了解 Java 易失性可见性

Question

I'm reading about the Java volatile keyword and have confusion about its 'visibility'.

A typical usage of volatile keyword is:

volatile boolean ready = false;
int value = 0;

void publisher() {
    value = 5;
    ready = true;
}

void subscriber() {
    while (!ready) {}
    System.out.println(value);
}

As explained by most tutorials, using volatile for ready makes sure that:

• change to ready on publisher thread is immediately visible to other threads (subscriber);
• when ready 's change is visible to other thread, any variable update preceding to ready (here is value 's change) is also visible to other threads;

I understand the 2nd, because volatile variable prevents memory reordering by using memory barriers, so writes before volatile write cannot be reordered after it, and reads after volatile read cannot be reordered before it. This is how ready prevents printing value = 0 in the above demo.

But I have confusion about the 1st guarantee, which is visibility of the volatile variable itself. That sounds a very vague definition to me.

In other words, my confusion is just on SINGLE variable's visibility, not multiple variables' reordering or something. Let's simplify the above example:

volatile boolean ready = false;

void publisher() {
    ready = true;
}

void subscriber() {
    while (!ready) {}
}

If ready is not defined volatile, is it possible that subscriber get stuck infinitely in the while loop? Why?

A few questions I want to ask:

• What does 'immediately visible' mean? Write operation takes some time, so after how long can other threads see volatile's change? Can a read in another thread that happens very shortly after the write starts but before the write finishes see the change?
• Visibility, for modern CPUs is guaranteed by cache coherence protocol (eg MESI) anyway, so what can volatile help here?
• Some articles say volatile variable uses memory directly instead of CPU cache, which guarantees visibility between threads. That doesn't sound a correct explain.

   Time : ---------------------------------------------------------->

 writer : --------- | write | -----------------------
reader1 : ------------- | read | -------------------- can I see the change?
reader2 : --------------------| read | -------------- can I see the change?

Hope I explained my question clearly.

Answer 1

Relevant bits of the language spec:

volatile keyword: https://docs.oracle.com/javase/specs/jls/se16/html/jls-8.html#jls-8.3.1.4

memory model: https://docs.oracle.com/javase/specs/jls/se16/html/jls-17.html#jls-17.4

The CPU cache is not a factor here, as you correctly said.

This is more about optimizations. If ready is not volatile, the compiler is free to interpret

// this
while (!ready) {}

// as this
if (!ready) while(true) {}

That's certainly an optimization, it has to evaluate the condition fewer times. The value is not changed in the loop, it can be "reused". In terms of single-thread semantics it is equivalent, but it won't do what you wanted.

That's not to say this would always happen. Compilers are free to do that, they don't have to.

Answer 2

Visibility, for modern CPUs is guaranteed by cache coherence protocol (eg MESI) anyway, so what can volatile help here?

That doesn't help you. You aren't writing code for a modern CPU, you are writing code for a Java virtual machine that is allowed to have a virtual machine that has a virtual CPU whose virtual CPU caches are not coherent.

Some articles say volatile variable uses memory directly instead of CPU cache, which guarantees visibility between threads. That doesn't sound a correct explain.

That is correct. But understand, that's with respect to the virtual machine that you are coding for. Its memory may well be implemented in your physical CPU's caches. That may allow your machine to use the caches and still have the memory visibility required by the Java specification.

Using volatile may ensure that writes go directly to the virtual machine's memory instead of the virtual machine's virtual CPU cache. The virtual machine's CPU cache does not need to provide visibility between threads because the Java specification doesn't require it to.

You cannot assume that characteristics of your particular physical hardware necessarily provide benefits that Java code can use directly. Instead, the JVM trades off those benefits to improve performance. But that means your Java code doesn't get those benefits.

Again, you are not writing code for your physical CPU , you are writing code for the virtual CPU that your JVM provides. That your CPU has coherent caches allows the JVM to do all kinds of optimizations that boost your code's performance, but the JVM is not required to pass those coherent caches through to your code and real JVM's do not. Doing so would mean eliminating a significant number of extremely valuable optimizations.

Answer 3

If ready is not defined volatile, is it possible that subscriber get stuck infinitely in the while loop?

Yes.

Why?

Because the subscriber may not ever see the results of the publisher's write.

Because ... the JLS does not require the value of an variable to be written to memory ... except to satisfy the specified visibility constraints.

What does 'immediately visible' mean? Write operation takes some time, so after how long can other threads see volatile's change? Can a read in another thread that happens very shortly after the write starts but before the write finishes see the change?

(I think) that the JMM specifies or assumes that it is physically impossible to read and write the same conceptual memory cell at the same time. So operations on a memory cell are time ordered. Immediately visible means visible in the next possible opportunity to read following the write.

Visibility, for modern CPUs is guaranteed by cache coherence protocol (eg MESI) anyway, so what can volatile help here?

1. Compilers typically generate code that holds variables in registers, and only writes the values to memory when necessary . Declaring a variable as volatile means that the value must be written to memory. If you take this into consideration, you cannot rely on just the (hypothetical or actual) behavior of cache implementations to specify what volatile means.

2. While current generation modern CPU / cache architectures behave that way, there is no guarantee that all future computers will behave that way.

Some articles say volatile variable uses memory directly instead of CPU cache, which guarantees visibility between threads.

Some people say that is incorrect ... for CPUs that implement a cache coherency protocol. However, that is beside the point, because as I described above, the current value of a variable may not yet have been written to the cache yet. Indeed, it may never be written to the cache.

 Time : ----------------------------------------------------------> writer : --------- | write | ----------------------- reader1 : ------------- | read | -------------------- can I see the change? reader2 : --------------------| read | -------------- can I see the change?

So lets assume that your diagram shows physical time and represents threads running on different physical cores, reading and writing a cache-coherent memory cell via their respective caches.

What would happen at the physical level would depend on how the cache-coherency is implemented.

I would expect Reader 1 to see the previous state of the cell (if it was available from its cache) or the new state if it wasn't. Reader 2 would see the new state. But it also depends on how long it takes for the writer thread's cache invalidation to propagate to the others' caches. And all sorts of other stuff that is hard to explain.

In short, we don't really know what would happen at the physical level.

But on the other hand, the writer and readers in the above picture can't actually observe the physical time like that anyway. And neither can the programmer.

What the program / programmer sees is that the reads and writes DO NOT OVERLAP. When the necessary happens before relations are present, there will be guarantees about visibility of memory writes by one thread to subsequent ¹ reads by another. This applies for volatile variables, and for various other things.

How that guarantee is implemented, is not your problem. And it really doesn't help if you do understand what it going on at the hardware level, because you don't actually know what code the JIT compiler is going to emit (today!) anyway.

---

^{1 - That is, subsequent according to the synchronization order ... which you could view as a logical time. The JLS Memory model doesn't actually talk about time at all.}

Answer 4

Answers to your 3 questions:

1. A change of a volatile write doesn't need to be 'immediately' visible to a volatile load. A correctly synchronized Java program will behave as if it is sequential consistent and for sequential consistency the real time order of loads/stores isn't relevant. So reads and writes can be skewed as long as the program order isn't violated (or as long as nobody can observe it). Linearizability = sequential consistency + respect real time order. For more info see this answer .

2. I still need to dig into the exact meaning of visible, but AFAIK it is mostly a compiler level concern because hardware will prevent buffering loads/stores indefinitely.

3. You are completely right about the articles being wrong. A lot of nonsense is written and 'flushing volatile writes to main memory instead of using the cache' is the most common misunderstanding I'm seeing. I think 50% of all my SO comments is about informing people that caches are always coherent. A great book on the topic is 'A primer on memory consistency and cache coherence 2e' which is available for free .

The informal semantics of the Java Memory model contains 3 parts:

• atomicity
• visibility
• ordering

Atomicity is about making sure that a read/write/rmw happens atomically in the global memory order. So nobody can observe some in between state. This deals with access atomicity like torn read/write, word tearing and proper alignment. It also deals with operational atomicity like rmw.

IMHO it should also deal with store atomicity; so making sure that there is a point in time where the store becomes visibly to all cores. If you have for example the X86, then due to load buffering, a store can become visible to the issuing core earlier than to other cores and you have a violation of atomicity. But I haven't seen it being mentioned in the JMM.

Visibility: this deals mostly with preventing compiler optimizations since the hardware will prevent delaying loads and buffering stores indefinitely. In some literature they also throw ordering of surrounding loads/stores under visibility; but I don't believe this is correct.

Ordering: this is the bread and butter of memory models. It will make sure that loads/stores issued by a single processor don't get reordered. In the first example you can see the need for such behavior. This is the realm of the compiler barriers and cpu memory barriers.

For more info see: https://download.oracle.com/otndocs/jcp/memory_model-1.0-pfd-spec-oth-JSpec/

Answer 5

I'll just touch on this part :

change to ready on publisher thread is immediately visible to other threads

that is not correct and the articles are wrong. The documentation makes a very clear statement here :

A write to a volatile field happens-before every subsequent read of that field.

The complicated part here is subsequent . In plain english this means that when someone sees ready as being true , it will also see value as being 5 . This automatically implies that you need to observe that value to be true , and it can happen that you might observe a different thing. So this is not "immediately".

What people confuse this with, is the fact that volatile offers sequential consistency , which means that if someone has observed ready == true , then everyone will also (unlike release/acquire , for example).