code block. That's the fairly well known aspect to synchronized. But synchronized also synchronizes memory. In fact synchronized synchronizes the whole of thread memory with "main" memory. So executing geti3() does the following:
The thread acquires the lock on the monitor for object this .
The thread memory flushes all its variables, i.e. it has all of its variables effectively read from "main" memory .
The code block is executed (in this case setting the return value to the current value of i3, which may have just been reset from "main" memory).
(Any changes to variables would normally now be written out to "main" memory, but for geti3() we have no changes.)
The thread releases the lock on the monitor for object this.
So where volatile only synchronizes the value of one variable between thread memory and "main" memory, synchronized synchronizes the value of all variables between thread memory and "main" memory, and locks and releases a monitor to boot. Clearly synchronized is likely to have more overhead than volatile.
http://javaexp.blogspot.com/2007/12/difference-between-volatile-and.html
There are 3 main issues with multithreading:
Race Conditions
Caching / stale memory
Compiler and CPU optimisations
volatile can solve 2 & 3, but can't solve 1. synchronized/explicit locks can solve 1, 2 & 3.
Elaboration:
Consider this thread unsafe code:
x++;
While it may look like one operation, it's actually 3: reading the current value of x from memory, adding 1 to it, and saving it back to memory. If few threads try to do it at the same time, the result of the operation is undefined. If x originally was 1, after 2 threads operating the code it may be 2 and it may be 3, depending on which thread completed which part of the operation before control was transferred to the other thread. This is a form of race condition.
Using synchronized on a block of code makes it atomic - meaning it make it as if the 3 operations happen at once, and there's no way for another thread to come in the middle and interfere. So if x was 1, and 2 threads try to preform x++ we know in the end it will be equal to 3. So it solves the race condition problem.
synchronized (this) {
x++; // no problem now
}
Marking x as volatile does not make x++; atomic, so it doesn't solve this problem.
In addition, threads have their own context - i.e. they can cache values from main memory. That means that a few threads can have copies of a variable, but they operate on their working copy without sharing the new state of the variable among other threads.
Consider that on one thread, x = 10;. And somewhat later, in another thread, x = 20;. The change in value of x might not appear in the first thread, because the other thread has saved the new value to its working memory, but hasn't copied it to the main memory. Or that it did copy it to the main memory, but the first thread hasn't updated its working copy. So if now the first thread checks if (x == 20) the answer will be false.
Marking a variable as volatile basically tells all threads to do read and write operations on main memory only. synchronized tells every thread to go update their value from main memory when they enter the block, and flush the result back to main memory when they exit the block.
Note that unlike data races, stale memory is not so easy to (re)produce, as flushes to main memory occur anyway.
The complier and CPU can (without any form of synchronization between threads) treat all code as single threaded. Meaning it can look at some code, that is very meaningful in a multithreading aspect, and treat it as if it’s single threaded, where it’s not so meaningful. So it can look at a code and decide, in sake of optimisation, to reorder it, or even remove parts of it completely, if it doesn’t know that this code is designed to work on multiple threads.
Consider the following code:
boolean b = false;
int x = 10;
void threadA() {
x = 20;
b = true;
}
void threadB() {
if (b) {
System.out.println(x);
}
}
You would think that threadB could only print 20 (or not print anything at all if threadB if-check is executed before setting b to true), as b is set to true only after x is set to 20, but the compiler/CPU might decide to reorder threadA, in that case threadB could also print 10. Marking b as volatile ensures that it won’t be reordered (or discarded in certain cases). Which mean threadB could only print 20 (or nothing at all). Marking the methods as syncrhonized will achieve the same result. Also marking a variable as volatile only ensures that it won’t get reordered, but everything before/after it can still be reordered, so synchronization can be more suited in some scenarios.
Note that before Java 5 New Memory Model, volatile didn’t solve this issue.
synchronized is method level/block level access restriction modifier. It will make sure that one thread owns the lock for critical section. Only the thread,which own a lock can enter synchronized block. If other threads are trying to access this critical section, they have to wait till current owner releases the lock.
volatile is variable access modifier which forces all threads to get latest value of the variable from main memory. No locking is required to access volatile variables. All threads can access volatile variable value at same time.
A good example to use volatile variable : Date variable.
Assume that you have made Date variable volatile. All the threads, which access this variable always get latest data from main memory so that all threads show real (actual) Date value. You don't need different threads showing different time for same variable. All threads should show right Date value.
Have a look at this article for better understanding of volatile concept.
Lawrence Dol cleary explained your read-write-update query.
Regarding your other queries
When is it more suitable to declare variables volatile than access them through synchronized?
You have to use volatile if you think all threads should get actual value of the variable in real time like the example I have explained for Date variable.
Is it a good idea to use volatile for variables that depend on input?
Answer will be same as in first query.
Refer to this article for better understanding.
I have been working on a certain server-type application for a while now, and I found that its design challenges the way I see memory coherence (so to speak) in Java.
This application uses NIO, therefore there is a limited amount of I/O threads (they only do network I/O and nothing else; they never terminate, but may get blocked waiting for more work).
Each connection is internally represented as an object of a specific type, let's call it ClientCon for the sake of this example. ClientCon has various session related fields, none of which are volatile. There is no synchronization of any kind in relation to getting/setting values for these fields.
Received data is made up of logical units with a fixed maximum size. Each such unit has some metadata that allows the handling type (class) to be decided. Once that is done, a new object of that type is created. All such handlers have fields, none of which are volatile. An I/O thread (a concrete I/O thread is assigned to each ClientCon) then calls a protected read method with remaining buffer contents (after metadata was read) on the new handler object.
After this, the same handler object is put into a special queue, which (the queue) is then submitted to a thread pool for execution (where each handler's run method is called to take actions based on the read data). For the sake of this example, we can say that TP threads never terminate.
Therefore, a TP thread will get its hands on an object it never had access to before. All fields of that object are non-volatile (and most/all are non-final, as they were modified outside the constructor).
The handler's run method may act based on session-specific fields in ClientCon as well as set them and/or act on handler object's own fields, whose values were set in the read method.
According to CPJ (Concurrent Programming in Java: Design and Principles):
The first time a thread accesses a field of an object, it sees either the initial value of the field or a value since written by some other thread.
A more comprehensive example of this quote can be found in JLS 17.5:
class FinalFieldExample {
final int x;
int y;
static FinalFieldExample f;
public FinalFieldExample() {
x = 3;
y = 4;
}
static void writer() {
f = new FinalFieldExample();
}
static void reader() {
if (f != null) {
int i = f.x; // guaranteed to see 3
int j = f.y; // could see 0
}
}
}
The class FinalFieldExample has a final int field x and a non-final
int field y. One thread might execute the method writer and another
might execute the method reader.
Because the writer method writes f after the object's constructor
finishes, the reader method will be guaranteed to see the properly
initialized value for f.x: it will read the value 3. However, f.y is
not final; the reader method is therefore not guaranteed to see the
value 4 for it.
This application has been running on x86 (and x86/64) Windows/Unix OSes (Linux flavors, Solaris) for years now (both Sun/Oracle and OpenJDK JVMs, versions 1.5 to 8) and apparently there have been no memory coherency issues related to received data handling. Why?
To sum it up, is there a way for a TP thread to see the object as it was initialized after construction and be unable to see all or some changes done by an I/O thread when it called the protected read method? If so, it would be nice if a detailed example could be presented.
Otherwise, are there some side-effects that could cause the object's field values to always be visible in other threads (e.g. I/O thread acquiring a monitor when adding the handler object to a queue)? Neither the I/O thread nor the TP thread synchronizes on the handler object itself. The queue does no such thing as well (not that it would make sense, anyway). Is this related to a concrete JVM's implementation details, perhaps?
EDIT:
It follows from the above definitions that:
An unlock on a monitor happens-before every subsequent lock on that
monitor. – Not applicable: monitor is not acquired on the handler object
A write to a volatile field (§8.3.1.4) happens-before every subsequent
read of that field. – Not applicable: no volatile fields
A call to start() on a thread happens-before any actions in the
started thread. – A TP thread might already exist when the queue with handler object(s) is submitted for execution. A new handler object might be added to queue amidst an execution on an existing TP thread.
All actions in a thread happen-before any other thread successfully
returns from a join() on that thread. – Not applicable: threads do not wait for each other
The default initialization of any object happens-before any other
actions (other than default-writes) of a program. – Not applicable: field writes are after default init AND after constructor finishes
When a program contains two conflicting accesses (§17.4.1) that are
not ordered by a happens-before relationship, it is said to contain a
data race.
and
Memory that can be shared between threads is called shared memory or
heap memory.
All instance fields, static fields, and array elements are stored in
heap memory. In this chapter, we use the term variable to refer to
both fields and array elements.
Local variables (§14.4), formal method parameters (§8.4.1), and
exception handler parameters (§14.20) are never shared between threads
and are unaffected by the memory model.
Two accesses to (reads of or writes to) the same variable are said to
be conflicting if at least one of the accesses is a write.
There was a write without forcing a HB relationship on field(s), and later there is a read, once again, not forcing a HB relationship on those field(s). Or am I horribly wrong here? That is, there is no declaration that anything about the object could have changed, so why would the JVM force-flush possibly cached values for these fields?
TL;DR
Thread #1 writes values to a new object's fields in a way that does not allow JVM to know that those values should be propagated to other threads.
Thread #2 acquires the object that was modified after construction by Thread #1 and reads those field values.
Why does the issue described in FinalFieldExample/JLS 17.5 NEVER happen in practice?
Why does Thread #2 never see only a default-initialized object (or, alternatively, the object as it was after construction, but before/in the mid of field value changes by Thread #1)?
I'm quite sure that when a thread pool starts a thread / runs a callable, it has a hapens-before semantics, so all changes before the happens-before are available to the thread.
The scenario you mentioned in CPJ is valid when you have more than one thread modifying data concurrently on the same object instance (e.g. 2 threads already running and modifying the same value (or values that happen to be next to each other in the heap).
It looks like in your case, there is no concurrent modification/read of the fields.
It might depend on what type of thread pool you are using. If it's an ExecutorService, then that class makes some strong guarantees about its task. From the documentation:
Memory consistency effects: Actions in a thread prior to the submission of a Runnable or Callable task to an ExecutorService happen-before any actions taken by that task, which in turn happen-before the result is retrieved via Future.get().
So when you initialize any object, plus any other objects, then submit that object to an ExecutorService, all those writes are made visible to the thread that will eventually handle your task.
Now, if you home-rolled your own thread pool, or you're using a thread pool with out these guarantees, then all bets are off. I'd say switch to something that has the guarantee though.
In practice one reason you will never see a violation here is "most are non-final" which means that there is at least one final field. The way HotSpot implements the guarantees given by the JLS when final fields are involved is to put a memory barrier at the end of the constructor, thereby granting the non-final fields the same visibility guarantees.
In theory now this is obviously not necessary, which means it depends on how you queue additional work in your thread pool. Generally I cannot imagine any design where there is not some synchronization going on when queueing the work, before it is executed - not only would it make working with this incredibly awkward (for the same reason that starting a thread invokes happens-before behavior), the way to implement such data structures also necessitates some synchronization.
Java's ThreadPoolExecutor.execute() for example does use a BlockingQueue internally which already gives you all the visibility and ordering guarantees you'd need.
I am wondering at the difference between declaring a variable as volatile and always accessing the variable in a synchronized(this) block in Java?
According to this article http://www.javamex.com/tutorials/synchronization_volatile.shtml there is a lot to be said and there are many differences but also some similarities.
I am particularly interested in this piece of info:
...
access to a volatile variable never has the potential to block: we're only ever doing a simple read or write, so unlike a synchronized block we will never hold on to any lock;
because accessing a volatile variable never holds a lock, it is not suitable for cases where we want to read-update-write as an atomic operation (unless we're prepared to "miss an update");
What do they mean by read-update-write? Isn't a write also an update or do they simply mean that the update is a write that depends on the read?
Most of all, when is it more suitable to declare variables volatile rather than access them through a synchronized block? Is it a good idea to use volatile for variables that depend on input? For instance, there is a variable called render that is read through the rendering loop and set by a keypress event?
It's important to understand that there are two aspects to thread safety.
execution control, and
memory visibility
The first has to do with controlling when code executes (including the order in which instructions are executed) and whether it can execute concurrently, and the second to do with when the effects in memory of what has been done are visible to other threads. Because each CPU has several levels of cache between it and main memory, threads running on different CPUs or cores can see "memory" differently at any given moment in time because threads are permitted to obtain and work on private copies of main memory.
Using synchronized prevents any other thread from obtaining the monitor (or lock) for the same object, thereby preventing all code blocks protected by synchronization on the same object from executing concurrently. Synchronization also creates a "happens-before" memory barrier, causing a memory visibility constraint such that anything done up to the point some thread releases a lock appears to another thread subsequently acquiring the same lock to have happened before it acquired the lock. In practical terms, on current hardware, this typically causes flushing of the CPU caches when a monitor is acquired and writes to main memory when it is released, both of which are (relatively) expensive.
Using volatile, on the other hand, forces all accesses (read or write) to the volatile variable to occur to main memory, effectively keeping the volatile variable out of CPU caches. This can be useful for some actions where it is simply required that visibility of the variable be correct and order of accesses is not important. Using volatile also changes treatment of long and double to require accesses to them to be atomic; on some (older) hardware this might require locks, though not on modern 64 bit hardware. Under the new (JSR-133) memory model for Java 5+, the semantics of volatile have been strengthened to be almost as strong as synchronized with respect to memory visibility and instruction ordering (see http://www.cs.umd.edu/users/pugh/java/memoryModel/jsr-133-faq.html#volatile). For the purposes of visibility, each access to a volatile field acts like half a synchronization.
Under the new memory model, it is still true that volatile variables cannot be reordered with each other. The difference is that it is now no longer so easy to reorder normal field accesses around them. Writing to a volatile field has the same memory effect as a monitor release, and reading from a volatile field has the same memory effect as a monitor acquire. In effect, because the new memory model places stricter constraints on reordering of volatile field accesses with other field accesses, volatile or not, anything that was visible to thread A when it writes to volatile field f becomes visible to thread B when it reads f.
-- JSR 133 (Java Memory Model) FAQ
So, now both forms of memory barrier (under the current JMM) cause an instruction re-ordering barrier which prevents the compiler or run-time from re-ordering instructions across the barrier. In the old JMM, volatile did not prevent re-ordering. This can be important, because apart from memory barriers the only limitation imposed is that, for any particular thread, the net effect of the code is the same as it would be if the instructions were executed in precisely the order in which they appear in the source.
One use of volatile is for a shared but immutable object is recreated on the fly, with many other threads taking a reference to the object at a particular point in their execution cycle. One needs the other threads to begin using the recreated object once it is published, but does not need the additional overhead of full synchronization and it's attendant contention and cache flushing.
// Declaration
public class SharedLocation {
static public volatile SomeObject someObject=new SomeObject(); // default object
}
// Publishing code
SharedLocation.someObject=new SomeObject(...); // new object is published
// Using code
// Note: do not simply use SharedLocation.someObject.xxx(), since although
// someObject will be internally consistent for xxx(), a subsequent
// call to yyy() might be inconsistent with xxx() if the object was
// replaced in between calls.
private String getError() {
SomeObject myCopy=SharedLocation.someObject; // gets current copy
...
int cod=myCopy.getErrorCode();
String txt=myCopy.getErrorText();
return (cod+" - "+txt);
}
// And so on, with myCopy always in a consistent state within and across calls
// Eventually we will return to the code that gets the current SomeObject.
Speaking to your read-update-write question, specifically. Consider the following unsafe code:
public void updateCounter() {
if(counter==1000) { counter=0; }
else { counter++; }
}
Now, with the updateCounter() method unsynchronized, two threads may enter it at the same time. Among the many permutations of what could happen, one is that thread-1 does the test for counter==1000 and finds it true and is then suspended. Then thread-2 does the same test and also sees it true and is suspended. Then thread-1 resumes and sets counter to 0. Then thread-2 resumes and again sets counter to 0 because it missed the update from thread-1. This can also happen even if thread switching does not occur as I have described, but simply because two different cached copies of counter were present in two different CPU cores and the threads each ran on a separate core. For that matter, one thread could have counter at one value and the other could have counter at some entirely different value just because of caching.
What's important in this example is that the variable counter was read from main memory into cache, updated in cache and only written back to main memory at some indeterminate point later when a memory barrier occurred or when the cache memory was needed for something else. Making the counter volatile is insufficient for thread-safety of this code, because the test for the maximum and the assignments are discrete operations, including the increment which is a set of non-atomic read+increment+write machine instructions, something like:
MOV EAX,counter
INC EAX
MOV counter,EAX
Volatile variables are useful only when all operations performed on them are "atomic", such as my example where a reference to a fully formed object is only read or written (and, indeed, typically it's only written from a single point). Another example would be a volatile array reference backing a copy-on-write list, provided the array was only read by first taking a local copy of the reference to it.
volatile is a field modifier, while synchronized modifies code blocks and methods. So we can specify three variations of a simple accessor using those two keywords:
int i1;
int geti1() {return i1;}
volatile int i2;
int geti2() {return i2;}
int i3;
synchronized int geti3() {return i3;}
geti1() accesses the value currently stored in i1 in the current thread.
Threads can have local copies of variables, and the data does not have to be the same as the data held in other threads.In particular, another thread may have updated i1 in it's thread, but the value in the current thread could be different from that updated value. In fact Java has the idea of a "main" memory, and this is the memory that holds the current "correct" value for variables. Threads can have their own copy of data for variables, and the thread copy can be different from the "main" memory. So in fact, it is possible for the "main" memory to have a value of 1 for i1, for thread1 to have a value of 2 for i1 and for thread2 to have a value of 3 for i1 if thread1 and thread2 have both updated i1 but those updated value has not yet been propagated to "main" memory or other threads.
On the other hand, geti2() effectively accesses the value of i2 from "main" memory. A volatile variable is not allowed to have a local copy of a variable that is different from the value currently held in "main" memory. Effectively, a variable declared volatile must have it's data synchronized across all threads, so that whenever you access or update the variable in any thread, all other threads immediately see the same value. Generally volatile variables have a higher access and update overhead than "plain" variables. Generally threads are allowed to have their own copy of data is for better efficiency.
There are two differences between volitile and synchronized.
Firstly synchronized obtains and releases locks on monitors which can force only one thread at a time to execute a code block. That's the fairly well known aspect to synchronized. But synchronized also synchronizes memory. In fact synchronized synchronizes the whole of thread memory with "main" memory. So executing geti3() does the following:
The thread acquires the lock on the monitor for object this .
The thread memory flushes all its variables, i.e. it has all of its variables effectively read from "main" memory .
The code block is executed (in this case setting the return value to the current value of i3, which may have just been reset from "main" memory).
(Any changes to variables would normally now be written out to "main" memory, but for geti3() we have no changes.)
The thread releases the lock on the monitor for object this.
So where volatile only synchronizes the value of one variable between thread memory and "main" memory, synchronized synchronizes the value of all variables between thread memory and "main" memory, and locks and releases a monitor to boot. Clearly synchronized is likely to have more overhead than volatile.
http://javaexp.blogspot.com/2007/12/difference-between-volatile-and.html
There are 3 main issues with multithreading:
Race Conditions
Caching / stale memory
Compiler and CPU optimisations
volatile can solve 2 & 3, but can't solve 1. synchronized/explicit locks can solve 1, 2 & 3.
Elaboration:
Consider this thread unsafe code:
x++;
While it may look like one operation, it's actually 3: reading the current value of x from memory, adding 1 to it, and saving it back to memory. If few threads try to do it at the same time, the result of the operation is undefined. If x originally was 1, after 2 threads operating the code it may be 2 and it may be 3, depending on which thread completed which part of the operation before control was transferred to the other thread. This is a form of race condition.
Using synchronized on a block of code makes it atomic - meaning it make it as if the 3 operations happen at once, and there's no way for another thread to come in the middle and interfere. So if x was 1, and 2 threads try to preform x++ we know in the end it will be equal to 3. So it solves the race condition problem.
synchronized (this) {
x++; // no problem now
}
Marking x as volatile does not make x++; atomic, so it doesn't solve this problem.
In addition, threads have their own context - i.e. they can cache values from main memory. That means that a few threads can have copies of a variable, but they operate on their working copy without sharing the new state of the variable among other threads.
Consider that on one thread, x = 10;. And somewhat later, in another thread, x = 20;. The change in value of x might not appear in the first thread, because the other thread has saved the new value to its working memory, but hasn't copied it to the main memory. Or that it did copy it to the main memory, but the first thread hasn't updated its working copy. So if now the first thread checks if (x == 20) the answer will be false.
Marking a variable as volatile basically tells all threads to do read and write operations on main memory only. synchronized tells every thread to go update their value from main memory when they enter the block, and flush the result back to main memory when they exit the block.
Note that unlike data races, stale memory is not so easy to (re)produce, as flushes to main memory occur anyway.
The complier and CPU can (without any form of synchronization between threads) treat all code as single threaded. Meaning it can look at some code, that is very meaningful in a multithreading aspect, and treat it as if it’s single threaded, where it’s not so meaningful. So it can look at a code and decide, in sake of optimisation, to reorder it, or even remove parts of it completely, if it doesn’t know that this code is designed to work on multiple threads.
Consider the following code:
boolean b = false;
int x = 10;
void threadA() {
x = 20;
b = true;
}
void threadB() {
if (b) {
System.out.println(x);
}
}
You would think that threadB could only print 20 (or not print anything at all if threadB if-check is executed before setting b to true), as b is set to true only after x is set to 20, but the compiler/CPU might decide to reorder threadA, in that case threadB could also print 10. Marking b as volatile ensures that it won’t be reordered (or discarded in certain cases). Which mean threadB could only print 20 (or nothing at all). Marking the methods as syncrhonized will achieve the same result. Also marking a variable as volatile only ensures that it won’t get reordered, but everything before/after it can still be reordered, so synchronization can be more suited in some scenarios.
Note that before Java 5 New Memory Model, volatile didn’t solve this issue.
synchronized is method level/block level access restriction modifier. It will make sure that one thread owns the lock for critical section. Only the thread,which own a lock can enter synchronized block. If other threads are trying to access this critical section, they have to wait till current owner releases the lock.
volatile is variable access modifier which forces all threads to get latest value of the variable from main memory. No locking is required to access volatile variables. All threads can access volatile variable value at same time.
A good example to use volatile variable : Date variable.
Assume that you have made Date variable volatile. All the threads, which access this variable always get latest data from main memory so that all threads show real (actual) Date value. You don't need different threads showing different time for same variable. All threads should show right Date value.
Have a look at this article for better understanding of volatile concept.
Lawrence Dol cleary explained your read-write-update query.
Regarding your other queries
When is it more suitable to declare variables volatile than access them through synchronized?
You have to use volatile if you think all threads should get actual value of the variable in real time like the example I have explained for Date variable.
Is it a good idea to use volatile for variables that depend on input?
Answer will be same as in first query.
Refer to this article for better understanding.