Okay, suppose I have a bunch of variables, one of them declared volatile:
int a;
int b;
int c;
volatile int v;
If one thread writes to all four variables (writing to v last), and another thread reads from all four variables (reading from v first), does that second thread see the values written to a, b and c by the first thread, even though they are not themselves declared volatile? Or can it possibly see stale values?
Since there seems to be some confusion: I'm not deliberately trying to do something unsafe. I just want to understand the Java memory model and the semantics of the volatile keyword. Pure curiosity.
I'm going to speak to what I think you may really be probing about—piggybacking synchronization.
The technique that it looks like you're trying to use involves using one volatile variable as a synchronization guard in concert with one or more other non-volatile variables. This technique is applicable when the following conditions hold true:
Only one thread will write to the set of values meant to be guarded.
The threads reading the set of values will read them only if the volatile guard value meets some criteria.
You don't mention the second condition holding true for your example, but we can examine it anyway. The model for the writer is as follows:
Write to all the non-volatile variables, assuming that no other thread will try to read them.
Once complete, write a value to the volatile guard variable that indicates that the readers' criteria is met.
The readers operate as follows:
Read the volatile guard variable at any time, and if its value meets the criteria, then
Read the other non-volatile variables.
The readers must not read the other non-volatile variables if the volatile guard variable does not yet indicate a proper value.
The guard variable is acting as a gate. It's closed until the writer sets it to a particular value, or set of values that all meet the criteria of indicating that the gate is now open. The non-volatile variables are guarded behind the gate. The reader is not permitted to read them until the gate opens. Once the gate is open, the reader will see a consistent view of the set of non-volatile variables.
Note that it is not safe to run this protocol repeatedly. The writer can't keep changing the non-volatile variables once it's opened the gate. At that point, multiple reader threads may be reading those other variables, and they can—though are not guaranteed—see updates to those variables. Seeing some but not all of those updates would yield inconsistent views of the set.
Backing up, the trick here is to control access to a set of variables without either
creating a structure to hold them all, to which an atomic reference could be swapped, um, atomically, or
using a lock to make writing to and reading from the entire set of variables mutually exclusive activities.
Piggybacking on top of the volatile guard variable is a clever stunt—not one to be done casually. Subsequent updates to the program can break the aforementioned fragile conditions, removing the consistency guarantees afforded by the Java memory model. Should you choose to use this technique, document its invariants and requirements in the code clearly.
Yes. volatile, locks, etc., setup the happens-before relationship, but it affects all variables (in the new Java Memory Model (JMM) from Java SE 5/JDK 1.4). Kind of makes it useful for non-primitive volatiles...
does that second thread see the values written to a, b and c by the first thread, even though they are not themselves declared volatile? Or can it possibly see stale values?
You will get stale reads, b/c you can't ensure that the values of a, b, c are the ones set after reading of v. Using state machine (but you need CAS to change the state) is a way to tackle similar issues but it's beyond the scope of the discussion.
Perhaps this part is unclear, after writing to v and reading first from v, you'd get the right results (non-stale reads), the main issue is that if you do
if (v==STATE1){...proceed...}, there is no guarantee some other thread would not be modifying the state of a/b/c. In that case, there will be state reads.
If you modify the a/b/c+v once only you'd get the correct result.
Mastering concurrency and and lock-free structures is a really hard one. Doug Lea has a good book on and most talks/articles of Dr. Cliff Click are a wonderful wealth, if you need something to start digging in.
Yes, volatile write "happens-before" next volatile read on the same variable.
While #seh is right on about consistency problems with multiple variables, there are use cases that less consistency is required.
For example, a writer thread updates some state variables; a reader thread displays them promptly. There's not much relation among the variables, we only care about reading the new values promptly. We could make every state variable volatile. Or we could use only one volatile variable as visibility guard.
However, the saving is only on the paper, performance wise there's hardly any difference. In either version, every state variable must be "flushed" by the writer and "loaded" by the reader. No free lunch.
Related
From the book Effective Java:
While the volatile modifier performs no mutual exclusion, it guarantees that any thread that reads the field will see the most recently written value
SO and many other sources claim similar things.
Is this true?
I mean really true, not a close-enough model, or true only on x86, or only in Oracle JVMs, or some definition of "most recently written" that's not the standard English interpretation...
Other sources (SO example) have said that volatile in Java is like acquire/release semantics in C++. Which I think do not offer the guarantee from the quote.
I found that in the JLS 17.4.4 it says "A write to a volatile variable v (§8.3.1.4) synchronizes-with all subsequent reads of v by any thread (where "subsequent" is defined according to the synchronization order)." But I don't quite understand.
There are quite some sources for and against this, so I'm hoping the answer is able to convince that many of those (on either side) are indeed wrong - for example reference or spec, or counter-example code.
Is this true?
I mean really true, not a close-enough model, or true only on x86, or only in Oracle JVMs, or some definition of "most recently written" that's not the standard English interpretation...
Yes, at least in the sense that a correct implementation of Java gives you this guarantee.
Unless you are using some exotic, experimental Java compiler/JVM (*), you can essentially take this as true.
From JLS 17.4.5:
A write to a volatile field (§8.3.1.4) happens-before every subsequent read of that field.
(*) As Stephen C points out, such an exotic implementation that doesn't implement the memory model semantics described in the language spec can't usefully (or even legally) be described as "Java".
The quote per-se is correct in terms of what is tries to prove, but it is incorrect on a broader view.
It tries to make a distinction of sequential consistency and release/acquire semantics, at least in my understanding. The difference is rather "thin" between these two terms, but very important. I have tried to simplify the difference at the beginning of this answer or here.
The author is trying to say that volatile offers that sequential consistency, as implied by that:
"... it guarantees that any thread.."
If you look at the JLS, it has this sentence:
A write to a volatile field (§8.3.1.4) happens-before every subsequent read of that field.
The tricky part there is that subsequent and it's meaning, and it has been discussed here. What is really wants to mean is "subsequent that observes that write". So happens-before is guaranteed when the reader observes the value that the writer has written.
This already implies that a write is not necessarily seen on the next read, and this can be the case where speculative execution is allowed. So in this regard, the quote is miss-leading.
The quote that you found:
A write to a volatile variable v (§8.3.1.4) synchronizes-with all subsequent reads of v by any thread (where "subsequent" is defined according to the synchronization order)
is a complicated to understand without a much broader context. In simple words, it established synchronizes-with order (and implicitly happens-before) between two threads, where volatile v variables is a shared variable. here is an answer where this has broader explanation and thus should make more sense.
It is not true. JMM is based on sequential consistency and for sequential consistency real time ordering isn't guaranteed; for that you need linearizability. In other words, reads and writes can be skewed as long as the program order isn't violated (or as long is it can't be proven po was violated).
A read of volatile variable a, needs to see the most recent written value before it in the memory order. But that doesn't imply real time ordering.
Good read about the topic:
https://concurrency-interest.altair.cs.oswego.narkive.com/G8KjyUtg/relativity-of-guarantees-provided-by-volatile.
I'll make it concrete:
Imagine there are 2 CPU's and (volatile) variable A with initial value 0. CPU1 does a store A=1 and CPU2 does a load of A. And both CPUs have the cacheline containing A in SHARED state.
The store is first speculatively executed and written to the store buffer; eventually the store commits and retires, but since the stored value is still in the store buffer; it isn't visible yet to the CPU2. Till so far it wasn't required for the cacheline to be in an EXCLUSIVE/MODIFIED state, so the cacheline on CPU2 still contains the old value and hence CPU2 can still read the old value.
So in the real time order, the write of A is ordered before the read of A=0, but in the synchronization order, the write of A=1 is ordered after the read of A=0.
Only when the store leaves the store buffer and wants to enter the L1 cache, the request for ownership (RFO) is send to all other CPU's which set the cacheline containing A to INVALID on CPU2 (RFO prefetching I'll leave out of the discussion). If CPU2 would now read A, it is guaranteed to see A=1 (the request will block till CPU1 has completed the store to the L1 cache).
On acknowledgement of the RFO the cacheline is set to MODIFIED on CPU1 and the store is written to the L1 cache.
So there is a period of time between when the store is executed/retired and when it is visible to another CPU. But the only way to determine this is if you would add special measuring equipment to the CPUs.
I believe a similar delaying effect can happen on the reading side with invalidation queues.
In practice this will not be an issue because store buffers have a limited capacity and need to be drained eventually (so a write can't be invisible indefinitely). So in day to day usage you could say that a volatile read, reads the most recent write.
A java volatile write/read provides release/acquire semantics, but keep in mind that the volatile write/read is stronger than release/acquire semantics. A volatile write/read is sequential consistent and release/acquire semantics isn't.
From Javadocs
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.
When changes made to a volatile variable are always visible to any other thread, then why volatile variable cant be used in case of multiple threads writing to that variable. Why is volatile only used for cases when one thread is writing or reading to that while the other thread is only reading the variable?
If changes are always visible to other threads, then suppose if thread B wants to write to that variable, it will see the new value(updated by thread A) and update it. And when the thread A again wants to write, it will again see the updated value by thread B and write to it.Where is the problem in that?
In short, i am not able to understand this.
if two threads are both reading and writing to a shared variable, then
using the volatile keyword for that is not enough. You need to use
synchronization in that case to guarantee that the reading and writing
of the variable is atomic.
There are plenty of purposes that volatile works fine for — but also plenty of purposes that it doesn't. For example, imagine that you have a field like this:
private volatile int i;
and two threads that both run ++this.i: reading this.i and then writing to it.
The problem is that ++this.i is a volatile read followed by a completely separate volatile write. Any number of things could have happened between the read and the write; in particular, you could get a situation where both threads read i before either thread writes to it. The net effect is that the value of i increases by only 1, even though two separate threads both incremented it.
AtomicInteger (and the other atomics) address this sort of problem by allowing you to simultaneously read and write in a single atomic (≈ volatile) step. (They do this by using a compare-and-swap instruction that performs the write only if the value that was read is still the current value. The increment-and-get method just runs a loop that retries this until the write actually succeeds.)
Think about what "atomicity" means. It means that two or more operations that happen in one thread appear to happen as an atomic unit as far as other threads can tell.
So if I declare some volatile int foobar, and I write code to perform some operations on it, how would the compiler know which of those operations are supposed to be the atomic unit?
When you write a synchronized block, the atomic unit is whatever you put inside the block.
So I am reading this book titled Java Concurrency in Practice and I am stuck on this one explanation which I cannot seem to comprehend without an example. This is the quote:
When thread A writes to a volatile
variable and subsequently thread B
reads that same variable, the values
of all variables that were visible to
A prior to writing to the volatile
variable become visible to B after
reading the volatile variable.
Can someone give me a counterexample of why "the values of ALL variables that were visible to A prior to writing to the volatile variable become visible to B AFTER reading the volatile variable"?
I am confused why all other non-volatile variables do not become visible to B before reading the volatile variable?
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.
Just for your reference, When is volatile needed ?
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.
From JLS §17.4.7 Well-Formed Executions
We only consider well-formed
executions. An execution E = < P, A,
po, so, W, V, sw, hb > is well formed
if the following conditions are true:
Each read sees a write to the same
variable in the execution. All reads
and writes of volatile variables are
volatile actions. For all reads r in
A, we have W(r) in A and W(r).v = r.v.
The variable r.v is volatile if and
only if r is a volatile read, and the
variable w.v is volatile if and only
if w is a volatile write.
Happens-before order is a partial
order. Happens-before order is given
by the transitive closure of
synchronizes-with edges and program
order. It must be a valid partial
order: reflexive, transitive and
antisymmetric.
The execution obeys
intra-thread consistency. For each
thread t, the actions performed by t
in A are the same as would be
generated by that thread in
program-order in isolation, with each
write wwriting the value V(w), given
that each read r sees the value
V(W(r)). Values seen by each read are
determined by the memory model. The
program order given must reflect the
program order in which the actions
would be performed according to the
intra-thread semantics of P.
The execution is happens-before consistent
(§17.4.6).
The execution obeys
synchronization-order consistency. For
all volatile reads r in A, it is not
the case that either so(r, W(r)) or
that there exists a write win A such
that w.v = r.v and so(W(r), w) and
so(w, r).
Useful Link : What do we really know about non-blocking concurrency in Java?
Thread B may have a CPU-local cache of those variables. A read of a volatile variable ensures that any intermediate cache flush from a previous write to the volatile is observed.
For an example, read the following link, which concludes with "Fixing Double-Checked Locking using Volatile":
http://www.cs.umd.edu/~pugh/java/memoryModel/DoubleCheckedLocking.html
If a variable is non-volatile, then the compiler and the CPU, may re-order instructions freely as they see fit, in order to optimize for performance.
If the variable is now declared volatile, then the compiler no longer attempts to optimize accesses (reads and writes) to that variable. It may however continue to optimize access for other variables.
At runtime, when a volatile variable is accessed, the JVM generates appropriate memory barrier instructions to the CPU. The memory barrier serves the same purpose - the CPU is also prevent from re-ordering instructions.
When a volatile variable is written to (by thread A), all writes to any other variable are completed (or will atleast appear to be) and made visible to A before the write to the volatile variable; this is often due to a memory-write barrier instruction. Likewise, any reads on other variables, will be completed (or will appear to be) before the
read (by thread B); this is often due to a memory-read barrier instruction. This ordering of instructions that is enforced by the barrier(s), will mean that all writes visible to A, will be visible B. This however, does not mean that any re-ordering of instructions has not happened (the compiler may have performed re-ordering for other instructions); it simply means that if any writes visible to A have occurred, it would be visible to B. In simpler terms, it means that strict-program order is not maintained.
I will point to this writeup on Memory Barriers and JVM Concurrency, if you want to understand how the JVM issues memory barrier instructions, in finer detail.
Related questions
What is a memory fence?
What are some tricks that a processor does to optimize code?
Threads are allowed to cache variable values that other threads may have since updated since they read them. The volatile keyword forces all threads to not cache values.
This is simply an additional bonus the memory model gives you, if you work with volatile variables.
Normally (i.e. in the absence of volatile variables and synchronization), the VM can make variables from one thread visible to other threads in any order it wants, or not at all. E.g. the reading thread could read some mixture of earlier versions of another threads variable assignments. This is caused by the threads being maybe run on different CPUs with their own caches, which are only sometimes copied to the "main memory", and additionally by code reordering for optimization purposes.
If you used a volatile variable, as soon as thread B read some value X from it, the VM makes sure that anything which thread A has written before it wrote X is also visible to B. (And also everything which A got guaranteed as visible, transitively).
Similar guarantees are given for synchronized blocks and other types of locks.
I was just wondering if someone could explain the meaning of this:
Operations like increment and
decrement (e.g. ++ and --) can't be
used on a volatile variable because
these operations are syntactic sugar
for a load, change and a store.
I think increment and decrement should just work fine for a volatile variable, the only difference would be every time you read or write you would be accessing from/writing to main memory rather than from cache.
volatile variable only ensures visibility . It does not ensure atomicity. I guess, that is how the statement should be interpreted.
I think you're taking the quote out of context.
Of course ++ and -- can be applied to volatile variables. They just won't be atomic.
And since volatile often implies that they must be handled in an atomic manner, this is counter to the goal.
The problem with ++ and -- is that they might feel like they are atomic, when indeed they are not.
Doing a = a + 1 makes it (somewhat) explicit that it is not an atomic operation, but one might (wrongly) think that a++ is atomic.
The Java Language Specification does not have atomic operations for the ++ and -- operators. In other words, when you write code in the following manner:
a++;
the Java compiler actually emits code that is similar to the set of steps below (the actual instructions will vary depending on the nature of the variable):
Load the operand onto the stack using one of the operations for loading data.
Duplicate the value of the operand on the stack (for the purpose of returning later). This usually accomplished using a dup operation.
Increment the value on the stack. Usually accomplished using the iadd operation in the VM.
Return the value (obtained in step 2).
As you can observe, there are multiple operations in the VM for what is commonly thought to be an atomic operation. The VM can ensure atomicity only upto the level of an individual operation. Any further requirement can be achieved only via synchronization or other techniques.
Using the volatile keyword, allows other threads to obtain the most recent value of a variable; all read operations on a variable will return the recently updated value on a per-instruction basis. For example, if the variable a were to be volatile in the previous example, then a thread reading the value of a would see different values if it were to read a after instruction 2 and after instruction 3. Use of volatile does not protect against this scenario. It protects against the scenario where multiple threads see multiple values for a after instruction 2 (for instance).
Volatile does not garanty atomicity in an opeartion that involves multiple steps.
Look at it this way it I am reading a value and that is all am doing, the read operation is an atomic operation. It is a single step and hence the use of volatile here will be fine. If however I am reading that value and changing that value before writing back, that is a multistep operation and for this volatile does not manage the atomicity.
The increment and decrement opeartions are multi-stepped and hence the use of the volatile modifier is not sufficient.
Nope -- you use "volatile" to indicate that the variable can be changed by an external entity.
This would typically be some JNI C code, or, a special register linked to some hardware such as a thermometer. Java cannot guarantee that all JVMs on all architectures can will be capable of incrementing these values in a single machine cycle. So it doesnt let you do it anywhere.
This may seem a very silly question.
Consider this:
I have a simple Boolean object with a getter and a setter. Now both of the methods are called from a lot of threads very frequently.
Do I need to have a synchronization for this boolean?
Also are Boolean assignments atomic operations?
[UPDATE]:
I know about Atomic Boolean already. I already have a lot of varied solutions, But I was specifically looking for answers and justification of answers for the above 2 question.
No, Boolean access is NOT atomic (on the level of machine code), although it does take "only 1 operation in Java".
Therefore, yes, you do need synchronization for Boolean.
Please see slides 4-6 of this presentation for code examples.
On a related note, you should not synchronize on a Boolean
Yes. But if it's a flag that is written from one thread only and you want to ensure visibility to all threads, then a cheap alternative is using volatile.
Yes - even though the internal representation of the object (i.e. the actual boolean flag inside the Boolean wrapper) were 64 bit and it could therefore get "split" in a concurrent situation, a boolean can only have one of the two values, right? So plain assignments (get or set) are atomic, but if you're doing anything else (like check-then-act), for instance x = !x, then it's of course not atomic unless synchronized.
From a technical perspective, synchronization is not required for writes in one thread to be perceived in another thread. What you do need is a happens-before edge. It is likely that either volatile or synchronized will be used to achieve the happens-before edge. Both of those techniques result in a synchronized-with edge. So, in practice, you will probably use synchronization to manage the state of your boolean.
Yes. Note that you are not changing the state of the Boolean object. You are only modifying the reference to the Boolean object. Section 17.7 of the language specification states that "writes to and reads of references are always atomic."
Update: Let me expound upon the need for a happens-before edge. Without a happens-before edge, then the changes that one thread makes to the variable are not guaranteed to ever be perceived by the other threads. It is not simply that the change may be perceived at a bad time such as in between a read and a write. The change may never be perceived.
Let's say that we have a boolean variable that we initialize to false. Then we start two threads. The first thread sets the variable to true and stop. The second thread continually checks the variable until it is true, after which it stops. There is no guarantee that the second thread will ever see the variable as true.
Use AtomicBoolean.
No, you don't. But declare the variable volatile so that the values are reflected in all threads that are accessing the boolean. If you look at AtomicBoolean's set(..) method, it doesn't have any synchronization either.
Yes, practically assignment is atomic. Just setting the value does not need synchronization. However, if you want to do something like:
if (!bool) {
bool = false;
}
then you need synchronization (or AtomicBoolean, which is more efficient than synchronization)
http://docs.oracle.com/javase/tutorial/essential/concurrency/atomic.html
says that read and write of primitive variables is atomic.
Hence it is possible to enforce strict alternation or happens after relationships using boolean (or volatile boolean in case of cache effects)
even if it was atomic there are still syncronization isues since you will propably check the value sometime
e.g.
if (boolVar==true)
-> other thread takes control
do_something();