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Why an instance variable can be assigned and not be referenced before its declaration in Java [duplicate]

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Java static class initialization
in what order are static blocks and static variables in a class executed?
When I run this code the answer is 1, I thought it would be 2.
What is the order of initialization and the value of k in each step?
public class Test {
static {k = 2;}
static int k = 1;
public static void main(String[] args) {
System.out.println(k);
}
}
Edit 1: As a follow up to "k is set to default value" then why this next code doesn't compile? Theres an error "Cannot reference a field before it's defined".
public class Test {
static {System.out.println(k);}
static int k=1;
public static void main(String[] args) {
System.out.println(k);
}
}
Edit 2: For some unknow to me reason it^ works when instead of "k" its "Test.k".
Thanks for all the answers. this will sufice :D

They are executed in the order that you write them. If the code is:
public class Test {
static int k = 1;
static {k = 2;}
public static void main(String[] args) {
System.out.println(k);
}
}
then the output becomes 2.
The order of initialization is: ..the class variable initializers and static initializers of the class..., in textual order, as though they were a single block.
And the values (for your code) are: k = 0 (default), then it's set to 2, then it's set back to 1.
You can check that it's actually set to 2 by running the following code:
private static class Test {
static {
System.out.println(Test.k);
k = 2;
System.out.println(Test.k);
}
static int k = 1;
public static void main(String[] args) {
System.out.println(k);
}
}

Short answer
When the initialization of the class starts, k will have initial value of 0.
The static block (since it precedes the assignment in the declaration) is then executed, and k will be assigned 2.
Then the initializer in the declaration is executed, and k will be assigned 1.
Long explanation
Let us use this example, since your example is a bit simple:
class TestInitOrder {
static {
System.out.println(TestInitOrder.stat1);
System.out.println(TestInitOrder.stat2);
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.str2);
str = "something";
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.str2);
System.out.println(TestInitOrder.lazy);
System.out.println(TestInitOrder.second);
}
private static final int stat1 = 10;
static final String str2 = "sdfff";
static String str = "crap";
private static int stat2 = 19;
static final Second second = new Second();
static final int lazy;
static {
lazy = 20;
}
static {
System.out.println(TestInitOrder.str2);
System.out.println(TestInitOrder.stat2);
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.lazy);
System.out.println(TestInitOrder.second);
}
public static void main(String args[]) {
}
}
class Second {
public Second() {
System.out.println(TestInitOrder.second);
}
}
According to Java Language Specification, from section 4.12.5:
Every variable in a program must have a value before its value is used:
Each class variable, instance variable, or array component is initialized with a default value when it is created
(The following lines from the specification specify the default value for all the types, basically some form of 0, such as 0, 0.0d, null, false, etc.)
So before the class is initialized (due to one of these reasons), the variables will hold an initial value.
According to the detailed initialization procedure (only the interesting steps are quoted here, and emphasis mine):
6.
[...] Then, initialize the final class variables and fields of interfaces whose values are compile-time constant expressions (§8.3.2.1, §9.3.1, §13.4.9, §15.28).
[...]
9.
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.
Let us look at step 6, with the 4 final class variables: stat1, str2, second, lazy.
Since 10 is constant expression, and so is "sdfff", and due to the order of execution, it is not possible to observe the initial value for str2 and stat1. In order to make an observation, the earliest you can do is in step 9.
The case of second demonstrate that when the right hand side is not a compile-time constant expression, so its initial value is visible.
The case of lazy is different, since the assignment is done in static block, and hence happen in step 9 - so it is possible to observe its initial value. (Well, the compiler checks carefully that lazy is assigned exactly once).
After the initialization of final class variables with compile-time constant expression comes the execution of static blocks and the rest of the initializers.
As you can see from the example, the static blocks and initialization happens according to textual order - demonstrated with the use of str variable - it is first printed out as null, then something, then crap.

public static void main(String[] args) throws Exception

What is the meaning of this?
public static void main(String[] args) throws Exception?
I don't know it. If someone know please help me.
I really want to know about "throws Exception".

public : it is a access specifier that means it can be accessed by
any other class in the program.
static : it is access modifier that means when the java program is
load then it will create the space in memory automatically.
void(return type) : it does not return any value.
main() : it is a method or a function name.(First method to execute by JVM)
string args[] : its a command line argument it is a collection of
variables in the string format.
throws Exception : Use exceptions to notify about things that should
not be ignored.

It's three completely different things:
public means that the method is visible and can be called from other objects of other types. Other alternatives are private, protected, package and package-private. See here for more details.
static means that the method is associated with the class, not a specific instance (object) of that class. This means that you can call a static method without creating an object of the class.
void means that the method has no return value. If the method returned an int you would write int instead of void.

It is simply the usual entry point method public static void main(String[]), except that it explicitly specifies an exception may be thrown. This is required by the compiler if any part of your code explicitly throws an exception without a try-catch block (excluding of course runtime-exceptions). For example, this will not compile:
public static void main(String[] args){
throw new Exception();
}

Those three keywords are pretty much different from one another. Public=This type can be called by any place from the program. not protected from other classes. Static=This type of methods does not have to be instantiated in order to invoke them. Void=This type of methods does not have a return value

The reason this main method is throwing an Exception outside of main is because the rest of the actual program's implementation doesn't actually care at all about what that Exception is about.
For example, if all I was doing in my program was printing something out onto the screen repeatedly every 5 seconds, I wouldn't really care too much about the InteruptedException being thrown due to the Thread.sleep() method. In such cases, we would then just throw it outside of main to just discard of its irrelevance instead of having to write some kind of code to handle it.
There's honestly nothing special about it at all. It's just a way to not have to fool with something meaningless.
public class IrrelevantExceptionExample {
/**
* Main method that repeatedly prints to the console every five seconds from a single Thread.
* #param args - Wise words from above.
* #throws InterruptedException - Uhm… There's only one Thread and it's sleeping! (IrrelevantException)
*/
public static void main(String[] args) throws InterruptedException {
System.out.println("Hey, at start up.");
for (int i = 5; i < 25; i += 5) {
Thread.sleep(5000);
System.out.printf("Hey, again, except %d seconds later!%n", i);
}
}
}

Public: it is an access specifier which can be access throughout the program.
Static: it is a keyword which can be used for single copy of that variable.
Void: it is an empty return type.

Why attempt to print uninitialized variable does not always result in an error message

Some may find it similar to the SO question Will Java Final variables have default values? but that answer doesn't completely solve this, as that question doesn't directly print the value of x within instance initializer block.
The problem arises when I try to print x directly inside the instance initializer block, while having assigned a value to x before the end of the block :
Case 1
class HelloWorld {
final int x;
{
System.out.println(x);
x = 7;
System.out.println(x);
}
HelloWorld() {
System.out.println("hi");
}
public static void main(String[] args) {
HelloWorld t = new HelloWorld();
}
}
This gives a compile time error stating that variable x might not have been initialized.
$ javac HelloWorld.java
HelloWorld.java:6: error: variable x might not have been initialized
System.out.println(x);
^
1 error
Case 2
Instead of directly printing, I am calling a function to print:
class HelloWorld {
final int x;
{
printX();
x = 7;
printX();
}
HelloWorld() {
System.out.println("hi");
}
void printX() {
System.out.println(x);
}
public static void main(String[] args) {
HelloWorld t = new HelloWorld();
}
}
This compiles correctly and gives output
0
7
hi
What is the conceptual difference between the two cases?

In the JLS, §8.3.3. Forward References During Field Initialization, its stated that there's a compile-time error when:
Use of instance variables whose declarations appear textually after the use is sometimes restricted, even though these instance variables
are in scope. Specifically, it is a compile-time error if all of the
following are true:
The declaration of an instance variable in a class or interface C appears textually after a use of the instance variable;
The use is a simple name in either an instance variable initializer of C or an instance initializer of C;
The use is not on the left hand side of an assignment;
C is the innermost class or interface enclosing the use.
The following rules come with a few examples, of which the closest to yours is this one:
class Z {
static int peek() { return j; }
static int i = peek();
static int j = 1;
}
class Test {
public static void main(String[] args) {
System.out.println(Z.i);
}
}
Accesses [to static or instance variables] by methods are not checked in this way, so the code above produces output 0, because the variable initializer for i uses the class method peek() to access the value of the variable j before j has been initialized by its variable initializer, at which point it still has its default value (§4.12.5 Initial Values of Variables).
So, to summarize, your second example compiles and executes fine, because the compiler does not check if the x variable was already initialized when you invoke printX() and when printX() actually takes place at Runtime, the x variable will be assigned with its default value (0).

Reading the JLS, the answer appears to be in section 16.2.2:
A blank final member field V is definitely assigned (and moreover is not definitely unassigned) before the block (§14.2) that is the body of any method in the scope of V and before the declaration of any class declared within the scope of V.
This means that when a method is called, the final field is assigned to its default value 0 before invoking it, so when you reference it inside the method, it compiles successfully and prints the value 0.
However, when you access the field outside of a method, it is considered unassigned, hence the compilation error. The following code will also not compile:
public class Main {
final int x;
{
method();
System.out.println(x);
x = 7;
}
void method() { }
public static void main(String[] args) { }
}
because:
V is [un]assigned before any other statement S of the block iff V is [un]assigned after the statement immediately preceding S in the block.
Since the final field x is unassigned before the method invocation, it is still unassigned after it.
This note in the JLS is also relevant:
Note that there are no rules that would allow us to conclude that V is definitely unassigned before the block that is the body of any constructor, method, instance initializer, or static initializer declared in C. We can informally conclude that V is not definitely unassigned before the block that is the body of any constructor, method, instance initializer, or static initializer declared in C, but there is no need for such a rule to be stated explicitly.

The difference is that in the first case you are calling System.out.println from initializer block so the block which is invoked before constructor. In the first line
System.out.println(x);
variable x is not yet initialized so that you get compilation error.
But in the second case you call instance method which doesn't know if variable has already been initialized so you don't have compilation error and you can see the default value for x

Ok, here is my 2 cents.
We all know that final variables can be initialized only While declaring or later on in constructors. Keeping that fact in mind, let see what happened here so far.
No errors Case:
So when you use inside a method, it have already a value.
1) If you initialize it, that value.
2) If not, the default value of data type.
Error case :
When you do that in an initialization block, which you are seeing errors.
If you look at the docs of initialization block
{
// whatever code is needed for initialization goes here
}
and
The Java compiler copies initializer blocks into every constructor. Therefore, this approach can be used to share a block of code between multiple constructors.
In compiler's eye, your code is literally equals to
class HelloWorld {
final int x;
HelloWorld() {
System.out.println(x); ------------ ERROR here obviously
x = 7;
System.out.println(x);
System.out.println("hi");
}
public static void main(String[] args) {
HelloWorld t = new HelloWorld();
}
}
You are using it before even initializing it.

Case 1 :
Gives you a compile-error,
Because at System.out.println(x);
you are trying to print x which was never initialized.
Case 2:
Works because you are not directly using any literal values, instead you are calling some method, which is correct.
General Rule is,
If you are trying to access any variable which is never initialized
then it will give a compilation error.

We deal here with initializer block. The Java compiler copies initializer blocks into every constructor.
The compiler error don't occure in second example, because printing x is in another Frame, please refer to spec.

In Java why would one initialize an int variable with 0 when it will be assigned 0 only by default when declared?

What purpose does it serve?
Just read an example in a book where the author has done so.
int numOfGuesses=0;

The automatic assignment to zero only applies to members, not to local variables. If it is a local variable and the = 0 is omitted then that variable has no value, not even zero. Attempting to use the value before it is assigned will result in a compile error. For example this code attempts to use an uninitialized local variable:
public class Program
{
public static void main(String[] args)
{
int numOfGuesses; // local variable
System.out.println(numOfGuesses);
}
}
and produces this compile error:
Program.java:6: variable numOfGuesses might not have been initialized
System.out.println(numOfGuesses);
Whereas this code using a member works and outputs zero:
public class Program
{
int numOfGuesses; // member variable
public void run()
{
System.out.println(numOfGuesses);
}
public static void main(String[] args)
{
new Program().run();
}
}
For members I tend to assign to zero explicilty if my code uses the fact that the initial zalue is zero, and omit the assignment if my code doesn't use the initial value (for example if it the value is assigned in the constructor or elsewhere). The result is the same either way, so it's just a style issue.

It's more explicit; some people like. Note that this applies only to fields -- local variables need to be initialized; there are no defaults.

The Java compilation and runtime differ.
When running the program, all classes are loaded with class loaders and they do the following:
This is done when class is used for the first time. Their execute order is defined by their order in the code.
run static blocks
static{
//do something here
}
initialize static variables
public static int number;
This will be initialized to zero 0;
The next group of initializations done is when creating an object.Their execute order is defined by their order in the code.
run non-static block
{
// do something here
}
initialize non-static(instance) variables
public int instance_number;
And this is when and why there is default initialization!
This is not true for methods because they don't have similar mechanism as classes.
So basically this means that you will have to initialize EXPLICITLY each method variable.enter code here

Explanation of how classloader loads static variables

Ok so this is a newbie question on java, but i can't seem to get my head around it.
I have the following code inside my class
private static final String [] LIST_CODE = gerarListCode();
private static final int [][] LIST_INTEGER = new int [][] {
{947,947}, {110,103},
{947,958}, {110,120},
{947,954}, {103,107},
{947,967}, {110,99,104}};
private static String [] gerarListCode()
{
String [] listCode = new String [LIST_INTEGER.length];
for (int i=0 ; i<LIST_INTEGER.length ; i++)
{
//do some stuff
}
return listaUnicode;
}
This code is giving me a initialization exception due to a nullpointerexception in the following line
String [] listCode = new String [LIST_INTEGER.length];
Seems the variable LIST_INTEGER is null at that time.
Can someone explain why? is the classloader process linear, in other words, does it invoke the method before fully loading all the other variables?

Yes, in short, it is linear.
"What the compiler actually does is to
internally produce a single class
initialization routine that combines
all the static variable initializers
and all of the static initializer
blocks of code, in the order that they
appear in the class declaration. This
single initialization procedure is run
automatically, one time only, when the
class is first loaded."
Taken from Java in a nutshell.
http://www.developer.com/java/other/article.php/2238491
You should define the variables and then initialize them in a static intitializer block in the correct order, or you could swap the order of the statements as follows:
private static final int [][] LIST_INTEGER = new int [][] { {947,947}, {110,103},
{947,958}, {110,120},
{947,954}, {103,107},
{947,967}, {110,99,104}};
private static final String [] LIST_CODE = gerarListCode();

The JVM will, indeed, initialize the static fields in the order it encounters them.
A class's static fields are initialized when the class is first encountered by the JVM. According to Java Puzzlers, puzzle 49 (which goes on to reference JLS 4.12.5), static fields are first set to their default values. Object variables are set to null, ints are set to 0, etc. After that, their initializers are executed in order of appearance.
So, in your example, LIST_CODE and LIST_INTEGER are first set to null. Then, LIST_CODE is initialized by calling gerarListCode(). LIST_INTEGER is still null when that method is executed. Only after that, LIST_INTEGER is initialized with the literal value you give in your example.