问题
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?
回答1:
Each thread can be on a different core with its own cache. This means that one thread can write to a value storing in a register, or its local cache, and this value is not visible to another thread for some time. (milli-seconds is not uncommon)
A more extreme example is that the reading thread's code is optimised with the assumption that since it never changes the value, it doesn't need to read it from memory. In this case the optimised code never sees the change performed by another thread.
In both cases, the use of volatile ensures that reads and write occur in a consistent order and both threads see the same value. This is sometimes described as always reading from main memory, though it doesn't have to be the case because the caches can talk to each other directly. (So the performance hit is much smaller than you might expect)
回答2:
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
回答3:
In substance that is correct. The main thing to take out of this is: unless you use some form of synchronization, there is no guarantee that a read that comes after a write in your program order sees the effect of that write, as the statements might have been reodered.
does it exist this situation (reads see writes that occur later) in real world? If it does, could you give me a real example?
From a wall clock's perspective, obviously, a read can't see the effect of a write that has not happened yet.
From a program order's perspective, because statements can be reordered if there isn't a proper synchronization (happens before relationship), a read that comes before a write in your program, could see the effect of that write during execution because it has been executed after the write by the JVM.
回答4:
It means if there is no synchronization mechanism in place, you may see the impact counter intuitively.
volatile int A = 0;
volatile int B = 0;
1: B = 1;
2: r2 = A; // r2 guarantees the value 1
3: A = 2;
4: r1 = B; // r1 guarantees the value 2
This is because volatile variables guarantees happens before relation. If A, and B are not volatile system can reorder the evaluation of variables and it may become counter intuitive.
来源：https://stackoverflow.com/questions/11970428/how-to-understand-happens-before-consistent