static non-static reference confusion

Question

UPDATE: I made some changes based on some answers I got and it worked. The correct code is-

public class Tsett {
        static HashSet<String> names=new HashSet<String>();
        static String name="";
        public static void main(String[] args) {
        Scanner input=new Scanner(System.in);

        while(2>1){

    String s=input.next();
    char choice=s.charAt(0);

    switch(choice){
    case '1': addname(); break;
    case '2': removename(); break;
    case '3': displayname(); break;
    case '4': System.exit(0); break;
    default:System.out.println("Invalid Choice Entered.");
    }
        }



}
public static void addname(){
Scanner b=new Scanner(System.in);       
System.out.println("Enter a name to be added.");
name=b.next();
names.add(name);    
}
public static void removename(){
    Scanner b=new Scanner(System.in);   
    System.out.println("Enter a name to be remove.");
    name=b.next();
    names.remove(name);
}

public static void displayname(){
System.out.println("The names are as follows-");
for(String newname:names)
    System.out.println(newname);
    }

}

Using the keyword 'static' solved the problem.

I wrote the following code for my menu driven program on 'names'.

package tut4;

import java.util.HashSet;
import java.util.Scanner;

public class Tsett {

    HashSet<String> names = new HashSet<String>();

    public static void main(String[] args) {

        while (2 > 1) {

            Scanner input = new Scanner(System.in);
            String s = input.next();
            char choice = s.charAt(0);

            switch (choice) {
                case '1':
                    addname();
                    break;
                case '2':
                    removename();
                    break;
                case '3':
                    displayname();
                    break;
                case '4':
                    System.exit(0);
                    break;
                default:
                    System.out.println("Invalid Choice Entered.");
            }
        }

    }

    public static void addname() {
        String name = "";
        Scanner b = new Scanner(System.in);
        System.out.println("Enter a name to be added.");
        name = b.next();
        names.add(name);
    }

    public static void removename() {
        String name = "";
        Scanner b = new Scanner(System.in);
        System.out.println("Enter a name to be remove.");
        name = b.next();
        names.remove(name);
    }

    public static void displayname() {
        System.out.println("The names are as follows-");
        for (String newname : names) {
            System.out.println(newname);
        }
    }

}

The problem arising is that

cannot make static reference to non-static field "names"

in the addname(), removename(), displayname() functions.

Can anyone explain to me why this error is occurring and how to fix it?

Answer 1

Get use to not relying on static references. They have there place and uses, but to many people become reliant on them. You should also get use to using your main method to only initialise the program to such a state that you can hand off operation to the main/core class of the application.

Instead, I would get rid of you static modifiers on your methods and create an method that could "run" the core logic of your class.

In you main method, I would create an instance of the Tsett class and call this method. This way you gain access to the instance field names and methods

Answer 2

Change the declaration of names to

static HashSet<String> names = new HashSet<String>();

Because the methods are static, the variables that they reference must be static as well.
This is because a non-static variable must exist in a particular instance and can be different between instances, while static methods can not. If they are asked to reference a variable that is non-static, there is no way to know which instance of the class to use.

Answer 3

You should create an instance of the class those methods are in.

String names;
Class class = new Class();  //Creating Instance of "Class" called "class".
names = class.addname(); //Access a method in "Class" using instance name "class" then method name and assign the value it returns to "names".

Than use names where ever you want to be displayed

  Switch(choice){
    case '1': names; break;

Answer 4

To fix your issue, you need to make the field 'names' static. The error you're getting is as a result of trying to call a non-static field in static context.

Try

private static HashSet<String> names = new HashSet<String>();

That will fix your problem.

Answer 5

Well, here are a few suggestions (you asked for them in your last comment..)

1. Change while (2 > 1) to while(true).

2. Put Scanner input = new Scanner(System.in); at class level.

3. Put declarations for : String s = input.next(); char choice = s.charAt(0); outside the loop.

4. No need for Scanner b = new Scanner(System.in); in all your methods.