Solved

Why isn't my array printing w/in function?

Posted on 2008-10-06
10
186 Views
Last Modified: 2012-05-05
Why isn't my array printing w/in the prinArray function?  It prints if in main, but I need it to reside in a function.  Please help.  Below is my code.


public class yamotdProtocol {
 

	static String[] message = new String[5];

	static Scanner is = null;

	static String m = null;

	private static String line;

	private static String msgget = "MSGGET";

	

	public static void main(String args[]) throws IOException {

	

	try {

		

		is = new Scanner(new FileInputStream("C:\\messages.txt"));

	}

	

	catch(FileNotFoundException e){}

	

	for(int i=0; i<message.length; i++)

	{

		line = is.nextLine();

		message[i] = line;

	}

	

	for(int n =0; n<message.length; n++)

		System.out.println(message[n]);

	}

	

	public void printArray(){

		

		for(int n =0; n<message.length; n++)

			System.out.println(message[n]);

	}

}

Open in new window

0
Comment
Question by:deliachang
  • 4
  • 4
  • 2
10 Comments
 
LVL 24

Expert Comment

by:sciuriware
ID: 22651345
Because 'printArray() is a separate method, that's never called.

;JOOP!
0
 
LVL 8

Accepted Solution

by:
mnrz earned 500 total points
ID: 22651401
it will, but you have to call that method
check this:


public class yamotdProtocol {

 

        static String[] message = new String[5];

        static Scanner is = null;

        static String m = null;

        private static String line;

        private static String msgget = "MSGGET";

        

        public static void main(String args[]) throws IOException {

        

        try {

                

                is = new Scanner(new FileInputStream("C:\\messages.txt"));

        }

        

        catch(FileNotFoundException e){}

        

        for(int i=0; i<message.length; i++)

        {

                line = is.nextLine();

                message[i] = line;

        }

        

        /*for(int n =0; n<message.length; n++)

                System.out.println(message[n]);

         */

        yamotdProtocol instance = new yamotdProtocol();

        instance.printArray()  ;

        }

        

        public void printArray(){

                

                for(int n =0; n<message.length; n++)

                        System.out.println(message[n]);

        }

}

Open in new window

0
 
LVL 24

Expert Comment

by:sciuriware
ID: 22651409
Change your code to something like this:

public class yamotdProtocol
{
   static String[] message = new String[5];
   static Scanner is = null;
   static String m = null;
   private static String line;
   private static String msgget = "MSGGET";

   public static void main(String args[]) throws IOException
   {
      try
      {
         is = new Scanner(new FileInputStream("C:\\messages.txt"));
      }
      catch(FileNotFoundException e){ /* No action */ }

      for(int i = 0; i < message.length; i++)
      {
         line = is.nextLine();
         message[i] = line;
      }

      for(int n = 0; n < message.length; n++)
         System.out.println(message[n]);
     
      printArray();  // This makes sense!!!!
   }
   
   public static void printArray()
   {

      for(int n = 0; n < message.length; n++)
         System.out.println(message[n]);
   }

}

;JOOP!
0
 
LVL 8

Expert Comment

by:mnrz
ID: 22651423
or you can declare that method as static:

also please note in Java we don't say function, method is used instead

public static void printArray(){

                

                for(int n =0; n<message.length; n++)

                        System.out.println(message[n]);

        }

Open in new window

0
 
LVL 24

Expert Comment

by:sciuriware
ID: 22651705
I disagree with mnrz: printArray() was embedded inside 'main()'
by bad positioning of { } and it was not called.

;JOOP!
0
Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

 

Author Comment

by:deliachang
ID: 22651749
mnrz, your solution work.  I do have another question and not sure if it can work.  Can I call this function from another class within the same package?

If I declare the an instance of the protocol class in another class, I get nothing but nulls.  Why is this?
public String printArray(String userInput){

		

		String theOutput = userInput;

				

		if (theOutput.equals("MSGGET"))

			theOutput = "It works." + message[0];

		

		return theOutput;

	}

Open in new window

0
 

Author Closing Comment

by:deliachang
ID: 31503472
Thank you so much.  I actually got it working - even calling it from another class.
0
 
LVL 24

Expert Comment

by:sciuriware
ID: 22651948
Thanks deliachang.

;JOOP!
0
 
LVL 8

Expert Comment

by:mnrz
ID: 22652042
sciuriware
no the closing bracket was in right place but the indent was not correct.

deliachang
since the method is declared as public you can call this method from another class but the process in which your message array is populating is placed in main() method and this method is static, so you have to call it before printArray()

check this example out:
 
0
 
LVL 8

Expert Comment

by:mnrz
ID: 22652117
sorry this is example:

public class YourClassName {

 

        String[] message = new String[5];

        Scanner is = null;

        String m = null;

        private String line;

        private String msgget = "MSGGET";

        

        public void populateArray() throws IOException {

        

        try {

                

                is = new Scanner(new FileInputStream("C:\\messages.txt"));

        }

        

        catch(FileNotFoundException e){}

        

        for(int i=0; i<message.length; i++)

        {

                line = is.nextLine();

                message[i] = line;

        }

      }

        

        public void printArray(){

                

                for(int n =0; n<message.length; n++)

                        System.out.println(message[n]);

        }
 

public String printArray(String userInput){

                

                String theOutput = userInput;

                                

                if (theOutput.equals("MSGGET"))

                        theOutput = "It works." + message[0];

                

                return theOutput;

        }

}
 

// in other class

public class AnotherClass {
 
 

        public static void main(String[] a) {

             YourClassName instance = new YourClassName();

             instance.populateArray();

             instance.printArray();

              instance.printArray("MSGGET");

         }

}

Open in new window

0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

For customizing the look of your lightweight component and making it look lucid like it was made of glass. Or: how to make your component more Apple-ish ;) This tip assumes your component to be of rectangular shape and completely opaque. (COD…
Java Flight Recorder and Java Mission Control together create a complete tool chain to continuously collect low level and detailed runtime information enabling after-the-fact incident analysis. Java Flight Recorder is a profiling and event collectio…
Viewers learn about the “for” loop and how it works in Java. By comparing it to the while loop learned before, viewers can make the transition easily. You will learn about the formatting of the for loop as we write a program that prints even numbers…
Viewers learn how to read error messages and identify possible mistakes that could cause hours of frustration. Coding is as much about debugging your code as it is about writing it. Define Error Message: Line Numbers: Type of Error: Break Down…

863 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

20 Experts available now in Live!

Get 1:1 Help Now