Can we call main() function inside any other function? I tried but did not come up with it.
If we can't call it then why?
Why main() is not like ordinary methods?
Yes why not try something like:
public class Main {
public static void main(String args[]) {
System.out.println("Hello World");
}
}
public class NewMain {
public static void main(String args[]) {
Main.main(args);
}
}
If you run:
java NewMain
the output is:
Hello World
Why main() is not like ordinary methods
Not like ordinary methods in what sense? It is like any other method. I'll try to explain why main looks like this, maybe it'll help you to understand what's going on.
It's void because when it finishes, it doesn't mean that the program finished. If it spawns a new thread it might be that these threads are still running.
It's public because it's called by the JVM, which is outside the scope of the project.
And of course it has to be static because when the JVM calls it, there's no object existing for the class being called.
Nothing special about it when you understand it, so yes.. it can be called like any other static method.
Sure you can, main() is like nay other method in that area.
public class A {
public static void main(String[] args) {
System.out.println("a's main()");
B.main(new String[0]);
}
}
public class B {
public static void main(String[] args) {
System.out.println("B's main()");
}
}
Running A's main() will produce:
a's main()
B's main()
All the actions that you do one other methods can also be done on the main method
The only difference between main method and other methods is that the main method servers as the starting point for running a class
The java command launches a Java application. It does this by starting a Java runtime environment, loading a specified class, and invoking that class's main method.
Except for this reason, there is nothing else.
Related
I'm still a bit knew to Java. Only done PHP and a bit of javascript most my life. I had a question concerning the order you have your classes/methods encapsulated inside each other. Does it matter? Is there a better professional way with doing this? Sorry for the newbie question.
Method 1 (class after main())
public class myClass {
public void main(String[] args) {
//Body Code Here
}
}
class myClass {
//Body Code Here
}
Method 2 (class inside main())
class mainClass {
public void main(String[] args) {
//Body Code Here
class myClass {
//Body Code Here
}
}
}
Method 3 ( class before main())
class myClass {
//Body Code Here
}
class mainClass {
public static main(String[] args) {
//Body Code Here
}
}
You need to have all your methods inside the body of your class. Outside it, you can only import some libraries for example java.util.Scanner(for reading from the console).
And you can insert methods before or after the Main method, both ways when you run the program they will be read.
Your main method, as any other method, must be within a class. Therefore only the first method would work.
Your first method will work.
as the main method should be in the class then only jvm will compile it.
1: I have a program like..
public class Test{
void dispose(){
System.out.println("disposing");
}
Test t=new Test();
public static void main(String[] args){
t.dispose();
}
}
why cant I call dispose method from main()? if its static and non static relation, why the below code works?
public class Test{
void dispose(){
System.out.println("disposing");
}
public static void main(String[] args){
Test t=new Test();
t.dispose();
}
}
2: should always the method call code shold be in method? because, the below code is not working..
public class Test{
void dispose(){
System.out.println("disposing");
}
Test t=new Test();
t.dispose();
public static void main(String[] args){
}
}
Please clarify me..
Example 1
You are in a static method (main) and trying to access a non-static variable t. You have declared the variable t as:
Test t=new Test();
This has created it as a member variable of the class. Instead you need to declare it as:
static Test t=new Test();
Now the static method can access it (although this is generally not a good way to do things).
Example 2
You are now declaring the variable t as a local variable inside the main method so it is valid to access it from within main.
Example 3
With the exception of initalizer blocks (which you don't need to worry about) all code must go inside a method.
I guess you come from a background in Procedural language like C.
Java is different. It's object-orriented.
Coming to your Question . . . .
Ans1: It's correct to say that you don’t necessarily have to create an instance of a class to use the methods of the class. If a method is declared with the static keyword, the method can be called without first creating an instance of the class. That’s because static methods are called from classes, not from objects.
BUT, you can not call non-static method from a static context (here as in static method main()). WHY?
Because you can't call something that doesn't exist. Since you haven't created an object, the non-static method doesn't exist yet. A static method (by definition) always exists.
However even that's not the exact case over here
You may feel that you have created an instance of the class at line 5 of your code but to to the compiler, it doesn't exists. It's outside the main() method, which is the first thing looked for in any run-able Java program. The compiler then ropes in other parts as required. You can't have executable code that is not in a method, look at your object initialization. In second block of code, the compiler sees the object initialization, so program executes.
Ans2: YES. As mentioned before, You can't have executable code that is not in a method
Illustration:
class DeclarationTest
{
int a = 5;
int b = 10;
int c = a + b;//it is Ok, this is a declaration statement for c
/*
int c;
c = a + b; ------> this is not Ok, you are performing an action here this must be inside a method!
*/
}
If that was the case it would make having methods a bit less useful. . . Think about it.
why cant I call dispose method from main()? if its static and non static relation, why the below code works?
Since you have a instance of Test, so you can use that even in static context.
should always the method call code shold be in method?
Yes. exactly. Either in methods, static block or in constructor. Other places not allowed.
Before starting everything let me clear what is an Class variable and an Object Variable
Class Variable : Static variable, which can be accessed without initializing the Class
Object Variable: non static Variable which can only be accessed on CLass instantiation
So in your case, when the main Gets Called, the Class is not initiated, so the object doesnt get initialized, so you cannot call the dispose method.
The reason the second block doesn't work is because of the static relation.
If a method is not static, then it must have an instance to be accessed. That is why, creating an instance of the class allows you to call the method.
If you made the dispose method static, then you could directly call the method since there is only a single instance of it.
Here is a link to a different question that explains it well:
calling non-static method in static method in Java
Hope this helps :)
why cant I call dispose method from main()? if its static and non
static relation, why the below code works?
non-static variable t cannot be referenced from a static context(compiler exception).
You should always remember that the jvm searches for the main() method and executes it. Methods and blocks are initialized after that.
note:- You always compile and run the class which contains the main method.
You are declaring the variable t as a local variable inside the main method so it is valid to access it from within main.
should always the method call code should be in method?
Yes method calls always need to be inside of a method or in the constructor or initialization block, even static block .
When a java program is being executed JVM looks for main method. Within the main if you don't write anything nothing will happen.
public class Test{
void dispose(){
System.out.println("disposing");
}
Test t=new Test(); //That's ok.
t.dispose(); //causes compilation error
public static void main(String[] args){
//Executed as soon as you run your program.
}
}
You want to call dispose(). What do you want? Call dispose() on object of test as t.dispose() or you can call it using Test.dispose();
Method 1:
public class Test{
void dispose(){
System.out.println("disposing");
}
public static void main(String[] args){
Test t=new Test(); //You need a reference to Test object
t.dispose(); //to call its methods
}
}
Is dispose() in Test static? No... So, you must call it by using Test object.
If dispose() in Test static? then...
Method 2:
public class Test{
static void dispose(){
System.out.println("disposing");
}
public static void main(String[] args){
Test.dispose(); // Since dispose() is static.
}
}
You can't call non static methods by using Class Reference. You must use Class object But you can call static methods by using class objects (not recommended). You should call it using Class Reference.
Method 3: (Not recommended)
public class Test{
static void dispose(){
System.out.println("disposing");
}
public static void main(String[] args){
Test t = new Test();
t.dispose(); //Static members should be accessed using class name
}
}
Yes, you can not call non-static object or variables inside static block. If you declare object as static then your code will work as follow.
public class Test{
void dispose(){
System.out.println("disposing");
}
static Test t=new Test();
public static void main(String[] args){
t.dispose();
}
}
You can also try something like below:
public class Test{
void dispose(){
System.out.println("disposing");
}
{
dispose();
}
public static void main(String[] args){
Test t=new Test();
}
}
Also, we can declare object outside method in class but we can not call method outside method or block.
I have a doubt regarding static methods. In the program written below, output will be: main. I understand this because main is a static method, so when class loads, it executes. If so, the same principle should apply for met() also, right? As it is also static. Why does only main executes whereas met() doesn't when the class loads?
public class Test {
static void met() {
System.out.println("method");
}
public static void main(String[] args) {
System.out.println("main");
}
}
No, this isn't correct.
Neither of these methods are called when the class is loaded.
main() is called when you execute the class Test.
Only static initialisers are called when the class is loaded. A static initialiser looks like this:
static
{
//code here
}
A class is loaded before the main() method is executed, and therefore its static initialiser is run before the main() method. The following snippet will make that clear.
public class TestA
{
static
{
System.out.println( "hello" );
}
public static void main( String[] args )
{
System.out.println( "bye" );
}
}
Let me explain it in detail
Types of methods There are basically two types of methods,
Instance methods
Class Methods
Instance Methods Belong to objects and you will always need an object/instance to call such methods.
static methods are the class methods and they can be called directly by class name, there is no need to have an instance of class to call them.
For example,
class Demo{
public void sayHello(){
System.out.println("Hello");
}
public static void sayHi(){
System.out.println("Hi")
}
public static void main(String args[]){
Demo.sayHi(); //Call to static/class method
Demo.sayHello(); //It will not work
Demo d = new Demo();
d.sayHello(); //It will work.
}
}
**But NONE of them gets called automatically when class loads.**
Main Difference between the two
In memory there is ONLY ONE copy of static methods which will be available for all the objects. But whenever an object is created a new copy of instance method is created for the object, so each object has its own instance method. Similar to instance & class variables.
Static methods are not meant to be run automatically, instead they are shared by all objects. Why main() method is called, because it is the entry point of the program.
Apart from them, there is static block which is called automatically only once when the class is loaded.
Example
class Main{
static{
System.out.println("static block");
}
public static void main(String args[]){
System.out.println("main");
}
}
Output will be
static block
main
main() method is not executed because it's static, it executes because it is the Entry Point for any Java program. If you want something to run, you'll need to call it from the main method. No other methods are automatically called when the class is executed.
Not at all. The main method will only run if that particular class is ran as entry point.
That met() method will not run until it has been called. The main difference it has with instance methods, is that you do not need to create an instance of the class in order to run it, you can simply run it through the class itself: Test.met();
What you mean is a static block:
private static String description;
static{
description = "this runs on loading the class";
}
You can use static block instead of static method, to print it before main method like this -
public class Test
{
static{
System.out.println("method");
}
public static void main(String[] args){
System.out.println("main");
}
}
met() is a static method, it will be in memory when the class is loaded, you need to call it.. You could use a static block to print "method".
If you want to execute on load , just intialise it as static block,
static{
System.out.println("method");
}
Because static blocks are executed once the class loads . And among other static methods main() has the high priority
No static method will get called when you call it only, you are mixing static initializer and static method
it prints main because when you run Java application it invokes main() method
there is a difference between static methods, static blocks and static variables. As you do not call the static method, it will not print
To make it print you will need to call Test.met ();
Alternatively you could have it set as a static block
as in
static {
System.out.println("static block");
}
This will be called as soon as Test is loaded.
Not all static method will be called by default when a program runs.
From Docs
The java tool launches a Java application. It does this by starting a Java runtime environment, loading a
specified class, and invoking that class's main method. The method declaration must look like the following:
**public static void main(String args[])**
So, main will be called by JVM and someone should call met() so that it is executed.
What you understood is wrong. Because whenever class loads JVM creates Class class object and int that class class object all static methods resides. main method is entry point for JVM thats why it is executing, JVM internally calling Main method. Whenever Class loads that time it only executes Static internalization blocks.
Main() is only executed because it is the entry point.
For more information you can read the documentation.
In an interview I have asked this question:Without using static and main how could we print message on console?Is it possible?
You could define a custom class loader that prints your message :
public class MyClassLoader extends ClassLoader {
public MyClassLoader(ClassLoader other) {
super(other);
System.out.println("Hi there");
System.exit(0);
}
}
Then run the java command :
java -Djava.system.class.loader=MyClassLoader
(don't need to add a class as parameter)
I have asked this question:Without using static and main how could we print
message on console?Is it possible?
Answer is No!
You cannot execute anything unless main() method is called. Prior to Java 7 classes were loaded before main() method was looked up. So you could print your data through static blocks(static block gets executed when classes are loaded) but from java 7 onward even that is not possible. So you always have to execute the main() method first.
Even in frameworks like Spring beans are generally initialized only when it's context is referenced(again main() is required to be executed first).So there is no way you can print something to console without invoking main() method or through static functions/blocks.
You always need to put syso in a block of code, perhaps:
public class example {
public void message(){
System.out.println("Hello");
}
}
Here the method is not static
public class Test {
public static PrintStream ps = System.out.printf("%s", "hello");
}
Pretty strange question for interview. It will print hello, and throw Exception in thread "main" java.lang.NoSuchMethodException.
possible with static block which executes before main method
The answer is definitely No.
At least either you need a static block or you need an empty main().
See the below examples:
1.
public class ABC {
static{
System.out.println("hai");
}
public static void main(String[] args) {}
}
OUTPUT:
hai
2.
public final class ABC {
static{
System.out.println("hai");
}
}
It will print 'hai' while running but after that one exception will also occur.
OUTPUT:
java.lang.NoSuchMethodError: main
hai
Exception in thread "main"
Java is a OOP language.
You can not create a program without creating a class and adding it a static main function.
Then, you can call System.out.println to print a line.
So, the answer is no.
# Aniket Thakur
As i am not able to comment because of my points. but following programs prints the message before it enters main.
NOTE: I used Java 7 and Java 8. both works fine and runs static block before main.
public class PrintBeforeMain {
private static int i = m1();
public static int m1(){
System.out.println("m1(): Before main() through static method...");
return 0;
}
static{
System.out.println("Inside standalone static{} block");
}
public static void main(String[] args) {
System.out.println("Inside main()");
}
}
I was wondering what the effect of creating extra main methods would do to your code.
For example,
public class TestClass {
public static void main (String[] args){
TestClass foo = new TestClass();
}
}
After the program initially starts up, foo will be created and it would have another public main method inside it. Will that cause any errors?
It will cause no errors. Just because you initialize an object, doesn't mean the main method gets executed. Java will only initially call the main method of the class passed to it, like
>java TestClass
However, doing something like:
public class TestClass
{
public static void main (String[] args)
{
TestClass foo = new TestClass();
foo.main(args);
}
}
Or
public class TestClass
{
public TestClass()
{
//This gets executed when you create an instance of TestClass
main(null);
}
public static void main (String[] args)
{
TestClass foo = new TestClass();
}
}
That would cause a StackOverflowError, because you are explicitly calling TestClass's main method, which will then call the main method again, and again, and again, and....
When in doubt, just test it out :-)
The main method is static, which means it belongs to the class rather than the object. So the object won't have another main method inside it at all.
You could call the main method on instances of the object, but if you do that it's literally just another way of calling TestClass.main() (and it's frowned upon by many, including me, to call a static method on an instance of an object anyway.)
If you're referring to multiple main methods in the same program, then this isn't a problem either. The main class is simply specified and its main method is executed to start the program (in the case of a jar file this is the main-class attribute in the manifest file.)
It won't have an additional main-method, as main is static. So it's once per class.
If you have multiple main-methods in your project, you will specify which one to launch when starting your application.
This is perfectly fine. Having multiple main methods doesn't cause any problems. When you first start a Java program, execution begins in some function called main in a class specified by the user or by the .jar file. Once the program has started running, all the other functions called main are essentially ignored or treated like other functions.
After searching for a Java Class with multiple main() methods or in plain words, overloaded main() methods, I came up with an example of my own. Please have a look
public class MultipleMain{
public static void main(String args[]){
main(1);
main('c');
main("MyString");
}
public static void main(int i){
System.out.println("Inside Overloaded main()");
}
public static void main(char i){
System.out.println("Inside Overloaded main()");
}
public static void main(String str){
System.out.println("Inside Overloaded main()");
}
}
I tested this Java Code on JDK 1.7 and works like a charm !
You need "public static void main(String args[])" to start with and then you can call overloaded main methods inside this main and it should work for sure.
Any comments and suggestion are highly appreciated. I am just a novice Java Developer willing to develop my Java skills.
Thanks,
PK
No, you can have any number of main-methods in a project. Since you specify which one you want to use when you launch the program it doesn't cause any conflicts.
You can have only one main method in one class, But you can call one main method to the another explicitly
class Expmain
{
public static void main(String[] ar)
{
System.out.println("main 1");
}
}
class Expmain1
{
public static void main(String[] ar)
{
System.out.println("main 2");
Expmain.main(ar);
}
}
when you run your Java class it will always look for the signature public static void main(String args[]) in the class. So suppose if you invoking by command line argument, it will look for the method Signature in the class and will not invoke other until if u explicitly inoke it by its class name.
class MainMethod1{
public static void main(String[] ags){
System.out.println("Hi main 1");
testig2 y = new testig2();
//in this case MainMethod1 is invoked/.......
// String[] a = new String[10];
// testig2.main(a);
}
}
class MainMethod2{
public static void main(String[] ags){
System.out.println("Hi main 2");
}
}
But when you try the same from eclipse it will ask for which class to compile. Means MainMethod1 or Mainmethod2. So if te class has the exact signature they can be used as individual entry point to start the application.
Coming to your question, If you remove the signature as u did above by changing the argument if main method. It will act as a normal method.
It is all about the execution engine of JVM. Remember, you write >java com.abc.MainClass on cmd prompt.
It explains everything. If main method is not found here it throws a run time Error:Main Method Not Found in class MainClass.
Now if main method is found here, it acts as the first point when Program Counters have to map and start executing the instructions. Referred classes are loaded then, referred methods may be called using the instances created inside. So, main is class specific though one class can have only one main method.
Please note, main method's signature never changes. You can have two overloaded main methods in same class, like
public static void main(String[] args) {}
public static void main() {} //overloaded in same class.
During Static binding, the original main is resolved and identified by execution engine.
Another interesting point to consider is a case where you have two different classes in of java file.
For example, you have Java file with two classes:
public class FirstClassMultiply {
public static void main (String args[]){
System.out.println("Using FirstClassMultiply");
FirstClassMultiply mult = new FirstClassMultiply();
System.out.println("Multiple is :" + mult.multiply(2, 4));
}
public static void main (int i){
System.out.println("Using FirstClassMultiply with integer argument");
FirstClassMultiply mult = new FirstClassMultiply();
System.out.println("Multiply is :" + mult.multiply(2, 5));
}
int multiply(int a, int b) {
return (a * b);
}
}
class SecondClass {
public static void main(String args[]) {
System.out.println("Using SecondClass");
FirstClassMultiply mult = new FirstClassMultiply();
System.out.println("Multiply is :" + mult.multiply(2, 3));
FirstClassMultiply.main(null);
FirstClassMultiply.main(1);
}
}
Compiling it with javac FirstClassMultiply.java will generate two .class files, first one is FirstClassMultiply.class and second one is SecondClass.class
And in order to run it you will need to do it for the generated .class files: java FirstClassMultiply and java SecondClass, not the original filename file.
Please note a couple of additional points:
You will be able to run SecondClass.class although it's class wasn't public in the original file!
FirstClassMultiply overloading of the main method
of is totally fine, but, the only entry point to your prog
will be the main method with String args[] argument.