Using the word super to access superclass fields - How?

Question

I just read that it is possible to access public members of the superclass by using the keyword super. To test it, I tried the following but it does not work. What is the proper way?
public class A {
     public String color = "Black";     
public class B extends A{
     public String color = "White";
public class test {
     public static void main(String[] args) {
          A x= new A();
          B y= new B();
          System.out.println(x.color); //returns Black
          System.out.println(y.color); //returns White
          System.out.println(y.super.color); ERROR!!!
}

Answer 1

Just trying to make this readable
package forums;
class Superclass {
  public String name = "Superclass";
class Subclass extends Superclass {
  public String name = "Subclass";
public class TheSuperSubClassTesterator {
  public static void main(String[] args) {
    Superclass superclass = new Superclass();
    System.out.println(superclass.name); //returns Superclass
    Subclass subclass = new Subclass();
    System.out.println(subclass.name);  //returns Subclass
    System.out.println(subclass.super.name); //ERROR!!!
compile
---------- build ----------
C:\Java\home\src\krc\tools>"C:\Program Files\Java\jdk1.6.0_12\bin\javac.exe" -Xlint -d C:\Java\home\classes -cp c:\java\home\src;.;C:\Java\home\classes C:\Java\home\src\krc\tools\TheSuperSubClassTesterator.java
C:\Java\home\src\krc\tools\TheSuperSubClassTesterator.java:20: cannot find symbol
symbol  : class subclass
location: class forums.TheSuperSubClassTesterator
    System.out.println(subclass.super.name); //ERROR!!!
                       ^
1 error
Output completed (1 sec consumed)