Java – Volatile variable in Java

concurrencyjavamultithreadingvolatile

So I am reading this book titled Java Concurrency in Practice and I am stuck on this one explanation which I cannot seem to comprehend without an example. This is the quote:

When thread A writes to a volatile
variable and subsequently thread B
reads that same variable, the values
of all variables that were visible to
A prior to writing to the volatile
variable become visible to B after
reading the volatile variable.

Can someone give me a counterexample of why "the values of ALL variables that were visible to A prior to writing to the volatile variable become visible to B AFTER reading the volatile variable"?

I am confused why all other non-volatile variables do not become visible to B before reading the volatile variable?

Best Answer

Declaring a volatile Java variable means:

The value of this variable will never be cached thread-locally: all reads and writes will go straight to "main memory".
Access to the variable acts as though it is enclosed in a synchronized block, synchronized on itself.

Just for your reference, When is volatile needed ?

When multiple threads using the same variable, each thread will have its own copy of the local cache for that variable. So, when it's updating the value, it is actually updated in the local cache not in the main variable memory. The other thread which is using the same variable doesn't know anything about the values changed by the another thread. To avoid this problem, if you declare a variable as volatile, then it will not be stored in the local cache. Whenever thread are updating the values, it is updated to the main memory. So, other threads can access the updated value.

From JLS §17.4.7 Well-Formed Executions

We only consider well-formed executions. An execution E = < P, A, po, so, W, V, sw, hb > is well formed if the following conditions are true:

Each read sees a write to the same variable in the execution. All reads and writes of volatile variables are volatile actions. For all reads r in A, we have W(r) in A and W(r).v = r.v. The variable r.v is volatile if and only if r is a volatile read, and the variable w.v is volatile if and only if w is a volatile write.

Happens-before order is a partial order. Happens-before order is given by the transitive closure of synchronizes-with edges and program order. It must be a valid partial order: reflexive, transitive and antisymmetric.

The execution obeys intra-thread consistency. For each thread t, the actions performed by t in A are the same as would be generated by that thread in program-order in isolation, with each write wwriting the value V(w), given that each read r sees the value V(W(r)). Values seen by each read are determined by the memory model. The program order given must reflect the program order in which the actions would be performed according to the intra-thread semantics of P.

The execution is happens-before consistent (§17.4.6).

The execution obeys synchronization-order consistency. For all volatile reads r in A, it is not the case that either so(r, W(r)) or that there exists a write win A such that w.v = r.v and so(W(r), w) and so(w, r).

Useful Link : What do we really know about non-blocking concurrency in Java?

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Java – What are the differences between a HashMap and a Hashtable in Java

There are several differences between HashMap and Hashtable in Java:

Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable does not allow null keys or values. HashMap allows one null key and any number of null values.
One of HashMap's subclasses is LinkedHashMap, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap for a LinkedHashMap. This wouldn't be as easy if you were using Hashtable.

Since synchronization is not an issue for you, I'd recommend HashMap. If synchronization becomes an issue, you may also look at ConcurrentHashMap.

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Best Answer

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