Private Static Void String
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private static void printCombos(String[][] strings) {
String[] sentence = new String[strings.length];
int[] indices = new int[strings.length];
int[] maxes = new int[strings.length];
for (int i = 0; i < strings.length; i++) {
System.out.println("Empty input array found. Exiting.");
System.out.println(String.join(" ", sentence));
while (!Arrays.equals(indices, maxes)) {
if (indices[setInd] <= maxes[setInd]) {
sentence[setInd] = strings[setInd][indices[setInd]];
sentence[setInd] = strings[setInd][0];
System.out.println(String.join(" ", sentence));
public static void main(String[] args) {
String[][] in = {{"tall", "short"}, {"fat", "skinny", "tubby"}, {"boy", "girl", "dog", "cat"}};
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Home » Java » public static void main(String[] args) – Java main method
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”?
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.
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.
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.
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.
We get below error when above program is compiled.
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.
And we try to run this program, it will throw an error that the main method is not found.
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.
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.
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.
I love Open Source technologies and writing about my experience about them is my passion.
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;
// represent difference between
// head position and track number
int distance = 0;
// true if track has been accessed
boolean accessed = false;
}
// 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)
// create array of objects of class node
node diff[] = new node[request.length];
// initialize array
for (int i = 0; i < diff.length; i++)
// 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);
// 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);
}
}
Help me determine the output for the given code snippet:
public class App
{
public static void main( String[] args )
{
for(;;) {
System.out.println(“Hi”);
}
}
}
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); } }
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.
In for loop section 2nd part must be condition either true or false
for loop syntax is wrong. u haven’t mentioned the condition in for loop
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
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?
1) Any name is fine. It will be exactly the same.
2) String St[] = String[] St. It’s the same thing.
3) None – probably easier to read code.
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.
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;
}
}
It’s not public main it’s public static void main
i think main method syntax is wrong
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)
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.
public int static main(String[ ] args)
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[]){
@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();
}
}
}
}
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{
System.out.println(“Shows Screen”);
}
}
public static void main(String[] args1) {
public static void main1(int args) {
JavaObject3 obj1=new JavaObject3();
System.out.println(“sucess shows”);
}
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.
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?
Both syntaxes are correct. But the recommended and conventional one is String [ ] args.
public is access specifier or access modifier . i think it is access specifier but in the explanation you given it as access modifier.
Both access modifier and specifier have same meaning.
nice explanation about each part thank you!……..
{
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….
“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”.
two main functions present when there should be only one
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.
What about adding throws Exception() in the definition, doesn’t it change the signature of the function?
No, throwing exceptions is not part of the signature.
/*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])?");
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
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.
awsome explanation sir,keep going ,,,,,,
Very Nice explanation about java main method…each part…….Great
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!
Thanks for noticing the typo. Java programming mandates to have a return type, but yes it’s not part of the method signature.
Sir I still haven’t understood the significance of String [] args. Please explain.
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?
It’s the syntax and to pass command line arguments to the main method.
Nice explanation on core main method.
//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);
//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.
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
*/
Sir I want to implement push notifications like your site. My website is built on spring and jsp. Please help me out of this.
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.
Excellent! Given more clarity on main method
Nice explanation on core main method.
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