Explain how JIT reordering works

问题

I have been reading a lot about synchronization in Java and all the problems that can occur. However, what I'm still slightly confused about is how the JIT can reorder a write.

For instance, a simple double check lock makes sense to me:

  class Foo {
    private volatile Helper helper = null; // 1
    public Helper getHelper() { // 2
        if (helper == null) { // 3
            synchronized(this) { // 4
                if (helper == null) // 5
                    helper = new Helper(); // 6
            }
        }
        return helper;
    }
}

We use volatile on line 1 to enforce a happens-before relationship. Without it, is entirely possible for the JIT to reoder our code. For example:

1. Thread 1 is at line 6 and memory is allocated to helper however, the constructor has not yet run because the JIT could reordering our code.

2. Thread 2 comes in at line 2 and gets an object that is not fully created yet.

I understand this, but I don't fully understand the limitations that the JIT has on reordering.

For instance, say I have a method that creates and puts a MyObject into a HashMap<String, MyObject> (I know that a HashMapis not thread safe and should not be used in a multi-thread environment, but bear with me). Thread 1 calls createNewObject:

public class MyObject {
    private Double value = null;

    public MyObject(Double value) {
        this.value = value;
    }
}

Map<String, MyObject> map = new HashMap<String, MyObject>();

public void createNewObject(String key, Double val){
    map.put(key, new MyObject( val ));
}

At the same time thread 2 calls a get from the Map.

public MyObject getObject(String key){
    return map.get(key);
}

Is it possible for thread 2 to receive an object from getObject(String key) that is not fully constructed? Something like:

1. Thread 1: Allocate memory for new MyObject( val )
2. Thread 1: Place object in map
3. Thread 2: call getObject(String key)
4. Thread 1: Finish constructing the new MyObject.

Or will map.put(key, new MyObject( val )) not put an object into the map until it's fully constructed?

I'd imagine that the answer is, it wouldn't put an object into the Map until it is fully constructed (because that sounds awful). So how can the JIT reorder?

In a nutshell can it only reorder when creating a new Object and assigning it to a reference variable, like the double checked lock? A complete rundown on the JIT might be much for a SO answer, but what I'm really curious about is how it can reorder a write (like line 6 on the double checked lock) and what stops it from putting an object into a Map that is not fully constructed.

回答1:

WARNING: WALL OF TEXT

The answer to your question is before the horizontal line. I will continue to explain deeper the fundamental problem in the second portion of my answer (which is not related to the JIT, so that's it if you are only interested in the JIT). The answer to the second part of your question lies at the bottom because it relies on what I describe further.

TL;DR The JIT will do whatever it wants, the JMM will do whatever it wants, being valid under the condition that you let them by writing thread unsafe code.

NOTE: "initialization" refers to what happens in the constructor, which excludes anything else such as calling a static init method after constructing etc...

---

"If the reordering produces results consistent with a legal execution, it is not illegal." (JLS 17.4.5-200)

If the result of a set of actions conforms to a valid chain of execution as per the JMM, then the result is allowed regardless of whether the author intended the code to produce that result or not.

"The memory model describes possible behaviors of a program. An implementation is free to produce any code it likes, as long as all resulting executions of a program produce a result that can be predicted by the memory model.

This provides a great deal of freedom for the implementor to perform a myriad of code transformations, including the reordering of actions and removal of unnecessary synchronization" (JLS 17.4).

The JIT will reorder whatever it sees fit unless we do not allow it using the JMM (in a multithreaded environment).

The details of what the JIT can or will do is nondeterministic. Looking at millions of samples of runs will not produce a meaningful pattern because reorderings are subjective, they depend on very specific details such as CPU arch, timings, heuristics, graph size, JVM vendor, bytecode size, etc... We only know that the JIT will assume that the code runs in a single threaded environment when it does not need to conform to the JMM. In the end, the JIT matters very little to your multithreaded code. If you want to dig deeper, see this SO answer and do a little research on such topics as IR Graphs, the JDK HotSpot source, and compiler articles such as this one. But again, remember that the JIT has very little to do with your multithreaded code transforms.

---

In practice, the "object that is not fully created yet" is not a side effect of the JIT but rather the memory model (JMM). In summary, the JMM is a specification that puts forth guarantees of what can and cannot be results of a certain set of actions, where actions are operations that involve a shared state. The JMM is more easily understood by higher level concepts such as atomicity, memory visibility, and ordering, those three of which are components of a thread-safe program.

To demonstrate this, it is highly unlikely for your first sample of code (the DCL pattern) to be modified by the JIT that would produce "an object that is not fully created yet." In fact, I believe that it is not possible to do this because it would not follow the order or execution of a single-threaded program.

So what exactly is the problem here?

The problem is that if the actions aren't ordered by a synchronization order, a happens-before order, etc... (described again by JLS 17.4-17.5) then threads are not guaranteed to see the side effects of performing such actions. Threads might not flush their caches to update the field, threads might observe the write out of order. Specific to this example, threads are allowed to see the object in an inconsistent state because it is not properly published. I'm sure that you have heard of safe publishing before if you have ever worked even the tiniest bit with multithreading.

You might ask, well if single-threaded execution cannot be modified by the JIT, why can the multithreaded version be?

Put simply, it's because the thread is allowed to think ("perceive" as usually written in textbooks) that the initialization is out of order due to the lack of proper synchronization.

"If Helper is an immutable object, such that all of the fields of Helper are final, then double-checked locking will work without having to use volatile fields. The idea is that a reference to an immutable object (such as a String or an Integer) should behave in much the same way as an int or float; reading and writing references to immutable objects are atomic" (The "Double-Checked Locking is Broken" Declaration).

Making the object immutable ensures that the state is fully initialized when the constructor exits.

Remember that object construction is always unsynchronized. An object that is being initialized is ONLY visible and safe with respect to the thread that constructed it. In order for other threads to see the initialization, you must publish it safely. Here are those ways:

"There are a few trivial ways to achieve safe publication:

1. Exchange the reference through a properly locked field (JLS 17.4.5)
2. Use static initializer to do the initializing stores (JLS 12.4)
3. Exchange the reference via a volatile field (JLS 17.4.5), or as the consequence of this rule, via the AtomicX classes
4. Initialize the value into a final field (JLS 17.5)."

(Safe Publication and Safe Initialization in Java)

Safe publication ensures that other threads will be able to see the fully initialized objects when after it finishes.

Revisiting our idea that threads are only guaranteed to see side effects if they are in order, the reason that you need volatile is so that your write to the helper in thread 1 is ordered with respect to the read in thread 2. Thread 2 is not allowed to perceive the initialization after the read because it occurs before the write to helper. It piggy backs on the volatile write such that the read must happen after the initialization AND THEN the write to the volatile field (transitive property).

To conclude, an initialization will only occur after the object is created only because another thread THINKS that is the order. An initialization will never occur after construction due to a JIT optimisation. You can fix this by ensuring proper publication through a volatile field or by making your helper immutable.

---

Now that I've described the general concepts behind how publication works in the JMM, hopefully understanding how your second example won't work will be easy.

I'd imagine that the answer is, it wouldn't put an object into the Map until it is fully constructed (because that sounds awful). So how can the JIT reorder?

To the constructing thread, it will put it into the map after initialization.

To the reader thread, it can see whatever the hell it wants. (improperly constructed object in HashMap? That is definitely within the realm of possibility).

What you described with your 4 steps is completely legal. There is no order between assigning value or adding it to the map, thus thread 2 can perceive the initialization out of order since MyObject was published unsafely.

You can actually fix this problem by just converting to ConcurrentHashMap and getObject() will be completely thread safe as once you put the object in the map, the initialization will occur before the put and both will need to occur before the get as a result of ConcurrentHashMap being thread safe. However, once you modify the object, it will become a management nightmare because you need to ensure that updating the state is visible and atomic - what if a thread retrieves an object and another thread updates the object before the first thread could finish modifying and putting it back in the map?

T1 -> get() MyObject=30 ------> +1 --------------> put(MyObject=31)
T2 -------> get() MyObject=30 -------> +1 -------> put(MyObject=31)

Alternatively you could also make MyObject immutable, but you still need to map the map ConcurrentHashMap in order for other threads to see the put - thread caching behavior might cache an old copy and not flush and keep reusing the old version. ConcurrentHashMap ensures that its writes are visible to readers and ensures thread-safety. Recalling our 3 prerequisites for thread-safety, we get visibility from using a thread-safe data structure, atomicity by using an immutable object, and finally ordering by piggybacking on ConcurrentHashMap's thread safety.

To wrap up this entire answer, I will say that multithreading is a very difficult profession to master, one that I myself most definitely have not. By understanding concepts of what makes a program thread-safe and thinking about what the JMM allows and guarantees, you can ensure that your code will do what you want it to do. Bugs in multithreaded code occur often as a result of the JMM allowing a counterintuitive result that is within its parameters, not the JIT doing performance optimisations. Hopefully you will have learned something a little bit more about multithreading if you read everything. Thread safety should be achieved by building a repertoire of thread-safe paradigms rather than using little inconveniences of the spec (Lea or Bloch, not even sure who said this).

来源：https://stackoverflow.com/questions/42079959/explain-how-jit-reordering-works