public static void main(String[] args) is the most important Java method. When you start learning java programming, this is the first method you encounter. Remember the first Java Hello World program you wrote that runs and prints “Hello World”?
Table of Contents
public static void main(String[] args)
Java main method is the entry point of any java program. Its syntax is always public static void main(String[] args). You can only change the name of String array argument, for example you can change args to myStringArgs.
Also String array argument can be written as String... args or String args[].
Let’s look at the java main method closely and try to understand each of its parts.
public
This is the access modifier of the main method. It has to be public so that java runtime can execute this method. Remember that if you make any method non-public then it’s not allowed to be executed by any program, there are some access restrictions applied. So it means that the main method has to be public. Let’s see what happens if we define the main method as non-public.
Copy
public class Test {
static void main(String[] args){
System.out.println("Hello World");
}
}
Copy
$ javac Test.java
$ java Test
Error: Main method not found in class Test, please define the main method as:
public static void main(String[] args)
or a JavaFX application class must extend javafx.application.Application
$
static
When java runtime starts, there is no object of the class present. That’s why the main method has to be static so that JVM can load the class into memory and call the main method. If the main method won’t be static, JVM would not be able to call it because there is no object of the class is present. Let’s see what happens when we remove static from java main method.
Copy
public class Test {
public void main(String[] args){
System.out.println("Hello World");
}
}
Copy
$ javac Test.java
$ java Test
Error: Main method is not static in class Test, please define the main method as:
public static void main(String[] args)
$
void
Java programming mandates that every method provide the return type. Java main method doesn’t return anything, that’s why it’s return type is void. This has been done to keep things simple because once the main method is finished executing, java program terminates. So there is no point in returning anything, there is nothing that can be done for the returned object by JVM. If we try to return something from the main method, it will give compilation error as an unexpected return value. For example, if we have the main method like below.
Copy
public class Test {
public static void main(String[] args){
return 0;
}
}
We get below error when above program is compiled.
Copy
$ javac Test.java
Test.java:5: error: incompatible types: unexpected return value
return 0;
^
1 error
$
main
This is the name of java main method. It’s fixed and when we start a java program, it looks for the main method. For example, if we have a class like below.
Copy
public class Test {
public static void mymain(String[] args){
System.out.println("Hello World");
}
}
And we try to run this program, it will throw an error that the main method is not found.
Copy
$ javac Test.java
$ java Test
Error: Main method not found in class Test, please define the main method as:
public static void main(String[] args)
or a JavaFX application class must extend javafx.application.Application
$
String[] args
Java main method accepts a single argument of type String array. This is also called as java command line arguments. Let’s have a look at the example of using java command line arguments.
Copy
public class Test {
public static void main(String[] args){
for(String s : args){
System.out.println(s);
}
}
}
Above is a simple program where we are printing the command line arguments. Let’s see how to pass command line arguments when executing above program.
Copy
$ javac Test.java
$ java Test 1 2 3
1
2
3
$ java Test "Hello World" "Pankaj Kumar"
Hello World
Pankaj Kumar
$ java Test
$
Java main method command line arguments in Eclipse
Below images show how to pass command line arguments when you are executing a java program in Eclipse.
That’s all for public static void main(String[] args), java main method.
DileeP Achar says
nice explanation about each part thank you!……..
ofek Shaltiel says
That’s amazing, thank you
raghavendra jha says
{
public static void main(int[] args)
{
System.out.println(“this is the overload of the method”);
}
public static void main(String[] args)
{
System.out.println(“this isn working anymore”);
}
output-this isn working anymore
can you explain me why this output will be show….
Jagadeesh Keerthi says
Refer https://stackoverflow.com/questions/3759315/can-we-overload-the-main-method-in-java
Sawan sharma says
“this isn working anymore”- this output came bacause JVM never accepts the int[] args in main method.
When you run a java program then JVM will search for main method in the class with String []args if it not present then it will not print anything. it will give an error like ” Main method not found”.
kamal says
two main functions present when there should be only one
Ellen Dares says
Hi, Great article, I will share my thoughts.
As per JLS (Java Language Specification), “A Java virtual machine starts execution by invoking the main() method of some specified class, passing it a single argument, which is an array of strings”.
Definition of your main method should be as below
public static void main(String[] args)
public – your method should be accessible from anywhere
static – Your method is static, you can start it without creating an instance. When JVM starts, it does not have any instance of the class having main method. So static.
void – It does not return anything.
Henceforth, the main() method is predefined in JVM to indicate as a starting point.
Hope that helps!
Samridhi Jain says
What about adding throws Exception() in the definition, doesn’t it change the signature of the function?
Pankaj says
No, throwing exceptions is not part of the signature.
Ranjan says
Sir i am new to java . I have compiled and run about 30 java programe but now in a specific folder java programe comiled but wont run showing error main method not found even i have main method psvm(S[] args).but when i run same programe in other folder it runs .why that only folder its not running .pls help me
Nanda says
can any one suggest me how to study or how to understand threads in java,,, because i studied so many times ,
but i’m unable to get the clarity about the concept and coding.
p.naveen says
awsome explanation sir,keep going ,,,,,,
Vijay says
Very Nice explanation about java main method…each part…….Great
Cacarian says
I think there is a small mistake in saying “Java programming mandates that every method signature provide the return type. “. Return type is not part of the method signature. Nice explanation though!
Pankaj says
Thanks for noticing the typo. Java programming mandates to have a return type, but yes it’s not part of the method signature.
Kanchan Gautam says
Sir I still haven’t understood the significance of String [] args. Please explain.
Duke says
Can you explain as to why do we need to have a String args[] array passed to main? Why does it not compile if we do not pass a String args[] array? Why does Java need a String args[] array?
Pankaj says
It’s the syntax and to pass command line arguments to the main method.
Naveen singh says
Nice explanation on core main method.
Priyanka says
not working
Pankaj says
what do you mean?
upendra says
//run this code and see the same error at run time. your idea fails
public class aggregation
{
String name;
int roll;
String bg;
int marks;
aggregation(String name, int roll, String bg, int marks)
{
this.name=name;
this.roll=roll;
this.bg=bg;
this.marks=marks;
}}
class students
{
int salary;
String gender;
aggregation info;
students(int salary, String gender, aggregation info)
{
this.salary=salary;
this.gender=gender;
this.info=info;
}
public static void main(String[] args)
{
aggregation a1=new aggregation(“Upendra Dhamala”, 48, “B+”,98);
students a2=new students(90000, “M”, a1);
System.out.println(a2.salary);
System.out.println(a2.gender);
System.out.println(a2.info.name);
System.out.println(a2.info.roll);
System.out.println(a2.info.bg);
System.out.println(a2.info.marks);
}
}
Raj says
//Very vague response. Try to avoid saying things like your idea fails. Made minor changes. Changed the first letter of the class to uppercase, but you might not have to.
package raj.basic.test;
public class Aggregation
{
String name;
int roll;
String bg;
int marks;
Aggregation(){};
Aggregation(String name, int roll, String bg, int marks)
{
this.name=name;
this.roll=roll;
this.bg=bg;
this.marks=marks;
}
}
class Students
{
int salary;
String gender;
Aggregation info;
Students(int salary, String gender, Aggregation info)
{
this.salary=salary;
this.gender=gender;
this.info=info;
}
public static void main(String[] args)
{
Aggregation a1=new Aggregation(“Upendra Dhamala”, 48, “B+”,98);
Students a2=new Students(90000, “M”, a1);
System.out.println(a2.salary);
System.out.println(a2.gender);
System.out.println(a2.info.name);
System.out.println(a2.info.roll);
System.out.println(a2.info.bg);
System.out.println(a2.info.marks);
}
}
//Output is as follows
/*
90000
M
Upendra Dhamala
48
B+
98
*/
Abhishek Tripathi says
Sir I want to implement push notifications like your site. My website is built on spring and jsp. Please help me out of this.
Pankaj says
I am using OneSignal push notification and they have a WordPress plugin, they also have Java API. Please look into their documentation for implementing it. You can also look at some other services for push notification.
Vinodkumar says
Excellent! Given more clarity on main method
Sambasiva says
Nice explanation on core main method.
Rahul says
Well said Sambasiva !!!