Main(String[] args) ...

I understand that args are passed when executing the program, but how exactly does it separate arguments?
Say I type: (for WinXP)
start myprogram.jar 123 -n -k
does it separate the "arguments" according to spaces? or does every character get its own String in the args array?

I am not at my computer, and this one does not have the JDK installed on it so I couldnt test it myself and I didnt feel like waiting until I got back to my house, im impatient =)
But thank you for your response!

Similar Messages

How to call java with public static void main(String[] args) throws by jsp?

how do i call this from jsp? <%spServicelnd temp = new spServicelnd();%> does not work because the program has a main. can i make another 2nd.java to call this spServiceInd.java then call 2nd.java by jsp? if yes, how??? The code is found below...
import java.net.MalformedURLException;
import java.io.IOException;
import com.openwave.wappush.*;
public class spServiceInd
     private final static String ppgAddress = "http://devgate2.openwave.com:9002/pap";
     private final static String[] clientAddress = {"1089478279-49372_devgate2.openwave.com/[email protected]"};
//     private final static String[] clientAddress = {"+639209063665/[email protected]"};
     private final static String SvcIndURI = "http://devgate2.openwave.com/cgi-bin/mailbox.cgi";
     private static void printResults(PushResponse pushResponse) throws WapPushException, MalformedURLException, IOException
          System.out.println("hello cze, I'm inside printResult");
          //Read the response to find out if the Push Submission succeded.
          //1001 = "Accepted for processing"
          if (pushResponse.getResultCode() == 1001)
               try
                    String pushID = pushResponse.getPushID();
                    SimplePush sp = new SimplePush(new java.net.URL(ppgAddress), "SampleApp", "/sampleapp");
                    StatusQueryResponse queryResponse = sp.queryStatus(pushID, null);
                    StatusQueryResult queryResult = queryResponse.getResult(0);
                    System.out.println("Message status: " + queryResult.getMessageState());
               catch (WapPushException exception)
                    System.out.println("*** ERROR - WapPushException (" + exception.getMessage() + ")");
               catch (MalformedURLException exception)
                    System.out.println("*** ERROR - MalformedURLException (" + exception.getMessage() + ")");
               catch (IOException exception)
                    System.out.println("*** ERROR - IOException (" + exception.getMessage() + ")");
          else
               System.out.println("Message failed");
               System.out.println(pushResponse.getResultCode());
     }//printResults
     public void SubmitMsg() throws WapPushException, IOException
          System.out.println("hello cze, I'm inside SubmitMsg");
          try
               System.out.println("hello cze, I'm inside SubmitMsg (inside Try)");
               //Instantiate a SimplePush object passing in the PPG URL,
               //product name, and PushID suffix, which ensures that the
               //PushID is unique.
               SimplePush sp = new SimplePush(new java.net.URL(ppgAddress), "SampleApp", "/sampleapp");
               //Send the Service Indication.
               PushResponse response = sp.pushServiceIndication(clientAddress, "You have a pending Report/Request. Please logIn to IRMS", SvcIndURI, ServiceIndicationAction.signalHigh);
               //Print the response from the PPG.
               printResults(response);
          }//try
          catch (WapPushException exception)
               System.out.println("*** ERROR - WapPushException (" + exception.getMessage() + ")");
          catch (IOException exception)
               System.out.println("*** ERROR - IOException (" + exception.getMessage() + ")");
     }//SubmitMsg()
     public static void main(String[] args) throws WapPushException, IOException
          System.out.println("hello cze, I'm inside main");
          spServiceInd spsi = new spServiceInd();
          spsi.SubmitMsg();
     }//main
}//class spServiceInd

In general, classes with main method should be called from command prompt (that's the reason for main method). Remove the main method, put the class in a package and import the apckage in your jsp (java classes should not be in the location as jsps).
When you import the package in jsp, then you can instantiate the class and use any of it's methods or call the statis methods directly:
<%
spServiceInd spsi = new spServiceInd();
spsi.SubmitMsg();
%>

Why can't created an object in the "public static void main(String args[])"

Example:
public Class Exp {
public static void main(String args[])
new Exp2();-------------> To Occur Error
Class Exp2 {
public Exp2()

You can't create an inner class within main, because
it is a static method. Inner classes can only be
created by true class methods.This is not correct. You can create an inner class object in a static method. You just have to specify an outer class object that the inner class object should be created within:
Exp2 exp2 = new Exp().new Exp2();

Can't find file when use the main(String[] args) method

I put a file under the DefaultDomain directory of Weblogic 11.1.1.5 on my local machine. Below is how I look up this file
FileInputStream fis = new FileInputStream("input.xml");
I started up the IntegratedWeblogicServer instance and I tested using the below 2 approaches.
1) When I test a web service program using the below approach, it CAN'T find the input.xml file
public static void main(String[] args)
2) When I package it into an EAR file and deploy it onto the server then log into the admin console and use the test client approach then it's able to find the file
How do I make it so it can find the file using #1 approach?
Thanks

That's how relative paths work, and it's nothing particular to Java.
When you type ls input.xml or dir input.xml does it list the file, or give an error? If it lists the file, then it's in your current directory, and it will work if you run Java from that directory. If it gives an error, then that file is not in your current working directory, and Java won't be able to find it either.

When does the Java main(String args[]) method exit?

Hi all,
I and my colleagues want to know, when does the main() method exit?
we wrote a small code as follows:
import java.awt.*;
public class Test {
public static void main(String[] args) {
Frame f = new Frame("hi");
f.show();
System.out.println("after show");
and ran the program, in which case I could see the printed message "after show", but although there is no code after the System.out.println(), the Java virtual machine does not exit and waits for the Frame to close.
My question is has the main method exited in this case? since it clearly shows that it does not have any more code to execute. Does java create the main method as a thread or as a process?
regards,
Harshad

To make your application terminate you need to add code to cause the AWT thread to handle the termination. In your code, before f.show(), try:
f.addWindowListener(new WindowAdapter() {
public void windowClosing(WindowEvent e) {
dispose();
public void windowClosed(WindowEvent e) {
System.exit(0);
});Now, at least you can click on the little "X" on the upper right-hand corner of the Frame (at least in Win95/98/NT) environments) to cause your application to receive a window closing event. It is also possible to add a menu with an option to exit; in that case I'd use an EventQueue to cause a window closing event to be sent.

Static main(String args[]) access modifiers possible

main(String args[]) can have private as access modifier?

hi,
yes of course this is possible, because main is at the end only a method like each other. But you have to know, that you cannot use this method from outside. so no application will start, if the main() of the Main-class has a private as modifier.
You can use this class as an applet.
So be carefull doing this
regards

What happens when main(String[] args)

Hi techies,
I am trying to understand, "how the JVM invokes public static void main(String[] args)".
I had gone through JVM tutorial, and found that there are 2 different types of classloaders 1. bootstrap loader 2. user defined class loader.
I had gone through java api for class loader class. i did not find the method main(String[] args).
I am having confusion whether bootstrap loader calls main(String[] args)(or) userdefined classloader calles main(String[] args).
Why do we have to give String[] args as argument to main??
can you guys guide to understand this .
regards,
Krishna

Hi techies,
I am trying to understand, "how the JVM
invokes public static void main(String[] args)".
I had gone through JVM tutorial, and
found that there are 2 different types of
classloaders 1. bootstrap loader 2. user defined
class loader.
I had gone through java api for class
loader class. i did not find the method
main(String[] args).
I am having confusion whether bootstrap
loader calls main(String[] args)(or) userdefined
classloader calles main(String[] args).
Why do we have to give String[] args as
argument to main??
can you guys guide to understand this
regards,
Krishnawell, you won't find main(String[] args) in the docs for ClassLoader, since it's not a method on ClassLoader!
the bootstrap loader loads, verifies, prepares and initializes the class for which you pass the fully.qualified.name into the JVM, and looks for the static method on that class called 'main' which has a single argument, an array of Strings, and no return type ( I'm not 100% sure of that last part actually - I'll try it ) and begins a thread of execution on that method. from there, your program is run
we have to give an array of Strings in order to pass arguments into the program
it can't be a user-defined classloader that does this, because most of the time there won't be a user-defined classloader!

Standard input - main(String[] args) -- how?

In eclipse, how do I feed the main with values?
public static void main(String[] args) {
     System.out.println("Enter value: ");
     args =
    if (args.length != 1) {
...Thanks,
george

You mean basically command line arguments?
Look in the run menu. For every run event there's a form that specifies how to invoke the program. Somewhere in there you can specifiy the command line arguments. I can't recall right now but I can look it up if you can't find it.
Standard input is a completely different thing from command line arguments. Standard input, from the java point of view, is what comes in on System.in.

Why should i write main(String[] args)?

Hello,
I don't know why i am obliged in the main function of the class to send an array of string as parameters?
Thanks

Actually I think this is a great question!
main()
Any java application has to have a main method to do anything, but as you gain further into java and object orientated practices you will start writing lots of stuff that do NOT have a main method but are *.java and *.class files, neither will they do anything unless they are attached to a program with a main. Andthen as you get DEEPER still you might even have 'private synchronised void main(String[] args)' - though not too often.
Why an array of strings? Hmmm ...pauses ...thinks, there is no obvious reason, other than the fact that if you're going to pass parameters from 'java ...' command line to the application. This is, logically, the most potentially useful data type to have. There may well be other reasons for this, so the answer to the second part of your question is a logical deduction - it may not be right.

Main(String args[]) cannot be overload ? True or False

main(String args[]) cannot be overload ? True or False

Although you can do this....
public class a
    public static void main(String[] args)
        System.out.println("Hello World - a!");
public class b extends a
    public static void main(String[] args)
        System.out.println("Hello World - b!");
}(I'm not sure if that is relevant to your actual question.... but then again your question isn't spot on either! ;-)

Private static void main(String args[]) ... ?????

even if the main method is declared
private static void main(String args[])
//Code Here
it works fine..can somebody explain the reason behind?

hi rkippen
so u mean, since the JVM is calling the main it can
break all kinds of access specifiers to any class..am
i right?
This could be the case.
However, I tried using 1.4.1 with no development environment (just java.exe) and it gave the following error:
C:\>java MainTest
Main method not public.
So, your development environment might be the cause (e.g. what you use to program your code (e.x. JBuilder)), or the Java version.
Your development environment might be using reflection to get around Java accessibility checks. For debugging purposes, it is possible to use reflection to obtain refereneces to private methods/fields and invoke those methods and access those fields.
E.g.
Vector v = new Vector();
Field f = v.getClass().getDeclaredField("elementCount");
f.setAccessible(true); // this is the trick
f.set(v, new Integer(100));In the presence of a security manager, the setAccessible method will fire a "suppressAccessChecks" permission check.

Detect shift key in main(String[] args)

Hello all,
I am trying to detect if the user is pressing the shift key while my application is starting.
I tried
Toolkit.getDefaultToolkit().addAWTEventListener(globalEventListener, AWTEvent.KEY_EVENT_MASK);
But it doesn't seem to detect the shift key.
Any help on this?

I don't think you can do it Java; the problem is that GUI never gets an event
from which it can check the modifiers. I thought maybe you could get around
this with a Robot creating an event, but with the Shift key down, it never gets
delivered (though with no keys pressed, it does). You might be able to do this
using some JNI, though, but of course, that will be platform specific.
Why not just make the admin functionality available off a menu item or
something like that?

How to pass (String[] args), as a parameter.

* StringMethods
* @author (Robin)
* @version (1)
public class StringMethods
    private String[] args;
    private int numArgs;
    public void printIt(String[] args)
        System.out.print ("You have entered " + numArgs + " Strings");
            if (numArgs > 10)
                int temp = numArgs;
                System.out.print (", the last " + (numArgs - 10) + " will not be processed.");
                numArgs = temp;
        for (int i = 0; (i < numArgs) && (i < 10); i++)
        System.out.println ("[" + i + "] : " + args[i] + " is " + (args).length() + " characters long.");
public void getArgs(String[] input)
numArgs = input.length;
arraycopy(input, 0, args, 0, numArgs-1);
arraycopy(input, 0, args, 0, numArgs-1)This is my method class for my main method StringArray.
What I want to know is how to pass command line arguments, i.e. {"example1", "example2", "example3"} from my main method to my method class?

* StringMethods
* @author (Robin)
* @version (1)
public class ArgMethods
private String[] args = new String[0];
private int numArgs;
public String[] input = {"null", "null"};
// constructor that takes string args to be manipulated by methods
public ArgMethods(String[] input)
numArgs = input.length;
String[] args = new String[numArgs];
System.arraycopy(input, 0, args, 0, input.length);
public void printIt()
if (args.length > 0)
System.out.print ("You have entered " + numArgs + " Strings");
if (numArgs > 10)
int temp = numArgs;
System.out.print (", the last " + (numArgs - 10) + " will not be processed.");
numArgs = temp;
for (int i = 0; (i < numArgs) && (i < 10); i++)
System.out.println ("[" + i + "] : " + args[i] + " is " + (args).length() + " characters long.");
else
System.out.println ("You must enter at least one String at the command line.");
public void reverseIt()
if(args.length > 0)
int next = args.length-1;
for (int index = 0; index < args.length/2; index++)
String temp = args[index];
args[index] = args[next];
args[next] = temp;
next--;
* StringArray
* @author (Robin)
* @version (1)
public class stringArray
public static void main (String[] args)
int numArgs = args.length;
// correct for output if strings less than 10 and greater than 0
if (numArgs > 0)
System.out.print ("You have entered " + numArgs + " Strings");
if (numArgs > 10)
int temp = numArgs;
System.out.print (", the last " + (numArgs - 10) + " will not be processed.");
numArgs = temp;
for (int i = 0; (i < numArgs) && (i < 10); i++)
System.out.println ("[" + i + "] : " + args[i] + " is " + (args[i]).length() + " characters long.");
else
System.out.println ("You must enter at least one String at the command line.");
ArgMethods test = new ArgMethods( args );
// test.printIt();
// test.reverseIt();
// test.printIt();
*almost finished -- posting this up here so i can get it at home ((only accessible e-mail down plus can't find pen drive))..

Newbie to java(String[] args and String args[])

what is the difference is it just that the arguments/parameters givven with the program are assigned to String (String[] args) or that they are assigned to args (String args[]) and why do you have to define the args part if yo don't assign something to it..
might seem a dumb question, but hey, when i say i'm new at it i mean new!!!!

Dear newbie,
when one issues a command on dos prompt like
c:\>java YourProgramName , then how would the program know where to begin executing the code. To make it standard and simple java.exe locates the method named main which has a particular signature that is , you can only define one main method as
public static void main(String [] args){
// remaining code
1. public [Access Specifier ]This method has to be declared public because it will be accessed from outside the class by jvm.
2. It is declared static because you are not allocating any memory to this class by the new operator for the main method to be executed by jvm. Only static methods and variables can be accessed without the need to create object of the class where they are defined. So when you say static, the jvm automatically allots memory to this method, accesses it and starts to execute the code declared in main method. All methods and variables (which are not defined static) have to be accessed by creating objects of the class in which they are defined.
So if you have a class which has static main(String args[]) and another method int multiply() which is not static, then to use it in main you will have to create an object of the class where it is defined.
public class yourClass{
public int multiply(int a, int b){
   return a*b;
   public static void main(String args){
       System.out.println("Product of 5*6 = "+ 5*6 );
        // you can not say
        // System.out.println(multiply(4,5)); because multiply is not a static method and main can not access it directly so what you have to do is
     yourClass yourObject=new yourClass();
   System.out.println("Product is "+ yourObject.multiply(4,5));
} 3 Return type is void because this method does not return any value.
4. (String [] args) is equivalent to (String args []). Java has provided this functionality and what one is doing here by either statement is making a String array named args. "[]" before or after args just implies that the Variable args is of Type String Array.
5. The Use of this signature for main method is that you can pass command line arguments which then your code can manipulate to give result, ie say if your main sums up two numbers
//when hard coded
public static void main(String args){
System.out.println("Product of 5*6 = "+ 5*6 );
//Using command line arguments
public static void main(String args){
int a=Integer.parseInt(args[0]);
int b=Integer.parseInt(args[1]);
System.out.println("Product Of "+a+"*"+b+" ="+ a*b);
}So you see the difference ? in case 1. you have to recompile the code to find the product of say 8 and 7 whereas in command line argument you just say
c:\>java findProduct 7 8 // for product of 7 and 8
c:\>java findProduct 15 8 // for product of 15 and 8 ...
ie the classs need not be recompiled....
hope that helps you
best wishes and regards,
gaurav

Using a the string args to input command line numerical input

Hello,
I am trying to allow the user to input a number from the command line to set up the number of elements in an array. I am having trouble converting the string info to int. Could someone post an example of how this is possible? Or a way to read a string for a numerical value?
Example:
nameofprogram 105 (so this means they want a 105 element array create when the program starts)
Message was edited by:
2disbetter

try this
public class Test {
     public static void main(String[] args)
          final int i = Integer.parseInt( args[0]);
          int[] a = new int;
          System.out.println(a.length);
if you have to have variable in size part of the array object creator it needs to be final.

Main(String[] args) ...

Similar Messages

Maybe you are looking for