volatile vs not volatile
Let's consider the following piece of code in Java int x = 0; int who = 1 Thread #1: (1) x++; (2) who = 2; Thread #2 while(who == 1); x++; print x; ( the value should be equal to 2 but, perhaps, it is not* ) (I don't know Java memory models- let assume that it is strong memory model- I mean: (1) and (2) will be doesn't swapped) Java memory model guarantees that access/store to the 32 bit variables is atomic so our program is safe. But, nevertheless we should use a attribute volatile because *. The value of x may be equal to 1 because x can be kept in register when Thread#2 read it. To resolve it we should make the x variable volatile. It is clear. But, what about that situation: int x = 0; mutex m; ( just any mutex) Thread #1: mutex.lock() x++; mutex.unlock() Thread #2 mutex.lock() x++; print x; // the value is always 2, why**? mutex.unlock() The value of x is always 2 though we don't make it volatile. Do I correctly understand that locking/unlocking mutex is connected with inserting memory barriers?
I'll try to tackle this. The Java memory model is kind of involved and hard to contain in a single StackOverflow post. Please refer to Brian Goetz's Java Concurrency in Practice for the full story. The value of x is always 2 though we don't make it volatile. Do I correctly understand that locking/unlocking mutex is connected with inserting memory barriers? First if you want to understand the Java memory model, it's always Chapter 17 of the spec you want to read through. That spec says: An unlock on a monitor happens-before every subsequent lock on that monitor. So yes, there's a memory visibility event at the unlock of your monitor. (I assume by "mutex" you mean monitor. Most of the locks and other classes in the java.utils.concurrent package also have happens-before semantics, check the documentation.) Happens-before is what Java means when it guarantees not just that the events are ordered, but also that memory visibility is guaranteed. We say that a read r of a variable v is allowed to observe a write w to v if, in the happens-before partial order of the execution trace: r is not ordered before w (i.e., it is not the case that hb(r, w)), and there is no intervening write w' to v (i.e. no write w' to v such that hb(w, w') and hb(w', r)). Informally, a read r is allowed to see the result of a write w if there is no happens-before ordering to prevent that read. This is all from 17.4.5. It's a little confusing to read through, but the info is all there if you do read through it.
Let's go over some things. The following statement is true: Java memory model guarantees that access/store to the 32 bit variables is atomic. However, it does not follow that the first pseudoprogram you listed is safe. Simply because two statements are ordered syntactically does not mean that the visibility of their updates are also so ordered as viewed by other threads. Thread #2 may see the update caused by who=2 before the increment in x is visible. Making x volatile would still not make the program correct. Instead, making the variable 'who' voliatile would make the program correct. That is because volatile interacts with the java memory model in specific ways. I feel like there is some notion of 'writing back to main memory' at the core of a common sense understanding of volatile which is incorrect. Volatile does not write back the value to main memory in Java. What reading from and writing to a volatile variable does is create what's called a happens-before relationship. When thread #1 writes to a volatile variable you're creating a relationship that ensures that any other threads #2 viewing that volatile variable will also be able to 'view' all the actions thread #1 has taken before that. In your example that means making 'who' volatile. By writing the value 2 to 'who' you are creating a happens-before relationship so that when thread #2 views who=2 it will similarly see an updated version of x. In your second example (assuming you meant to have the 'who' variable too) the mutex unlocking creates a happens-before relationship as I specified above. Since that means other threads viewing the unlock of the mutex (ie. they are able to lock it themselves) they will see the updated version of x.
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