Help. Going crazy with subclass.

Question

I am extending class Client with myClient. Why do I get a symbol not found in the myClient() constructor?
// File: Client.java
package testproj;
public class Client {
    public Client(String conn){
        String xyz=conn;
// File: myClient.java
package testproj;
public class myClient extends Client {
// Gives error on the myClient Constructor as follows:
// C:\App\testproj\src\testproj\myClient.java:5: cannot find symbol
// symbol  : constructor Client() location: class testproj.Client
    public myClient(String conn){  
        int x=1;
}

Answer 1

Typically when you extend a class it is because you do want to initialize, and probably use, what is in that class. This rings doubly true when you created the parent class specifically for extending it in this project.
For example, why on earth would you override foo with the exact same method? I realize both are empty and you probably intend there to be different code in each, but as it currently stands, there is no reason to override foo, the foo in MyClient is just simply not needed.
But in my opinion your real problem here is you haven't figured out what constructors and methods are for. Constructors are for initializing a class, methods are for java code. Your example falls into some wierd gray area the myVar=xyz definitely fits and belongs in the constructor, but adding the additional text should probably be handled by a method since it appears that getting the different strings presumably for output is the crux of what you are trying to do.
The reason you don't want to initialize is because you think " take care" will somehow get added in if you run it the way you have it, which is exactly why this sort of stuff should not be in constructors. But rest assured, for as you have it, the string will not contain "Loser take care" because as soon as you set it to that you turn around and reset it to "Loser terminate his a**". Any other concerns you have about initialization the parent are equally unfounded in the code you provided.
Plus you don't seem to be aware that there can be more than one constructor. If you just added the following to class Client just above your current constructor;
public Client(){
} then you can remove the (xyz) in your super call and not have to worry about anything.
JSG