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How to understand happens-before consistent

Asked 2012-08-15T13:45:32.750
16

In chapter 17 of JLS, it introduce a concept: happens-before consistent.

A set of actions A is happens-before consistent if for all reads r in A, where W(r) is the write action seen by r, it is not the case that either hb(r, W(r)) or that there exists a write w in A such that w.v = r.v and hb(W(r), w) and hb(w, r)"

In my understanding, it equals to following words: ..., it is the case that neither ... nor ...

So my first two questions are:

  • is my understanding right?
  • what does "w.v = r.v" mean?

It also gives an Example: 17.4.5-1

Thread 1 Thread 2

B = 1; A = 2; 

r2 = A; r1 = B; 

In first execution order:

1: B = 1;

3: A = 2;

2: r2 = A;  // sees initial write of 0

4: r1 = B;  // sees initial write of 0

The order itself has already told us that two threads are executed alternately, so my third question is: what does left number mean?

In my understanding, the reason of both r2 and r1 can see initial write of 0 is both A and B are not volatile field. So my fourth quesiton is: whether my understanding is right?

In second execution order:

1: r2 = A;  // sees write of A = 2

3: r1 = B;  // sees write of B = 1

2: B = 1;

4: A = 2;

According to definition of happens-before consistency, it is not difficult to understand this execution order is happens-before consistent(if my first understanding is correct). So my fifth and sixth questions are: does it exist this situation (reads see writes that occur later) in real world? If it does, could you give me a real example?

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The Java Memory Model defines a partial ordering of all your actions of your program which is called happens-before.
To guarantee that a thread Y is able to see the side-effects of action X (irrelevant if X occurred in different thread or not) a happens-before relationship is defined between X and Y.
If such a relationship is not present the JVM may re-order the operations of the program.
Now, if a variable is shared and accessed by many threads, and written by (at least) one thread if the reads and writes are not ordered by the happens before relationship, then you have a data race.
In a correct program there are no data races.
Example is 2 threads A and B synchronized on lock X.
Thread A acquires lock (now Thread B is blocked) and does the write operations and then releases lock X. Now Thread B acquires lock X and since all the actions of Thread A were done before releasing the lock X, they are ordered before the actions of Thread B which acquired the lock X after thread A (and also visible to Thread B).
Note that this occurs on actions synchronized on the same lock. There is no happens before relationship among threads synchronized on different locks

answered 2012-08-15T14:05:18.343

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