Improper use of the super keyword?

Hi guys,
I'd appreciate some help on this... I'm extending JTable but i'm getting an error on the overloaded super call... Here;s the simple code:
import javax.swing.JTable;
import javax.swing.table.TableModel;
public class SpecialisedJTable
    public SpecialisedJTable(TableModel model)
        super(model);
    public SpecialisedJTable()
        super();
}And this is the error message i get:
ucl/webmodel/gui/SpecialisedJTable.java [12:1] Object() in java.lang.Object cannot be applied to (javax.swing.table.TableModel)
super(model);
^
1 error
Errors compiling SpecialisedJTable.
Any ideas why the error is being generated? Thanks...

You say you're extending JTable but I see no hint of that.
public class SpecialisedJTable extends JTable
    public SpecialisedJTable(TableModel model)
        super(model);
    public SpecialisedJTable()
        super();
}

Similar Messages

Warning on use of the 'subtable' keyword

I have previously given advice on this list about the use of the "subtable" keyword which could lead to kern pairs being unavailable. If you have used this keyword, please review the following.
Background.
Class pairs in a kern feature are compiled as a lookup with a single subtable. The representation of the class pairs for a subtable looks like a big Excel spreadsheet, with all the left side pairs as the titles of the rows, and all the right-side pairs as the titles of the columns, and the kern pair values in the fields. As a result, every left side pair you name is kerned against every right side pair you name. Obviously, most of the fields are 0; there are non-zero values only where you specified a class pair kern value. Less obviously, all glyphs not named in a right side class are automatically assigned to right side class 0. This means that every left side class in the subtable is kerned against every glyph in the font ( with most of the kern values being 0).
If you have a lot of left and right side class pairs, this subtable can get so large some of the offset values in the compiled lookup exceed the limit of 64K, and the lookup can't be built. The 'subtable' keyword is provided to help with this problem. It forces the class pair lookup to start a new subtable at the point where you put the 'subtable' keyword. If you have any degree of order in your class kern pair definitions, each of the subtables will then have fewer left and right classes, and the overall size of the lookup is smaller.
My advice on using this keyword is:
a) use it only if you have to because of a size oveflow problem, and
b) there is not much use in trying to be smart about where you put it; it is usually simplest to put the first keyword more or less in the middle of the list of the class kern pairs. If this does not shrink the kern lookup enough, then put another 'subtable' keyword in the middle of each half of the class kern pair list, and so on.
However, I omitted an important warning:
c) any left side class can be used only within a single subtable.
This is because each left side class in a subtable is by definition kerned against every right side class, and hence every glyph in the font because of class 0. It follows that a program looking for a kern pair will never look past the first subtable with a left side class containing the left side glyph. Moral: any left side class can be used only within a single subtable. If you put kern class pairs with that left side class in a subsequent subtable ( e.g; after a subtable break), those kern class pairs will never be seen by any program.
My thanks to Karsten Luecke, whose struggle with an exanple of this problem showed that my previous advice on this list was incomplete.

You say you're extending JTable but I see no hint of that.
public class SpecialisedJTable extends JTable
    public SpecialisedJTable(TableModel model)
        super(model);
    public SpecialisedJTable()
        super();
}

Inheriting and the "super" keyword

I found a case where "super" is not working as I understand it should...
Can someone explain me the underlying theory that makes JAVA behave this way??
public class Base {
     public void method1() {
          System.out.println("method1 from \"Base\" class. About to call method2.");
          method2();
     public void method2() {
          System.out.println("method2 from \"Base\" class. About to call method3.");
          method3();
     public void method3() {
          System.out.println("method3 from \"Base\" class.");
public class Extension extends Base {
     @Override
     public void method1(){
          System.out.println("method1 from \"Extension\" class. About to call super.method2.");
          super.method2();
     @Override
     public void method2(){
          System.out.println("method2 from \"Extension\" class. About to call method3.");
          method3();
     @Override
     public void method3(){
          System.out.println("method3 from \"Extension\" class.");
public class Main {
     public static void main(String args[]) {
          System.out.println("In \"The Java Programming Language,\n" +
                    "Fourth Edition By Ken Arnold, James Gosling, David Holmes\"\n"+
                    "chapter: 3.3: \"Inheriting and Redefining Members\""+
                    " says:");
          System.out.println("Using super is the only case in which the type of\n" +
                    "the reference governs selection of the method implementation\n" +
                    "to be used. An invocation of super.method always uses the\n" +
                    "implementation of method the superclass defines (or\n" +
                    "inherits). It does not use any overriding implementation of\n" +
                    "that method further down the class hierarchy.\n");
          System.out.println("But by running this code, i get:");
          System.out.println("------------------------------------------");
          Extension ext = new Extension();
          ext.method1();
          System.out.println("------------------------------------------");
          System.out.println("\nWHY??? I was expecting:\n"+
                    "method1 from \"Extension\" class. About to call super.method2.\n" +
                    "method2 from \"Base\" class. About to call method3.\n" +
                    "method3 from \"Base\" class.");
THANKS!!

IgnacioKriche wrote:
But I used "super", so this means to use the methods of the super class and I understand that a method called within another, is part of this last one. this is:
if inside method 'x', method 'y' is called, then 'y' is part of 'x' . Agree? And therefore since I am using super I'm saying use 'x' method from super class since 'y' is part of 'x' then super applies for 'y' as well so what is method from the extended class doing here? It's like we have a mix between the 'x' from super and 'y' from the extended. Something like super only applies for the first level not at a deeper level (a method called from another)No. Just because the base class method2() invokes method3() does NOT mean it will always invoke base's version of method3. If it did that, then polymorphism would definitely be broken.
You explicitly invoked super.method2(), so that told the compiler to explicitly use the base class version of that method. But then method2 in the base class invokes method3(). That ends up invoking the overridden version in the subclass, by design, because afterall the actual object whose methods are being executed is a subclass instance. If it didn't do that, then polymorphism would be broken for everyone else.
If you think you need the behavior you are looking for, you are just designing it wrong in the first place.

Inserting data into subtypes by making use of the super type

Dear All,
Let's say I have the following statements defining two object types and create two tables for them:
create type t1 as object(x number) not final;
create type t2 under t1(y number);
create table t1t of t1;
create table t2t of t2;
I know I can use insert into t2t values(t2(12,12)) to get a new row.
However, let's say I have run an insert into t1t values(t1(12)) and I want to make use of that object in the insert for t2t (something like
insert into t2t values(t1value,12)). Is this possible?
Jim

>
That works, but do we need to repeat all the data in t1's constructor? In the trivial example it is easy because t1 has a single argument but if you are meant to be typing everythinh again, you might as well do the full insert. Is there not something that will give you the value of the object you need (something like select value(e) from t1...)?
Another idea that I am going to try is various member methods so that if you know the ID of the superclass objects you get the information; still can not beleive there is no function that will give me the values of the superclass.
>
You are using the default constructor.
Just create your own constructors. Create several of them if you want: one that takes a T1 instance. Then you can provide that T1 instance statically or from a query and the constructor will access the values without you needing to specify them.
You could also create your own constructor so that if you only provide one value the constructor calls 'super' and creates some 'default' T1.x value.

Using the Super Word in a tutorial

ive created two classes:
public class Polygon
int sides;
double area;
public Polygon()
System.out.println("Inside no-arg constructor");
public String NumberOfSides()
System.out.println("The Number of Sides = " + sides);
return sides;
public String toString()
System.out.println("Inside Polygon toString")
String rep = "Sides" + sides + "area" + area;
return rep;
public Polygon(int s)
sides = s;
System.out.println("Inside One-Arg Constructor");
public class Triangle extends Polygon
int base;
int height;
public Triangle(int b, int h)
base = b;
height = h;
The tutorial i am using says, 'Add a constructor to Triangle that takes in two int's for base and height. The constructor needs to use super() to invoke the constructor in Polygon, passing in 3 for the number of sides. Print out the message "Inside Triangle Constructor"
as you can see i've created the triangle constructor which takes in the two int's but i am unsure how to use the super keyword to use the constructor within the polygon class to pass in 3 as the number of sides.
can someone help me in solving this problem.
Thanks
Richard.

if i was to change the order to that would it not
take the NumberOfSides() and the toString() methods
out of the no-argument constructor?You cannot put methods inside methods.
That makes no sense in Java.
i deliberately
put them there because i only want them for the
no-arg constructor. or must methods be seperate
entities from constructors?Yes.
And in fact, you have 2 more problems:
your "sides" and "area" variable are uninitialized
if people use your "no-arg" constructor,
so the compiler will complain.
You need to assign them some value in your constructor
(perhaps "slides=0; area=0;")

Shoukd the 'final' keyword should be used on all input parameters?

In my very limited experience as a java programmer have never seen this done any where. Except in the java.lang.Math class which has a final variable, of type double, called PI. As pi is not a value that should change during the execution of a program, though this is not passed in as a parameter.
However, on a new project, a co-worker is proposing the use of the 'final' keyword on all input parameters as a standard. He has about 5 years experience of maintaining java systems and is putting this suggestion forward as in the past people have overwritten values passed in as parameters causing confusion. Personally I don't see why this is such a major issue if one reads the code of the called programs.
I would be grateful if anyone has seen this done before, or has any views on the advantages or disadvantages of this.
Thanks in advance,
LS

Lord_Sidcup wrote:
In my very limited experience as a java programmer have never seen this done any where. Except in the java.lang.Math class which has a final variable, of type double, called PI. As pi is not a value that should change during the execution of a program, though this is not passed in as a parameter.
I'm sorry, but this point is irrelevant. There are lots of final fields (but not that many final parameters) in Java, BTW. Every field you declare in an interface is automatically final, for example.
However, on a new project, a co-worker is proposing the use of the 'final' keyword on all input parameters as a standard. He has about 5 years experience of maintaining java systems and is putting this suggestion forward as in the past people have overwritten values passed in as parameters causing confusion. Personally I don't see why this is such a major issue if one reads the code of the called programs.
IMHO, it's definately not a major issue. It is simply being restrictive, to be restrictive. The guy may have "five years experience", but he sounds lazy, to me. Besides, the first thing I would do in that case (if final were the default), would be something like this:
public void myMethod (final String a, final int b) {
String myA = a;
String myB = b;
}And now, what has he won by making it final? It simply added some up front work for some people, with no real added value. Also, if I had to retroactively change my code, I would change it as follows:
//Original Code
public void myMethod (String myA, int myB) {
String myA = a;
String myB = b;
// New code
public void myMethod (final String a, final int b) {
String myA = a;
String myB = b;
}That way, you wouldn't even have to worry about changing anything else about your method, as the variables actually used in the rest of the body will remain the same.
I would be grateful if anyone has seen this done before, or has any views on the advantages or disadvantages of this.The only thing declaring the parameter final does, is it prevents you from assigning a new value to the variable. The instance that parameter references (as long as it is not immutable to begin with) can still be manipulated. I.E. except in some very narrow, and specific, cases, no value added, simply a restriction that exists to be restrictive.
>
Thanks in advance,
LSEdit: And man, am I slow and long-winded.

Using the 'this' keyword

What is the use of the this keyword:
import java.util.*;
public abstract class LegalEntity implements HasAddress{
private String address;
private String area;
private String registrationDate;
private ArrayList accounts;
public LegalEntity()
     accounts = new ArrayList();
public void addAddress(String address){this.address = address;}
public String getAddress(){return address;}
public void addAccount(Account account)
     accounts.add(account);
public ArrayList returnAccounts()
     return accounts;
}

This can be used when there are lots of static functions and you want to be sure you are invoking the current objects methods/members.

Querstion regarding the "this" keyword

hi everyone,
i've doubt regarding the usage of this keyword in the inner classes & static methods. like the following:
class A {
    int val1 = 0;
    int val2 = 0;
    A() {
        B b = new B();
        C c = new C();
        for(int i=1; i < 11; i++) {
            b.add(i);
            c.add(i);
            System.out.println("Val 1 : " + val1);
            System.out.println("Val 2 : " + val2);
    public static void main(String args[]) {
        new A();
    class B {
        public void add(int val) {
            // the this keyword is used as static variable
            A.this.val1 += val;
    static class C {
        public void add(int val) {
            // the "this" keyword cannot be used
            //A.this.val2 += val;
}the code doesn't matter what its meant for, i wrote code to explain u my doubt with example.
so any explanation from ur side, i'm not clear with the this keyword usage.
thanx

I am in agreement with trinta, although I think your doubt arises from a lack of understanding of inner classes, not from the use of 'this'. You seem very adept in the use of 'this'.
Realise that there are 4 categories of nested classes:
1> Top-level or static nested classes
These are declared by the static keyword (as in class C in your example). They can be instantiated with or without an instance of the containing class. i.e. using the class names from your example.
C c = new A.C();
or
C c = new instanceOfA.C();
Since a static nested class does not require an instance of a class, there is no use of the 'this' keyword. There may not be any instance created, if if there was, which one since you don't have to specify which when creating it (see first example code above).
2> Member Inner Classes
As in class B in your example. The name should indicate that we think about these classes as members, or part of, the containing instance. Use these classes when it makes no sense for the inner class to exist outside the containing class. Usually these inner classes are designed to be helper classes and often hidden from view. When they aren't, the only way to create new instances of them is through an actual instance of the containing class. i.e. using the class names from your example,
B b = new instanceOfA.B();
Note that this code
B b = new A.B();
will yield an error. There must be an associated containing class.
For this reason, the use of 'this' is only appropriate for member classes.
3> Local Inner Classes
4> Anonymous Inner Classes
Not really relevant to this question.

My itunes 11 Home sharing is not working with my windows 7 PC and iPad it shows up on iPad and when I go to share the music app does loading for 3 mins and crashes I am using a virgin super hub router please please please help

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How to reference a class without using the new keyword

I need to access some information from a class using the getter, but I don't want to use the new keyword because it will erase the information in the class. How can I reference the class without using the new keyword.
ContactInfo c = new ContactInfo(); // problem is here because it erases the info in the class
c.getFirstName();

quedogf94 wrote:
I need to access some information from a class using the getter, but I don't want to use the new keyword because it will erase the information in the class. How can I reference the class without using the new keyword.
ContactInfo c = new ContactInfo(); // problem is here because it erases the info in the class
c.getFirstName();No.
Using new does not erase anything. There's nothing to erase. It's brand new. It creates a new instance, and whatever that constructor puts in there, is there. If you then change the contents of that instance, and you want to see them, you have to have maintained a reference to it somewhere, and access that instance's state through that reference.
As already stated, you seem to be confused between class and instance, at the very least.
Run this. Study the output carefully. Make sure you understand why you see what you do. Then, if you're still confused, try to rephrase your question in a way that makes some sense based on what you've observed.
(And not that accessing a class (static) member through a reference, like foo1.getNumFoos() is syntactically legal, but is bad form, since it looks like you're accessing an instance (non-static) member. I do it here just for demonstration purposes.)
public class Foo {
private static int numFoos; // class variable
private int x; // instance varaible
public Foo(int x) {
    this.x = x;
    numFoos++;
// class method
public static int getNumFoos() {
    return numFoos;
// instance method
public int getX() {
    return x;
public static void main (String[] args) {
    System.out.println ("Foo.numFoos is " + Foo.getNumFoos ());
    System.out.println ();
    Foo foo1 = new Foo(42);
    System.out.println ("Foo.numFoos is " + Foo.getNumFoos ());
    System.out.println ("foo1.numFoos is " + foo1.getNumFoos ());
    System.out.println ("foo1.x is " + foo1.getX ());
    System.out.println ();
    Foo foo2 = new Foo(666);
    System.out.println ("Foo.numFoos is " + Foo.getNumFoos ());
    System.out.println ("foo1.numFoos is " + foo1.getNumFoos ());
    System.out.println ("foo1.x is " + foo1.getX ());
    System.out.println ("foo2.numFoos is " + foo2.getNumFoos ());
    System.out.println ("foo2.x is " + foo2.getX ());
    System.out.println ();
}

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Using the LIKE keyword in a SQL query

Does anyone have an example using the Database Connectivity Execute Query vi and the LIKE keyword? I would like to do something like:
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It works fine if I say LIKE "Cyclic Voltammetry" (without the *) but if I put the * in there, I get no records.
I'm using Microsofts SQL Server Desktop Engine (comes with Access).
Thanks,
Paul

Paul-
Thank you for contacting National Instruments. I don't have an example program, but I did find some information for you.
LIKE - similar to.
Using the LIKE command, you can use the wildcard symbol %. To find names starting with J : names LIKE 'J%'; Ending in s ? name LIKE '%S'; You can use more than one % charactor too : names LIKE '%sam%'; Also, LIKE is not case sensitive.
What you have written, may work if you change the wildcard syntax from * to %.
-Erik

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Hi There,
Can you tell me the Operating System and Captivate version you are using?
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Regards,
Mayank

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Model: MacBookPro11,1
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System diagnostics
   2014-09-20 PluginProcess spin
   2014-09-20 PluginProcess spin
   2014-09-20 iTunes spin
   2014-09-22 PluginProcess spin
   2014-09-22 PluginProcess spin
   2014-09-25 PluginProcess spin
   2014-09-26 Google Chrome spin
   2014-09-26 Google Chrome spin
User diagnostics
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   2014-09-22 DashboardClient crash
Kernel messages
   Sep 26 17:21:23   IOPPF: Sent gpu-internal-plimit-notification last value 4 (rounded time weighted average 4)
   Sep 26 17:21:35   IOPPF: Sent cpu-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:21:35   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 3 (rounded time weighted average 3)
   Sep 26 17:21:35   IOPPF: Sent gpu-internal-plimit-notification last value 3 (rounded time weighted average 3)
   Sep 26 17:21:47   IOPPF: Sent cpu-plimit-notification last value 8 (rounded time weighted average 7)
   Sep 26 17:21:47   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 3 (rounded time weighted average 3)
   Sep 26 17:21:47   IOPPF: Sent gpu-internal-plimit-notification last value 3 (rounded time weighted average 3)
   Sep 26 17:22:01   IOPPF: Sent cpu-plimit-notification last value 5 (rounded time weighted average 7)
   Sep 26 17:22:01   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 2 (rounded time weighted average 3)
   Sep 26 17:22:01   IOPPF: Sent gpu-internal-plimit-notification last value 2 (rounded time weighted average 3)
   Sep 26 17:22:21   IOPPF: Sent cpu-plimit-notification last value 10 (rounded time weighted average 9)
   Sep 26 17:22:21   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 4 (rounded time weighted average 3)
   Sep 26 17:22:21   IOPPF: Sent gpu-internal-plimit-notification last value 4 (rounded time weighted average 3)
   Sep 26 17:22:34   IOPPF: Sent cpu-plimit-notification last value 8 (rounded time weighted average 8)
   Sep 26 17:22:34   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:22:34   IOPPF: Sent gpu-internal-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:22:47   IOPPF: Sent cpu-plimit-notification last value 6 (rounded time weighted average 6)
   Sep 26 17:22:47   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:22:47   IOPPF: Sent gpu-internal-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:23:01   IOPPF: Sent cpu-plimit-notification last value 7 (rounded time weighted average 7)
   Sep 26 17:23:01   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:23:01   IOPPF: Sent gpu-internal-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:23:12   IOPPF: Sent cpu-plimit-notification last value 6 (rounded time weighted average 6)
   Sep 26 17:23:12   IOPPF: Sent gpu-internal-single-slice-plimit-notification last value 5 (rounded time weighted average 5)
   Sep 26 17:23:12   IOPPF: Sent gpu-internal-plimit-notification last value 5 (rounded time weighted average 5)
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Improper use of the super keyword?

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