The Java main
method is usually the first method you learn about when you start programming in Java because its the entry point for executing a Java program. The main
method can contain code to execute or call other methods, and it can be placed in any class that’s part of a program. More complex programs usually have a class that contains only the main
method. The class that contains the main
method can have any name, although typically you can just call the class Main
.
In the examples that follow, the class that contains the main
method is called Test
:
In this article you’ll learn what each component of the main
method means.
The syntax of the main
method is always:
You can change only the name of the String
array argument. For example, you can change args
to myStringArgs
. The String
array argument can be written as String... args
or String args[]
.
public
The access modifier of the main
method needs to be public
so that the JRE can access and execute this method. If a method isn’t public, then access is restricted. In the following example code, the main
method is missing the public
access modifier:
When you compile and run the program, the following error occurs because the main
method isn’t public and the JRE can’t find it:
OutputError: 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 the Java program starts, there is no object of the class present. The main
method has to be static
so that the JVM can load the class into memory and call the main
method without creating an instance of the class first. In the following example code, the main
method is missing the static
modifier:
When you compile and run the program, the following error occurs because the main
method isn’t static
:
OutputError: Main method is not static in class Test, please define the `main` method as:
public static void main(String[] args)
void
Every Java method must provide the return type. The Java main
method return type is void
because it doesn’t return anything. When the main
method is finished executing, the Java program terminates, so there is no need for a returned object. In the following example code, the main
method attempts to return something when the return type is void
:
When you compile the program, the following error occurs because Java doesn’t expect a return value when the return type is void
:
OutputTest.java:5: error: incompatible types: unexpected return value
return 0;
^
1 error
main
The Java main
method is always named main
. When a Java program starts, it always looks for the main
method. The following example code shows a main
method renamed to myMain
:
When you compile and run the program, the following error occurs because the JRE can’t find the main
method in the class:
OutputError: 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. Each string in the array is a command line argument. You can use command line arguments to affect the operation of the program, or to pass information to the program, at runtime. The following example code shows how to print the command line arguments that you enter when you run the program:
When you compile the program and then run it with a few command line arguments separated by spaces, the arguments get printed in the terminal:
Output1
2
3
Testing the main method
In this article you learned about each component of the Java main
method. Continue your learning with more Java tutorials.
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Nice explanation on core main method.
- Sambasiva
Well said Sambasiva !!!
- Rahul
Excellent! Given more clarity on main method
- Vinodkumar
Sir I want to implement push notifications like your site. My website is built on spring and jsp. Please help me out of this.
- Abhishek Tripathi
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.
- Pankaj
not working
- Priyanka
what do you mean?
- Pankaj
//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); } }
- upendra
//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 */
- Raj
Nice explanation on core main method.
- Naveen singh
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?
- Duke
It’s the syntax and to pass command line arguments to the main method.
- Pankaj
Sir I still haven’t understood the significance of String [] args. Please explain.
- Kanchan Gautam
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!
- Cacarian
Thanks for noticing the typo. Java programming mandates to have a return type, but yes it’s not part of the method signature.
- Pankaj
Very Nice explanation about java main method…each part…Great
- Vijay
awsome explanation sir,keep going ,
- p.naveen
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
- Ranjan
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.
- Nanda
check your package
- Jilson
What about adding throws Exception() in the definition, doesn’t it change the signature of the function?
- Samridhi Jain
No, throwing exceptions is not part of the signature.
- Pankaj
/*My program below is returning “Solution.java:44: error: method main(String[]) is already defined in class Solution public static void main(String[] args) { */ //I can’t get to understand why though am new to programming// import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*; public class Solution { public static void main(String args[]) { Scanner scan = new Scanner(System.in); int n =scan.nextInt(); scan.close(); String result =”“; if (n100){ result=”“; } else{ if(n%2==1){ result = “Weird”; } else{ if(n>=2 && n>=5){ result =“Not Weird”; } else{ if(n>=6 && n<=20){ result =“Weird”; } else{ result =“Not Weird”; } System.out.println(result); } } } } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip(”(\r\n|[\n\r\u2028\u2029\u0085])?"); scanner.close(); } }
- Julie
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!
- Ellen Dares
Awesome explanation
- Thomas L Dean
{ 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…
- raghavendra jha
Refer https://stackoverflow.com/questions/3759315/can-we-overload-the-main-method-in-java
- Jagadeesh Keerthi
“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".
- Sawan sharma
ok
- qwerty
two main functions present when there should be only one
- kamal
That’s amazing, thank you
- ofek Shaltiel
nice explanation about each part thank you!..
- DileeP Achar
public is access specifier or access modifier . i think it is access specifier but in the explanation you given it as access modifier.
- satyanarayana
Both access modifier and specifier have same meaning.
- kamal
Why do you sometime type: public static void mymain(String args[ ]) and other times: public static void mymain(String [ ] args) ? What is the correct sintax? Thanks, Miguel Ferreira
- Miguel Ferreira
Both syntaxes are correct. But the recommended and conventional one is
String [ ] args
.- Pankaj
Hello Sir, I have a query about 2 main method in java? I was written 2 main method in single main class,But i am not getting 2 output like(Show screen,sucess shows). you can see the program below? --------------------------------------------------------------------------------- class Module{ public void Dress() { System.out.println(“Shows Screen”); } } public class JavaObject3 { public static void main(String[] args1) { Module m=new Module(); m.Dress(); } public static void main1(int args) { JavaObject3 obj1=new JavaObject3(); System.out.println(“sucess shows”); } }
- arshad khan
When you execute a java class, method with the signature
public static void main(String[] args)
gets executed. The other method is just having a name as “main” and it’s overloading the main method but it won’t get executed unless you call it explicitly.- Pankaj
help a lot…thank you so much.
- SITI NUR ADILAH SULAIMAN
I was trying to run this program so that class Lion and class Dog can extend Class Animal to override animalSound and return “Roar” and “Woof”, no compilation error and no output. Help public abstract class Animal{ public abstract void animalSound(); public static void main(String args[]){ class Lion extends Animal{ @Override public void animalSound(){ System.out.println(“Roar”); Lion obj = new Lion(); obj.animalSound(); } } class Dog extends Animal{ @Override public void animalSound(){ System.out.println(“Woof”); Dog obj = new Dog(); obj.animalSound(); } } } }
- Ben
are these correct? public static main(String[ ] args) public static void[ ] main(String[ ] args) public int static main(String[ ] args) public static void mian (String args)
- yon
public static main(String[ ] args) - invalid, return type is missing. public static void[ ] main(String[ ] args) - invalid, void[] is an invalid return type. public int static main(String[ ] args) - invalid, static should come before return type. “public static int main(String[ ] args)” is fine, although it’s not a java main method. public static void mian (String args) - valid method but it’s not a java main method.
- Pankaj
public int static main(String[ ] args)
- Ram
public class Main { private String isbn, author, title, publisher, publishdate; private int price; private String borrower,category; private int payment; private int numberofdays; public Main(){ isbn=“unassigned”; author=“unnasigned”; title=“unnasigned”; publisher=“unnasigned”; publishdate=“unnasigned”; price= 0; } public void setNumberofdays(int nd){ numberofdays = nd; } public void setBorrower(String b){ category = b; } public void setCategory (String c){ category= c; } public void setBookTitle(String t){ title = t; } public void setPublisher(String p){ publisher = p; } public void setAuthor(String a){ author = a; } public void setPublishDate(String pd){ publishdate = pd; } public void setpayment(int pm){ payment = pm; } public void setISBN(String i){ isbn = i; } public void setPrice(int p){ price = p; } public void setPayment (int nd){ payment = nd*5; } public int getNumberofdays(){ return numberofdays; } public String getAuthor(){ return author; } public String getISBN(){ return isbn; } public String getBorrower(){ return borrower; } public String getCategory(){ return category; } public String getBookTitle(){ return title; } public int getpayment(){ return payment; } public String getPublisher(){ return publisher; } public String getPublishDate(){ return publishdate; } public int getPrice(){ return price; } public int getPayment(){ return payment; } } Why my code is not working??
- Ricardo
It’s not public main it’s public static void main
- sdfsdf
i think main method syntax is wrong
- Ramya
My teacher writes as public static void main(String St[]) and it is executed. My question is that (1) what are the maximum possibilities to write in different string names? (2) Why we need to use array in different names? (3) What is the purpose of using different name for String?
- Fareha
- ccxvvxc
OK sir, thak you. By the way, I was preparing for my java programming exams, and have done paper in month of march. But I glad that you removed my confusion. Sorry English is not strong.
- Fareha
class Test { public static void main (String args[]) { String s = “Hello Java” ; for(int i=0; 0 ; i++) { System.out.println( s ); break; } } } Explain why the above code give the compile-time error? Correct the above code.
- kim ka
In for loop section 2nd part must be condition either true or false
- Haribabu
for loop syntax is wrong. u haven’t mentioned the condition in for loop
- Ramya
You need to mention the condition in the mid statement of the loop, as compiler is unable to run the loop without knowing the condition true or false
- Ali
please help me find the errors in this code import java.util.Scanner; public class StringLength { public void main(String[] args) { Scanner scan = new Scanner(System); String message = "Enter a sentence: "; boolean valid = false; String input = “”; int indexOfFullStop = 0; int indexOfQuestion = 0; int indexOfExclamation = 0; int lastIndex = 0; do ( System.out.print(message); input = scan.nextLine(); indexOfFullStop = input.indexOf(‘.’); indexOfQuestion = input.indexOf(‘?’); indexOfExclamation = input.indexOf(‘!’); if((indexOfFullStop != 0) && (indexOfQuestion != 0) && (indexOfExclamation != 0)) { //the sentence ended correctly valid = true; //get the index of the final character that ends the senten ce if(indexOfFu llStop != ‐1) { lastIndex = indexOfFullStop; } else { if(indexOfQuestion == ‐1) lastIndex = indexOfQuestion; } else { lastIndex = indexOfExclamation; } } } else { //the sentence did not end with one of the correct characters message = “A sentence must end with a . OR ! OR ?” + "\nEnter a sentence again: "; } }while(valid) int length = lastIndex; message = “The length of the sentence is " + length + " characters”; System.out.println(message); } }
- nhlanhla
Help me determine the output for the given code snippet: public class App { public static void main( String[] args ) { for(;;) { System.out.println(“Hi”); } } }
- Dia
it will print hi infinite times.
- coder
hi, can someone please assist me I try to run this program but its saying “Error: Could not find or load main class Main” import java.io.File; import java.util.Scanner; class node { // represent difference between // head position and track number int distance = 0; // true if track has been accessed boolean accessed = false; } public class SSTF { // Calculates difference of each // track number with the head position public static void calculateDifference(int queue[], int head, node diff[]) { for (int i = 0; i < diff.length; i++) diff[i].distance = Math.abs(queue[i] - head); } // find unaccessed track // which is at minimum distance from head public static int findMin(node diff[]) { int index = -1, minimum = Integer.MAX_VALUE; for (int i = 0; i diff[i].distance) { minimum = diff[i].distance; index = i; } } return index; } public static void shortestSeekTimeFirst(int request[], int head) { if (request.length == 0) return; // create array of objects of class node node diff[] = new node[request.length]; // initialize array for (int i = 0; i < diff.length; i++) diff[i] = new node(); // count total number of seek operation int seek_count = 0; // stores sequence in which disk access is done int[] seek_sequence = new int[request.length + 1]; for (int i = 0; i < request.length; i++) { seek_sequence[i] = head; calculateDifference(request, head, diff); int index = findMin(diff); diff[index].accessed = true; // increase the total count seek_count += diff[index].distance; // accessed track is now new head head = request[index]; } // for last accessed track seek_sequence[seek_sequence.length - 1] = head; System.out.println("Total number of seek operations = " + seek_count); System.out.println(“Seek Sequence for SSTF is”); // print the sequence for (int i = 0; i < seek_sequence.length; i++) System.out.println(seek_sequence[i]); } public static void main(String[] args) throws Exception { //read from file File file = new File(“num.txt”); Scanner sc = new Scanner (file); Scanner sn = new Scanner (file); int c = 0; while (sc.hasNextLine()) { c++; sc.nextLine(); } int [] arr = new int [c]; int i = 0; while (sn.hasNextInt()) arr[i++] = sn.nextInt(); //get head position int rwhead; System.out.println("Enter current position of R/W head: "); Scanner s = new Scanner(System.in); rwhead = s.nextInt(); shortestSeekTimeFirst(arr, rwhead); } }
- Ackeem
You have to create an object under public static void main(String[] args){} and then access your method through the main method
- Shubham Ray
Arrays in java are all supposed to have a fixed size, but the array of strings that’s passed to main (String[] args) doesn’t. Is this a hard-coded exception to the rule? Are there any other such examples?
- Jackson
Just because Arrays in java have a fixed size, doesnt mean you have to pass in a specific array length. all that means is that once you instantiate a new array in java, you cannot change the size, except reallocating.
- Owen
public class Main { public static void main(String[] args) { int a[]=new int[]{12,2,6,7,11}; int b[]=new int[]{2,6,7,11}; int i=0,j; int way=0; int f; int c[]=new int[12]; for(i=1;i<=12;i++) { f=0; for(j=0;j<4;j++) { if(i==a[j]) f=1; } if(f==0) c[i-1]=i; else c[i-1]=0; if(i%2==0) { if(c[i-1]!=0 && c[i-2]!=0) way++; } } for(i=0;i<10;i++) { if(c[i]!=0 && c[i+2]!=0) way++; } System.out.println(“”+way); } }
- RK
import java.util.Scanner; public class Main { public static void main (String[] args){ Scanner input= new Scanner(System.in); boolean isOnRepeat=true; while(isOnRepeat){ System.out.println (“play the currentsong”); System.out.println(“do you want to change the song, If so answer yes”); String userInput = input.next(); if(userInput.equals(“yes”)) { isOnRepeat=false; } } System.out.println (“play the next song”); } }
- Jeyaharini