Why are we allowed to have a final main method in java? - java

Can anyone tell me the use of making main method as final in java.
while this is allowed in java
public static final void main(String[] args) {
}
I dont see any use of making it final. anyways it is static so we can not override it.

Adding final to a static method can actually make a difference. Consider the following code:
class A {
public static void main(String[] args) {
System.out.println("A");
}
}
class B extends A {
public static void main(String[] args) {
System.out.println("B");
}
}
class C extends B {
}
public class Test {
public static void main(String[] args) {
C.main(args); // Will invoke B.main
}
}
Adding final to A.main would prevent accidental hiding of A.main. In other words, adding final to A.main guarantees that B.main is not allowed, and that C.main therefore prints "A" as opposed to for instance "B".
Why are we allowed to have a final main method in java?
Beside the above corner case, adding final to a static method doesn't make much difference, so I don't see a big point in adding a rule for disallowing it.
More information available here: Behaviour of final static method

Related

Why can an instance only be seen by main methods

public static void main(String[] args)
{
GUI TestGUI = new GUI();
TestGUI.setVisible(true);
}
public void blahh()
{
TestGUI.setVisible(true);
}
Can not find symbol for TestGUI in blahh, but can be seen in the main method.
How can I access TestGUI from other methods?
This is a scope issue. You could solve this by passing your GUI object to the blahh() method. Currently your blahh method has no way of reaching that variable.
public void blahh(GUI testGui) {
...
}
You can then call this method like this:
blahh(testGui);
Here is some reading you can do on scope, hopefully it will be helpful
Alternatively, you could declare your testGui variable as a field, and it will be accessible from anywhere in the class (make sure to make it static if you must access it in a static method). However, this will offer you less privacy with that variable even though it might seem more convenient.
because you declared TestGUI inside main method as a local varaible to a method , declare it as a class property
static GUI TestGUI;
public static void main(String[] args)
{
Test = new GUI();
TestGUI.setVisible(true);
}
public void blahh()
{
TestGUI.setVisible(true);
}

how to initialize static final variable from other class in java

I want to initialize Final.value in Main method.
Is it possible to initialize static final constant in other class than
in its deceleration class?
public class Main {
public static void main(String[] args) {
//I want to initialize Final.value in Main method.
}
}
class Final {
//here is the static final variable which can be assigned vai a static block
//but how do I initialize it in the main method if I don't use the static block?
static final int value;
}
You cannot. Your perception might be that main happens before everything else, so it is safe to initialise things there, but that is incorrect.
Consider the following code.
class Scratch
{
static
{
System.out.println(Foo.i);
}
public static void main(String[] args)
{
Foo.i = 100;
}
}
class Foo
{
static int i;
}
It does not print 100. It prints 0 because there are other things which happen before main.
Making the field final does not change that fact.
You have two options for static initialization. In a static initializer block, like you showed, or in-line:
static final int value = 421
Java prevents you from doing what you want to do for a good reason: because it is likely to lead to bugs.

Access static variable from a different class

So I have two classes:
A Main class with the public static void main(String args[]) method
and a Voice class that accesses a static variable from that class.
Within the main class are methods that are used by itself, and are required to be static along with some of its variables.
So I have a static variable within the Main class (that's created/filled in the public static void main(String args[]) method. That's why this case is special) which the other class should be able to access.
Here is an example of what's happening:
public class Main(){
public static int variable;
/*
Unrelated methods go here.
*/
public static void main(String args[]){
Voice v = new Voice();//This is just here for the code to make sense.
variable = 5;
v.doSomething();
}
}
public class Voice(){
public void doSomething(){
System.out.println(Main.variable);
}
}
Upon calling the doSomething() method in Voice, it leads to a nullPointerException.
I could fix this by passing on the variable variable to the Voice class, but is there a more easy way to fix this in the long run, if for instance, I needed to use more than one static variable from the Main class?
Your code is having syntax error. You can use this
class Main{
public static int variable;
/*
Unrelated methods go here.
*/
public static void main(String args[]){
Voice v = new Voice();//This is just here for the code to make sense.
variable = 5;
v.doSomething();
}
}
class Voice{
public void doSomething(){
System.out.println(Main.variable);
}
}
Output will be 5
You should do as follows
public class Main{
public static int variable;
/*
Unrelated methods go here.
*/
public static void main(String args[]){
Voice v = new Voice();//This is just here for the code to make sense.
variable = 5;
v.doSomething();
}
}
class Voice{
public void doSomething(){
System.out.println(Main.variable);
}
}

Invoke java class static initialization via VM option

Is there any way to force static initialization of some class B before entering the main() method of class A, without changing class A, using only VM options?
I'm not aware of any way to do it without code. In code it's easy of course.
public class C {
static {
// Fetch and throw away an instance of B to make sure it is loaded
B.getInstance();
}
static void main(String[] args) {
// Do stuff, B is initialized
A.main(args);
}
}
In fact you could just do
public class C {
static void main(String[] args) {
B.getInstance();
A.main(args);
}
}
Your request doesn't make a lot of sense though. Ask a question about the problem you are trying to solve by doing this and you will hopefully get a much more useful answer.
you could create a class that initializes other classes and then calls the real main method, e.g:
public static void main(String[] args) throws Exception {
Class<?> mainClass = Class.forName(System.getProperty("mainClass"));
for (String className : System.getProperty("initClasses").split(";")) {
Class.forName(className);
}
Method main = mainClass.getMethod("main", String[].class);
main.invoke(null, new Object[] { args });
}
Then you would start the application with that class as the main class and specify the real main class and the classes to be initialized via properties.

public static void main () access non static variable

It is said that non-static variables cannot be used in a static method. But public static void main does. How is that?
No, it doesn't.
public class A {
int a = 2;
public static void main(String[] args) {
System.out.println(a); // won't compile!!
}
}
but
public class A {
static int a = 2;
public static void main(String[] args) {
System.out.println(a); // this works!
}
}
or if you instantiate A
public class A {
int a = 2;
public static void main(String[] args) {
A myA = new A();
System.out.println(myA.a); // this works too!
}
}
Also
public class A {
public static void main(String[] args) {
int a = 2;
System.out.println(a); // this works too!
}
}
will work, since a is a local variable here, and not an instance variable. A method local variable is always reachable during the execution of the method, regardless of if the method is static or not.
Yes, the main method may access non-static variables, but only indirectly through actual instances.
Example:
public class Main {
public static void main(String[] args) {
Example ex = new Example();
ex.variable = 5;
}
}
class Example {
public int variable;
}
What people mean when they say "non-static variables cannot be used in a static method" is that non-static members of the same class can't be directly accessed (as shown in Keppils answer for instance).
Related question:
Accessing non-static members through the main method in Java
Update:
When talking about non-static variables one implicitly means member variables. (Since local variables can't possible have a static modifier anyway.)
In the code
public class A {
public static void main(String[] args) {
int a = 2;
System.out.println(a); // this works!
}
}
you're declaring a local variable (which typically is not referred to as non-static even though it doesn't have a static modifier).
The main method does not have access to non-static members either.
final public class Demo
{
private String instanceVariable;
private static String staticVariable;
public String instanceMethod()
{
return "instance";
}
public static String staticMethod()
{
return "static";
}
public static void main(String[] args)
{
System.out.println(staticVariable); // ok
System.out.println(Demo.staticMethod()); // ok
System.out.println(new Demo().instanceMethod()); // ok
System.out.println(new Demo().instanceVariable); // ok
System.out.println(Demo.instanceMethod()); // wrong
System.out.println(instanceVariable); // wrong
}
}
This is because by default when you call a method or variable it is really accessing the this.method() or this.variable. But in the main() method or any other static method(), no "this" objects has yet been created.
In this sense, the static method is not a part of the object instance of the class that contains it. This is the idea behind utility classes.
To call any non-static method or variable in a static context, you need to first construct the object with a constructor or a factory like your would anywhere outside of the class.
More depth:
Basically it's a flaw in the design of Java IMO which allows static members (methods and fields) to be referenced as if they were instance members. This can be very confusing in code like this:
Thread newThread = new Thread(runnable);
newThread.start();
newThread.sleep(1000);
That looks like it's sending the new thread to sleep, but it actually compiles down into code like this:
Thread newThread = new Thread(runnable);
newThread.start();
Thread.sleep(1000);
because sleep is a static method which only ever makes the current thread sleep.
Indeed, the variable isn't even checked for non-nullity (any more; it used to be, I believe):
Thread t = null;
t.sleep(1000);
Some IDEs can be configured to issue a warning or error for code like this - you shouldn't do it, as it hurts readability. (This is one of the flaws which was corrected by C#...)
**Here you can see table that clear the access of static and non-static data members in static and non-static methods. **
You can create non-static references in static methods like :
static void method() {
A a = new A();
}
Same thing we do in case of public static void main(String[] args) method
public class XYZ
{
int i=0;
public static void increament()
{
i++;
}
}
public class M
{
public static void main(String[] args)
{
XYZ o1=new XYZ();
XYZ o2=new XYZ();
o1.increament();
XYZ.increament(); //system wont be able to know i belongs to which object
//as its increament method(static method)can be called using class name system
//will be confused changes belongs to which object.
}
}

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