String[] args == String args[] ?

Why are there two ways of declarating an array?
e.g.: int[] bla;
or int bla[];
What's the difference? And what's Suns recommendation?
e.g.: public static void main(String[] args)
or public static void main(String args[])
I always made it the first way...

Why are there two ways of declarating an array?
e.g.: int[] bla;
or int bla[];
What's the difference? And what's Suns
recommendation?Apart from being textually different, nothing. They generate the same byte codes.
Sun does not have a recommendation as far as I know. My recommendation is to use which ever you think makes your code easier to read.

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();
%>

Req help in conversion of string to array of args

Hi,
i need help in converting a string into array of args just like command line args
i have a string s1 = 12 13 56 etc
now i want to copy that into an array a[]
a[0] = 12
a[1] = 13 etc
Thanks for help

U can use String tokenizer.Why prefer that over split?Yah Bobby u can use split too..
J2SE 1.4 added the split() method to simplify the task of breaking a string into substrings, or tokens.
Thanks BigDaddy.. I realized after you told.
try this too..
String str = "Your string";
String[] arr = str.split (" ");
for (int i=0; i < arr.length; i++)
System.out.println (arr);

Need of String args[] in main()

Hai all, i had a small doubt regarding the usage of String args[] in main function. When we are not passing any parameters to the program also, we have to give this String args[]. why should we do so. I mean when we pass some arguements to the program then there is the usage of String args[] but when we are not passing also, we have to give String args[] in the main function. why is it made mandatory?

There is only one method signature that the JVM will call. This is part of the Java specification.
The String[] args parameter represents the command line arguments. If there were none, the array will be empty.
Think about it: imagine there were two versions, one with and one without the String[] args parameter, and your application doesn't expect any command line arguments, so it implements only the version with no parameter at all. What would happen if the user decides to provide some command line arguments just for fun?

What is the use of passing String[] args in main() method?

what is the use of passing String[] args in main() method?
Is there any specific use for this ?

actually my sir asked me the same question & I gave
the same reply as given by you........but he further
asked what is the use of this also??
ie accepting cmd line args at runtime??is there any
specific purpose ??Apart from the one you just mentioned? No

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))..

Error 1 occurred at Scan From String (arg 1)

I am communicating with the AT Trainer 5000 via serial port to USB converter.
I installed all drivers of converter successfully and it is shown in MAX as com port. I also installed VISA software.
I am using labview 2013 version.
When I connect the hardware to PC via usb to serial converter, try to run the program for seeing of real time graphs, I am receiving this error Error 1 occurred at Scan From String (arg 1) whereas there is no waveform shown in waveform chart.
Also read buffer didn't show an incoming data at read buffer.
Other details for this error is
Possible reason(s):
LabVIEW: An input parameter is invalid. For example if the input is a path, the path might contain a character not allowed by the OS such as ? or @.
Also pics of error is:
Program is also attached. Please checked the also.
Expert please help me how to get rid of this error.
Attachments:
Project.vi ‏23 KB

Since you are using the termination character, you should not use the Bytes At Port property node. Just set the number of bytes to read for the VISA Read to something really high (larger than what you expect in a single transmission). The VISA Read will stop reading once the termination character is reached.
Now, what is the format of the data coming from the instrument? I see you are converting an ASCII Hex string into a number while trying to also turn it into a double. That is likely why you are getting your error.
There are only two ways to tell somebody thanks: Kudos and Marked Solutions

Is it possible to pass a string representing a class name in java as an arg

Hi, this is probably a bit of a stupid question, but one that has me quite confused all the same!
Is it possible to pass a string or class name etc. representing a type of class in java, to a method so that instead of having to redefine a method with say the following args:
public SolarPanels[] bestPVPanels(int budget, int percent) {
HashMap<Integer, SolarPanels> panelsMap = new HashMap<Integer, SolarPanels>();
}As at present needing to create a methods bestWindTurbine() and many others exactly the same, but for the type, I would instead like to be able to create a method like:
public Object[] bestRenewable(int budget, int percent, String aClassName) {
HashMap<Integer, aClassName > renewableMap = new HashMap<Integer, aClassName >();
}But cant sus how to do this as passing a String is off course going to cause problems unless its it possible to cast a string to a class name, any help or advance would be much appreciated.
Thanks in advance
Pat Nevin

pNev wrote:
But cant sus how to do this as passing a String is off course going to cause problems unless its it possible to cast a string to a class nameIt's not. You can do things like
Class.forName(classNameInString);But that will only return a Class object.
And as Java generics are erased at runtime, passing the class name to the method is useless as it is too late to use it anyways.
Ah, Peter's way is what I was thinking of. Too early in the morning.
Edited by: Kayaman on 23.6.2010 10:02

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

Please what does 'String[] args' in main is set to in a normal no parameter

Please,
what does 'String[] args' in main is set to in a normal no parameters static main function??
I have a frame with a static main function in it, in the main i want to do one thing if the function got parameters and another if it didnt.
if i check for (if (args[0] == "Load") i get an exception
and args is not null, i know, how do i know if the array has parameters??
p.s
I new to java

Yes, like jsalonen said.
And args[0]=="Load" won't work, you have to use
args[0].equals("Load");

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.

Help:String arg[ ] problems

Hi:
I am having problems with my encryption program and I want to check that the contents at the command line a being read properly.
ex: java Monoalphabet -k feather in.txt out.txt
-k(indicates encryption)
feather(keyword)
How do I check to make sure that the input in being read into String args[]properly?
Thanks.

Print the args out using System.out.println() in a loop at the beginning of your main method.

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.

Please help, urgently Error 1 occurred at Scan From String (arg 1).

I am communicating with the AT Trainer 5000 via serial port to USB converter.
I installed all drivers of converter successfully and it is shown in MAX as com port. I also installed VISA software.
I am using labview 2013 version.
When I connect the hardware to PC via usb to serial converter, try to run the program for seeing of real time graphs, I am receiving this error Error 1 occurred at Scan From String (arg 1) whereas there is no waveform shown in waveform chart.
Also read buffer didn't show an incoming data at read buffer.
Other details for this error is
Possible reason(s):
LabVIEW: An input parameter is invalid. For example if the input is a path, the path might contain a character not allowed by the OS such as ? or @.
Also pics of error is:
Program is also attached. Please checked the also.
Expert please help me how to get rid of this error.
Attachments:
Project.vi ‏23 KB

duplicate post http://forums.ni.com/t5/Instrument-Control-GPIB-Serial/Error-1-occurred-at-Scan-From-String-arg-1/td...

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.

String[] args == String args[] ?

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