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Static variable initialization during declaration in multi-threading

I want to add the following code to my class:
static private final ILogic_P logicInstanceI =
(ILogic_P)Factory.CreateAnon("some.path.ILogic_P" + (SomeClass.isIMDB() ? "1" : "2"));
public static ILogic_P getLogicInstanceI(){
return logicInstanceI;
}
I can't figure it out if the initialization of the static variable is thread safety or not.
Is there a chance that two threads will try to initialize this attribute simultaneously?

The answer is given by the Java Language Specification §12.4.2:
Because the Java programming language is multithreaded, initialization of a class or interface requires careful synchronization, since some other thread may be trying to initialize the same class or interface at the same time. There is also the possibility that initialization of a class or interface may be requested recursively as part of the initialization of that class or interface; for example, a variable initializer in class A might invoke a method of an unrelated class B, which might in turn invoke a method of class A. The implementation of the Java Virtual Machine is responsible for taking care of synchronization and recursive initialization by using the following procedure. […]
Note the last sentence starting with “The implementation of the Java Virtual Machine is responsible for taking care …”
So you are not responsible for synchronization in the case of class initialization, and assigning initial values to static variables is part of the class initialization, as specified in §8.3.2:
8.3.2. Field Initialization
If a declarator in a field declaration has a variable initializer, then the declarator has the semantics of an assignment (§15.26) to the declared variable.
If the declarator is for a class variable (that is, a static field), then the following rules apply to its initializer:
…
At run time, the initializer is evaluated and the assignment performed exactly once, when the class is initialized (§12.4.2).

Why is a subclass' static initializer not invoked when a static method declared in its superclass is invoked on the subclass?

Given the following classes:
public abstract class Super {
protected static Object staticVar;
protected static void staticMethod() {
System.out.println( staticVar );
}
}
public class Sub extends Super {
static {
staticVar = new Object();
}
// Declaring a method with the same signature here,
// thus hiding Super.staticMethod(), avoids staticVar being null
/*
public static void staticMethod() {
Super.staticMethod();
}
*/
}
public class UserClass {
public static void main( String[] args ) {
new UserClass().method();
}
void method() {
Sub.staticMethod(); // prints "null"
}
}
I'm not targeting at answers like "Because it's specified like this in the JLS.". I know it is, since JLS, 12.4.1 When Initialization Occurs reads just:
A class or interface type T will be initialized immediately before the first occurrence of any one of the following:
...
T is a class and a static method declared by T is invoked.
...
I'm interested in whether there is a good reason why there is not a sentence like:
T is a subclass of S and a static method declared by S is invoked on T.

Be careful in your title, static fields and methods are NOT inherited. This means that when you comment staticMethod() in Sub , Sub.staticMethod() actually calls Super.staticMethod() then Sub static initializer is not executed.
However, the question is more interesting than I thought at the first sight : in my point of view, this shouldn't compile without a warning, just like when one calls a static method on an instance of the class.
EDIT: As #GeroldBroser pointed it, the first statement of this answer is wrong. Static methods are inherited as well but never overriden, simply hidden. I'm leaving the answer as is for history.

I think it has to do with this part of the jvm spec:
Each frame (§2.6) contains a reference to the run-time constant pool (§2.5.5) for the type of the current method to support dynamic linking of the method code. The class file code for a method refers to methods to be invoked and variables to be accessed via symbolic references. Dynamic linking translates these symbolic method references into concrete method references, loading classes as necessary to resolve as-yet-undefined symbols, and translates variable accesses into appropriate offsets in storage structures associated with the run-time location of these variables.
This late binding of the methods and variables makes changes in other classes that a method uses less likely to break this code.
In chapter 5 in the jvm spec they also mention:
A class or interface C may be initialized, among other things, as a result of:
The execution of any one of the Java Virtual Machine instructions new, getstatic, putstatic, or invokestatic that references C (§new, §getstatic, §putstatic, §invokestatic). These instructions reference a class or interface directly or indirectly through either a field reference or a method reference.
...
Upon execution of a getstatic, putstatic, or invokestatic instruction, the class or interface that declared the resolved field or method is initialized if it has not been initialized already.
It seems to me the first bit of documentation states that any symbolic reference is simply resolved and invoked without regard as to where it came from. This documentation about method resolution has the following to say about that:
[M]ethod resolution attempts to locate the referenced method in C and its superclasses:
If C declares exactly one method with the name specified by the method reference, and the declaration is a signature polymorphic method (§2.9), then method lookup succeeds. All the class names mentioned in the descriptor are resolved (§5.4.3.1).
The resolved method is the signature polymorphic method declaration. It is not necessary for C to declare a method with the descriptor specified by the method reference.
Otherwise, if C declares a method with the name and descriptor specified by the method reference, method lookup succeeds.
Otherwise, if C has a superclass, step 2 of method resolution is recursively invoked on the direct superclass of C.
So the fact that it's called from a subclass seems to simply be ignored. Why do it this way? In the documentation you provided they say:
The intent is that a class or interface type has a set of initializers that put it in a consistent state, and that this state is the first state that is observed by other classes.
In your example, you alter the state of Super when Sub is statically initialized. If initialization happened when you called Sub.staticMethod you would get different behavior for what the jvm considers the same method. This might be the inconsistency they were talking about avoiding.
Also, here's some of the decompiled class file code that executes staticMethod, showing use of invokestatic:
Constant pool:
...
#2 = Methodref #18.#19 // Sub.staticMethod:()V
...
Code:
stack=0, locals=1, args_size=1
0: invokestatic #2 // Method Sub.staticMethod:()V
3: return

The JLS is specifically allowing the JVM to avoid loading the Sub class, it's in the section quoted in the question:
A reference to a static field (§8.3.1.1) causes initialization of only the class or interface that actually declares it, even though it might be referred to through the name of a subclass, a subinterface, or a class that implements an interface.
The reason is to avoid having the JVM load classes unnecessarily. Initializing static variables is not an issue because they are not getting referenced anyway.

The reason is quite simple: for JVM not to do extra work prematurely (Java is lazy in its nature).
Whether you write Super.staticMethod() or Sub.staticMethod(), the same implementation is called. And this parent's implementation typically does not depend on subclasses. Static methods of Super are not supposed to access members of Sub, so what's the point in initializing Sub then?
Your example seems to be artificial and not well-designed.
Making subclass rewrite static fields of superclass does not sound like a good idea. In this case an outcome of Super's methods will depend on which class is touched first. This also makes hard to have multiple children of Super with their own behavior. To cut it short, static members are not for polymorphism - that's what OOP principles say.

According to this article, when you call static method or use static filed of a class, only that class will be initialized.
Here is the example screen shot.

for some reason jvm think that static block is no good, and its not executed
I believe, it is because you are not using any methods for subclass, so jvm sees no reason to "init" the class itself, the method call is statically bound to parent at compile time - there is late binding for static methods
http://ideone.com/pUyVj4
static {
System.out.println("init");
staticVar = new Object();
}
Add some other method, and call it before the sub
Sub.someOtherMethod();
new UsersClass().method();
or do explicit Class.forName("Sub");
Class.forName("Sub");
new UsersClass().method();

When static block is executed Static Initializers
A static initializer declared in a class is executed when the class is initialized
when you call Sub.staticMethod(); that means class in not initialized.Your are just refernce
When a class is initialized
When a Class is initialized in Java After class loading, initialization of class takes place which means initializing all static members of class. A Class is initialized in Java when :
1) an Instance of class is created using either new() keyword or using reflection using class.forName(), which may throw ClassNotFoundException in Java.
2) an static method of Class is invoked.
3) an static field of Class is assigned.
4) an static field of class is used which is not a constant variable.
5) if Class is a top level class and an assert statement lexically nested within class is executed.
When a class is loaded and initialized in JVM - Java
that's why your getting null(default value of instance variable).
public class Sub extends Super {
static {
staticVar = new Object();
}
public static void staticMethod() {
Super.staticMethod();
}
}
in this case class is initialize and you get hashcode of new object().If you do not override staticMethod() means your referring super class method
and Sub class is not initialized.

Order of initialization of data fields in java

In the code below, what is the order of initialization of data fields? What is the general rule followed by java for data member and member functions?
public class TestClass
{
int j=10;
static int h=5;
public static void main(String[] args)
{
TestClass obj= new TestClass();
}
}

In general:
1) static field members (static initializers in general)
2) non-static field members
3) constructor
However you can test it with a snippet of code like this:
public class TestClass {
int i = 10;
static int j = 20;
public TestClass() {
// TODO Auto-generated constructor stub
System.out.println(i);
i = 20;
System.out.println(i);
}
public static void main(String[] args) {
new TestClass();
}
}
Quoting from the great "Thinking In Java":
Within a class, the order of initialization is determined by the order that the variables are
defined within the class. The variable definitions may be scattered throughout and in between method definitions, but the variables are initialized before any methods can be
called—even the constructor.
...................................
There’s only a single piece of storage for a static, regardless of how many objects are created.
You can’t apply the statickeyword to local variables, so it only applies to fields. If a field is a
staticprimitive and you don’t initialize it, it gets the standard initial value for its type. If it’s
a reference to an object, the default initialization value is null.
To summarize the process of creatingan object, consider a class called Dog:
Even though it doesn't explicitly use the static keyword, the constructor is actually a
static method. So the first time an object of type Dog is created, or the first time a
static method or static field of class Dog is accessed, the Java interpreter must
locate Dog.class, which it does by searching through the class path.
As Dog.class is loaded (creating a Class object, which you’ll learn about later), all of
its static initializers are run. Thus, static initialization takes place only once, as the
Class object is loaded for the first time.
When you create a new Dog( ), the construction process for a Dog object first
allocates enough storage for a Dog object on the heap.
This storage is wiped to zero, automatically setting all the primitives in that Dog
object to their default values (zero for numbers and the equivalent for boolean and
char) and the references to null.
Any initializations that occur at the point of field definition are executed.
Constructors are executed.This might actually involve a fair amount of activity, especially when inheritance is involved.

Here is the order.
Static fields initialized and static block executed at class loading time, they will be initialized in the same order they declared.
Set fields to default initial values (0, false, null)
Call the constructor for the object (but don't execute the body of the constructor yet)
Invoke the constructor of the superclass
Initialize fields using initializers and initialization blocks
Execute the body of the constructor

I think you're just missing section 12.4.2 of JLS, which includes:
Next, execute either the class variable initializers and static initializers of the class, or the field initializers of the interface, in textual order, as though they were a single block.
The "in textual order" part is the important bit.
If you change m from being static variable to an instance variable, then the field won't be initialized by class initialization - it'll only be initialized by instance initialization (i.e. when an instance is constructed). At the moment, that'll cause a stack overflow - creating one instance requires creating another instance, which requires creating another instance, etc.
So Finally Here is Order which you are searching for:
Initialization blocks run in the order they appear in the program.
Static initialization blocks run when the class is first loaded into JVM
Instance initialization blocks run whenever a new instance of class is created.

What does "When a Class is loaded" actually mean?

It is said that static blocks in java run only once when that class is loaded. But what does it actually mean? At which point is a class loaded by JVM (Java Virtual Machine)?
Is it when the main method in that class is called? And is it that all the super-classes of the same class are also loaded when the main method starts execution?
Consider that A extends B and B extends C. All have static blocks. If A has the main method, then what will be the sequence of execution of static blocks?

This is described in the Execution section of the JLS. Namely:
Initialization of a class consists of executing its static initializers and the initializers for static fields declared in the class. Initialization of an interface consists of executing the initializers for fields declared in the interface.
Before a class is initialized, its direct superclass must be initialized, but interfaces implemented by the class need not be initialized. Similarly, the superinterfaces of an interface need not be initialized before the interface is initialized.
So in your example, the static block of the "topmost" class (C) runs first, then that of B, then the most-derived one.
See that documentation for a detailed description of all the steps that go into loading a class.
(Classes get loaded when they are first actively used.)

I think the following example will solve all of your problems:
Before a class is initialized, its superclasses are initialized, if they have not previously been initialized.
Thus, the test program:
class Super {
static { System.out.print("Super "); }
}
class One {
static { System.out.print("One "); }
}
class Two extends Super {
static { System.out.print("Two "); }
}
class Test {
public static void main(String[] args) {
One o = null;
Two t = new Two();
System.out.println((Object)o == (Object)t);
}
}
prints:
Super Two false
The class One is never initialized, because it not used actively and therefore is never linked to. The class Two is initialized only after its superclass Super has been initialized.
For more details visit this link
Edit details: Removed confusing lines.

From the Java Language Specification:
Initialization of a class consists of executing its static initializers and the initializers for static fields (class variables) declared in the class. Initialization of an interface consists of executing the initializers for fields (constants) declared there.
Before a class is initialized, its superclass must be initialized, but interfaces implemented by the class are not initialized. Similarly, the superinterfaces of an interface are not initialized before the interface is initialized.
The process is described in more detail in the Java Virtual Machine Specification.

Is static code always executed when we use a class for the first time?

Our application is using initialization code that depends on the order static code is executed and I'm wondering if this order will be consistent across all JVMs.
Here is a sample of what I mean:
public class Main {
static String staticVar = "init_value";
public static void main(String[] args) {
System.out.println(A.staticVar);
staticVar = "mainValue";
System.out.println(A.staticVar);
}
}
public class A {
static String staticVar = Main.staticVar;
}
will give:
init_value
init_value
and
public class Main {
static String staticVar = "init_value";
public static void main(String[] args) {
// System.out.println(A.staticVar);
staticVar = "mainValue";
System.out.println(A.staticVar);
}
}
public class A {
static String staticVar = Main.staticVar;
}
will give (on my environment):
mainValue
To summarize, across all JVMs, is static code always executed when we use a class for the first time?

EDIT: Despite all the reassurances below, if you're thinking of relying on this sort of thing, I would try hard to refactor your code so that it doesn't crop up. While it is guaranteed to work, it's also likely to make your code very brittle. The fact that static initializers get called "invisibly" makes them relatively hard to reason about and debug.
Yes, this is guaranteed by the language specification. From section 8.7 of the spec:
Any static initializers declared in a class are executed when the class is initialized and, together with any field initializers (§8.3.2) for class variables, may be used to initialize the class variables of the class (§12.4).
StaticInitializer: static Block
It is a compile-time error for a static initializer to be able to complete abruptly (§14.1, §15.6) with a checked exception (§11.2). It is a compile-time error if a static initializer cannot complete normally (§14.21).
The static initializers and class variable initializers are executed in textual order.
And from section 12.4:
Initialization of a class consists of
executing its static initializers and
the initializers for static fields
declared in the class. Initialization
of an interface consists of executing
the initializers for fields declared
in the interface.
Before a class is
initialized, its direct superclass
must be initialized, but interfaces
implemented by the class need not be
initialized. Similarly, the
superinterfaces of an interface need
not be initialized before the
interface is initialized.
A class or interface type T will be
initialized immediately before the
first occurrence of any one of the
following:
T is a class and an instance of T is
created.
T is a class and a static
method declared by T is invoked.
A
static field declared by T is
assigned.
A static field declared by T
is used and the field is not a
constant variable (§4.12.4).
T is a
top-level class, and an assert
statement (§14.10) lexically nested

Static initialisers (e.g. your staticVar declarations) are always executed when you use a class for the first time.
Static methods are only executed when they are called. In your case, this is happening because the static void main(String[] args) is the entry point to your application. But if you defined a different static method then it would not be called.
It is also possible to create a static initialiser block that will also be called automatically before the class is first used, e.g.
static {
// some initialisation code here
}

That should be all the same on all platforms. See JVM Specification : http://java.sun.com/docs/books/jvms/second_edition/html/Concepts.doc.html#19075

A quote from Java specification:
Initialization of a class consists of
executing its static initializers and
the initializers for static fields
declared in the class. Initialization
of an interface consists of executing
the initializers for fields declared
in the interface.

Yes, according to the Java Language Specification static code is always executed in the same order on all (compliant) JVM's.
8.7 Static Initializers
The static initializers and class variable initializers are executed in textual order.

Dittos to other posters, but let me add that a static initializer that depends on the state of other classes like in your example seems to me to make for very fragile and difficult-to-maintain code. I'll use static intializers to populate internal class data -- to fill up an array with default values or that sort of thing. But I don't think I've ever had one peek at the data in another class. Technically legal, and maybe there are odd cases where it's a good idea, but yuck.

Yes, I believe that would be the point by definition.

Static Block will always run first not just the first time... Before executing any code block, JVM execute the static code block first, and then only it run the code block as it has been designed...