Java method showing error

Question

2.00/5 (1 vote)

See more:

I'm getting error for the following Java code. I tried but unable to correct. Any help is appreciated. I started learning Java just a week ago.

C#

public class class1 {

    public static void main(String[] args) {


        int assigner()

        {
                int t = 7;
                return t;

        }


        int nom = assigner();
        System.out.println(nom) ;


    }

}

Posted 22-Sep-14 5:21am

IAM_Dhruv

Add a Solution

Comments

[no name] 22-Sep-14 11:25am

Take your function declaration out of your main method.

Sergey Alexandrovich Kryukov 22-Sep-14 11:54am

Do what Wes Aday suggested. Move assigner method out of main method.
When asking question, provide exact information on the error, especially — show in which line.
—SA

IAM_Dhruv 22-Sep-14 12:00pm

Thanks. I also had to make the function declaration static. It worked. I will keep in mind the suggestion to give extra information about the error.

Sergey Alexandrovich Kryukov 22-Sep-14 16:10pm

You don't have to make it static, but, in this very case, static is the most appropriate, because the function does not use "this" pointer (does not access any instance members). Code analyzers would mark this instance method as a minor performance leak. Do you understand that?

I hope you understand that this method is very redundant as it does next to nothing. Also, the name "assigner" would be misleading, as it does not assign anything, only its call may or may not assign. In practice, it would be much better to replace 7 (immediate constant) with explicitly declared constant.

—SA

IAM_Dhruv 26-Sep-14 23:05pm

Got the point. I'm going through the concepts again. Thank you very much.

Sergey Alexandrovich Kryukov 27-Sep-14 0:24am

You are welcome.
—SA

Richard MacCutchan 22-Sep-14 12:54pm

I would suggest working through the tutorials at http://download.oracle.com/javase/tutorial/index.html.

Sergey Alexandrovich Kryukov 22-Sep-14 16:10pm

Best advice so far...
—SA

IAM_Dhruv 26-Sep-14 23:05pm

Yes I'm going through the official Java documentation. Thanks. Got the point.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

TorstenH. · Accepted Answer · 2014-09-22T22:42:00

As the solution is already given in the comments, here a clean code for that.
That might make things a little easier to understand.

Java

public class Project1 { // Class names always start with upper case! be careful with the wording, using keywords can ruin your day!
 
  public static void main(String[] args) {
    // create instance from class to break out of static context
    // 0 args constructor is always given, even if not written out
    Project1 project = new Project1(); 

    // call function from that class
    int nom = project.assigner(); 
    System.out.println("nom is: " + nom);
  }

  // private method, can only be used in this class
  private int assigner() 
  {
      int t = 7;
      return t;
  }
}

have fun!