Java Error- cannot find symbol- Variable is

Question

I tried to make a "shortcut" to the print, println, and the scanner. So I created a different class, and that's what it had in it:

import java.util.*;
public class ShortCutClass {
    Scanner input = new Scanner(System.in);
    public void p (String text){
        System.out.print (text);
    }
    public void pl (String text){
        System.out.println (text);
    }
    public String is (){
        String a= input.nextLine();
        return a;
    }
    public double id (){
        double b= input.nextDouble();
        return b;
    }
    public int ii (){
        int c= input.nextInt();
        return c;
    }
}

So, I tried to check if it works- using this:

import java.util.*;
class ShortcutTest{
    Scanner input = new Scanner(System.in);
    public static void main(String args[]){
        ShortCutClass scc = new ShortCutClass();
        scc.pl("So ummmm... How ya doin?");
        scc.p("Connected1 ");
        scc.p("Connected2 ");
        String a=scc.is;
        double b=scc.id;
        int c=scc.ii;
        scc.pl (a);
        scc.pl (b);
        scc.pl (c);        
    }
}

and it gives me the following Error: "cannot find symbol- variable is

What am I doing wrong? How can I fix this?

*** Edit:

I fixed it, and ran into another problem... I can not print out b or c because the print I wrote requires a String, and b and c can only be double and int, so that won't work. is there any way of changing the print (scc.p and scc.pr) so it will be able to receive any kind of parameter?

() is missing. double b=scc.id; should be double b=scc.id();. — Eng.Fouad
– Eng.Fouad, Commented Nov 27, 2012 at 18:51
These comments should really be answers. A good comment would be something like this: "Why would you ever want to make your code this hard to read?" — Erick Robertson
– Erick Robertson, Commented Nov 27, 2012 at 18:53

Jon Skeet · Accepted Answer · 2012-11-27 18:51:01Z

3

Here:

String a=scc.is;
double b=scc.id;
int c=scc.ii;

... you're trying to refer to methods as if they were variables. You need to call the methods:

String a=scc.is();
double b=scc.id();
int c=scc.ii();

As a side-note, I would strongly discourage you from doing all of this to start with. Even if you want a convenience class, please give the methods sensible names - writeInt, readInt etc. Using abbreviations like this makes the code very hard to read.

answered Nov 27, 2012 at 18:51

Jon Skeet

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6 Comments

Rohit Jain Over a year ago

Actually it's 7377 more than that. ;)

user1857509 Over a year ago

Thank you for your answer! the thing is, i ran into another problem... please read the the Extra part i just wrote there... thanks in ahead!

Jon Skeet Over a year ago

@user1857509: Well it sounds like you want overloads for print to take double and float parameters...

user1857509 Over a year ago

ummm.... New to Java... Overloads? whats that and how do I do that? Thanks in ahead!

Jon Skeet Over a year ago

@user1857509: Did you try searching before asking? If you're new to Java, you should be learning from a book or well-structured tutorial - did you see what that had to say about overoading?

|

jco.owens · Accepted Answer · 2012-11-27 18:51:13Z

1

Thats because "is" is a method and you are attempting to access it as a public variable.

Try:

String a = scc.is();

answered Nov 27, 2012 at 18:51

jco.owens

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Comments

AlexWien · Accepted Answer · 2012-11-27 18:52:11Z

0

In line

String a=scc.is;
double b=scc.id;

the () are missing:

correct woul dbe:

String a=scc.is();
double b=scc.id();

Further Tipp: Please remove the short cuts, that code is less readable

answered Nov 27, 2012 at 18:52

AlexWien

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