If I have three classes as follows:
package com.Bob.Marley;
public class SuperClass{
protected int x = 0;
}
package com.Bob.Marley;
public class SubClass extends SuperClass{
protected int x = 1;
}
package com.Bob.Marley;
public class TestClass{
public static void main (String[] args){
SubClass s = new SubClass();
//print 1
System.out.println(s.x);
//how do I print the superclass variable?
//I know inside SubClass I can access it with plain old super.x
//but what about outside the subclass with a new object.
}
}
So the question is how would I print out 0 from the superclass of the new object s created in a separate class. System.out.println(s.super.x); does not work. I don't think it changes anything but I am using java 8.
The expression s.super.x is invalid here. Whenever you prefix a super.x with something, it should be a type name, not a variable name, e.g. SuperClass.super.x. However, this would be valid only inside the subclass for accessing the superclass of the enclosing class, which does not exist here.
Cast x to be a SuperClass so you can access the x declared in Superclass.
System.out.println( ((SuperClass) s).x);
or
SuperClass sc = (SuperClass) s;
System.out.println(sc.x);
This works because variable access is statically binded. The type of the variable or expression determines the scope searched for variable access.
TL;DR: if you introduce a new field in a subclass, don't re-use a field name from the parent class. You gain nothing, only confusion and problems.
If I understand correctly, you want SubClass instances to have two fields, one inherited from the SuperClass (for the discussion, let's rename that to superX to make things clearer), and one from the subclass itself (let's rename that to subX).
For a given SubClass instance, you want to be able to access both fields, superX and subX (of course, using different expressions). What makes things difficult in your code sample, is the fact that you chose to give both of them the same name x.
So, if you really want your instances to carry both fields, I'd recommend to rename them, so you don't have to use ugly tricks like casting to the SuperClass.
public class SuperClass{
protected int superX = 0;
}
public class SubClass extends SuperClass{
protected int subX = 1;
}
But, if x stands for the same property with the same meaning for both the super and the sub class, just with different initial values, then it doesn't make sense to have two different fields, and you should change the code to become:
public class SuperClass{
protected int x = 0;
}
public class SubClass extends SuperClass{
// constructor initializes the x field with 1.
public SubClass(){
x = 1;
}
}
Of course, then it's impossible to get two different values from a single instance of SubClass.
Related
This question already has answers here:
Difference between "this" and"super" keywords in Java
(9 answers)
Closed 4 years ago.
I am trying to find proof for the statement - keyword super is the reference to parent class just like keyword this is the reference to current class.
I am trying multilevel Inheritance in Java A->B->C: class A is grand parent, class B is parent, class C is child.
I have a variable X declared in all three classes with values respectively (A:x=100,B:x=200,C:x=300)
In the child class constructor I am printing values. However the casting isn't working for super keyword whereas it's working for this keyword.
((A)super).x is not working, but ((A)this).x is working.
class A {
int x = 100;
}
class B extends A {
int x = 200;
}
public class C extends B {
int x = 300;
public C () {
System.out.println(this.x); //OP = 300
System.out.println(super.x); // OP = 200
System.out.println(((A)this).x);// OP = 100
System.out.println(((A)super).x); // Giving Compile time Error.. Why?
B reftoB = new B();
System.out.println(((A)reftoB).x); // OP = 100
}
public static void main(String[] args) {
C t1= new C();
}
}
I expect the output of System.out.println(((A)super).x) is 100, but it is giving a compile time error.
So my question is if super is a reference to the parent class then why isn't type casting working on it?
Use of Super Keyword:
super() can be used to refer immediate parent class instance
variable.
super() can be used to invoke immediate parent class method.
super() can be used to invoke immediate parent class constructor.
Your compiler is a compilation error for (A)super).x , because it is not a valid statement, moreover we don't use it this way, ontop of all It violates encapsulation, you should not able to bypass parent class. In every definition of super() you will find something mentioned as current parent class, but what you are trying to do here is bypassing current parent.
Now coming to your problem:
x // Field x in class C
this.x // Field x in class C
super.x // Field x in class B
((B)this).x // Field x in class B
((A)this).x // Field x in class A
super.super.x // Illegal; does not refer to x in class A
((A)super).x // Illegal as well as compilation error
If you still want to access variables like what you intented then use something like below:
t1.x // Field x of class C
((B)t1).x // Field x of class B
((A)t1).x // Field x of class A
Note: t1 is your class C instance.
The answer to why this can be cast but super cannot is the same answer to the question why this can be passed as a method argument but super cannot: namely, because this is defined by the JLS as a Primary Expression but super is not.
The super keyword refers to a class parent, and this keyword refers to the class you are in. Let's create a parent class to start the demonstration:
public class ParentClass{
private int justANumber;
public ParentClass(int justANumber){
this.justANumber = justANumber;
}
}
Notice how the this keyword is used here, which is telling "Hey, assign the justANumber value to THIS class attribute also called justANumber". Let's create a subclass for this parent class now:
public class Subclass extends ParentClass{
// You don't need to declare the justANumber variable, cause it's from the parent
public Subclass(int justANumber){
super(justANumber);
}
public showNumber(){
return this.justANumber;
}
}
The super() method calls the parent constructor, which need an int value on the first place, so you pass it as an argument. Now see how the showNumber() method returns this.justANumber? Why? that's because when the super() method is called, the parent class automatically delegates his variables to the subclass, so in this case, Subclass can now say that justANumber is HIS variable, being able to use the this keyword. Hope you now understand de difference.
Hopefully this question hasn't already been asked. I've had a look around but haven't found a similar post.
Experimenting in Java I've noticed that there is no restriction on having duplicate method signatures in a nested class, which seems counter-intuitive.
For example, if I create class A containing a method with the signature int x() and later add a nested class B containing an identical method, the compiler seems to have no problem with it. My initial assumption was that it would complain that x is already defined. Perhaps I'm missing something obvious that explains why this is allowed?
class A {
int x() {
return 1;
}
class B {
int x() {
return 2;
}
}
}
Subsequently, is there any way to access class A's method x from within the scope of class B, or is it permanently hidden by the method x of the local scope?
Edit: I appreciate that the core of the question is the same as this post, however, I was more interested in understanding why this behaviour is allowed as it wasn't immediately clear to me.
Where a class is defined doesn't matter so much. Keep in mind, in the end you have
class A { int x()
and
class A.B { int x()
Two (almost) independent classes. The only relationship that we have here is that any instance of B needs an "enclosing" instance of A to which it belongs (because it is a non-static inner class).
And of course, you can access the "enclosing" A "stuff" from within B, for example using A.this.x().
This should be available somewhere in the JLS - but ultimately it boils down to scope. Each of them has a different scope - thus the compiler does not complain.
Why would the compiler complain? They are two different classes, they just happen to be nested.
Subsequently, is there any way to access class A's method x from within the scope of class B, or is it permanently hidden by the method x of the local scope?
A.this.x()
At first method signature is not the combination of return type and method name it is method name and parameters.
if you call x() it will run x() inside the B
A.this.x(); it will run x() in A
Its scope, it's similar to have an instance variable called foo and then a local one in the method called foo as well.
Its worth reading about scope in java
Inner class in java can access all the members (i.e variables and methods) including private one, but outer class can not access member of inner class directly.
To access x() method of class B inside class A you can either create B's instance and call it or call like this A.this.x();
To access x() method outside class A you can do something like this:
B b = a.new B();
b.x();
If you are using non-static nested class (inner class) then it wouldn't be able to access to x method of B class from other classes. But if you are using static nested class you will be able to access that method from other classes. Example:
public class A {
int x() {
new A.B().x();
return 1;
}
static class B {
int x() {
new A().x();
return 2;
}
}
}
public static void main(String[] args) {
A a = new A();
a.x();
B b = new A.B();
b.x();
}
I hope this example answering your question... ☺
B is nested inside A. According to scope rules we can do the following:
class A {
int x() {
return 1;
}
class B {
int x() {
return 2;
}
int xOfA(){
return A.this.x();
}
}
public static void main(String[] args) {
final A objA = new A();
final B objB = objA.new B();
System.out.println(objA.x());
System.out.println(objB.x());
System.out.println(objB.xOfA());
}
}
that is because B is visible from an instance of A.
Moreover B can reference methods in the containing class through their full paths.
For example, if I create class A containing a method with the signature int x() and later add a nested class B containing an identical method, the compiler seems to have no problem with it. My initial assumption was that it would complain that x is already defined. Perhaps I'm missing something obvious that explains why this is allowed?
When you define something in a nest scope which is otherwise in an outer scope, this is hiding.
What is important is that -: The version of a method that is executed will NOT be determined by the object that is used to invoke it. In fact it will be determined by the type of reference variable used to invoke the method
Subsequently, is there any way to access class A's method x from within the scope of class B, or is it permanently hidden by the method x of the local scope?
So as it is clear from above we can access it using class name, eg A.x
Hope this helps!!
I want to create data structures to capture the following ideas:
In a game, I want to have a generic Skill class that captures general information like skill id, cool down time, mana cost, etc.
Then I want to have specific skills that define actual interaction and behaviours. So these would all extend from base class Skill.
Finally, each player will have instances of these specific skills, so I can check each player's skill status, whether a player used it recently, etc.
So I have an abstract superclass Skill that defines some static variables, which all skills have in common, and then for each individual skill that extends Skill, I use a static block to reassign the static variables. So I have the following pattern:
class A {
static int x = 0;
}
class B extends A {
static {
x = 1;
}
}
...
// in a method
A b = new B();
System.out.println(b.x);
The above prints 1, which is exactly the behaviour I want. My only problem is that the system complains about I'm accessing static variable in a non-static way. But of course I can't access it in that way, because I only want to treat the skill as Skill without knowing exactly which subclass it is. So I have to suppress the warning every time I do this, which leads me to think whether there is a better/neater design pattern here.
I have thought about making the variables in question non-static, but because they should be static across all instances of the specific skill, I feel like it should be a static variable...
You should generally avoid such use of global state. If you know for sure that the field x will be shared across all instances of all subtypes of the base class, then the correct place to put such a field is probably somewhere other than the base class. It may be in some other configuration object.
But even with your current configuration, it just does't make sense since any subclass that modifies the static variable will make the variable visible to all classes. If subclass B changes x to 1, then subclass C changes it to 2, the new value would be visible to B as well.
I think that the way you described in the question, every subclass should have its own separate static field. And in the abstract base class, you can define a method to be implemented by each subclass in order to access each field:
abstract class A {
public abstract int getX();
}
class B extends A {
public static int x = 1;
public int getX() {
return x;
}
}
class C extends A {
public static int x = 2;
public int getX() {
return x;
}
}
As already pointed out by some answers and comments, your approach won't work the way you want because every static block changes the static variable for all classes extending A.
Use an interface and instance methods instead:
public interface A {
int getX();
}
-
public class B implements A {
private static final int X = 1;
#Override
public int getX() {
return X;
}
}
-
A myInstance = new B();
System.out.println(myInstance.getX()); // prints "1"
Let me know if someone does not understand the question. I tried my best to frame the question below.
I have a common method in parent class for generating pattern. For re usability, I thought to retain this method under parent class.
Inside the method, I use several variables. So I thought it is better to pass the object as the parameter to the method generatePattern. But since variables cannot be overridden, how can I use respective variables from these sub classes? Any other feature in java other than using subclass? Will "Generic types" as parameter work in such case?
class Parent {
int var1;
String[] values;
void generatePattern(Parent obj1) {
// This will not make use of respective values of
// subclass object that is passed I guess.
newPattern(obj1.values, obj1.var1);
}
}
class AscendSubClass extends Parent {
int var1 = 5;
String[] values = {"S", "R"};
}
class DescendSubClass extends Parent {
int var1 = 10;
String[] values = {"N", "D"};
}
I may pass either AscendSubClass or DescendSubClass above to generatePattern().
Inside this method, I need to use the variables var1, values and many other variables of subclasses.
These variables are of same type and have same name, but the values are different and depends on the subclass. How can I refer these variables now in generatePattern() so that method does not change?
It is possible to achieve this by making variables as a parameters to methods or by if/else statements, but I have several variables to pass and it is big inconvenience.
Add a getter method and override it in either subclass.
public int getVar() {
return var1;
}
Consider the int a variables in these classes:
class Foo {
public int a = 3;
public void addFive() { a += 5; System.out.print("f "); }
}
class Bar extends Foo {
public int a = 8;
public void addFive() { this.a += 5; System.out.print("b " ); }
}
public class test {
public static void main(String [] args){
Foo f = new Bar();
f.addFive();
System.out.println(f.a);
}
}
I understand that the method addFive() have been overridden in the child class, and in class test when the base class reference referring to child class is used to call the overridden method, the child class version of addFive is called.
But what about the public instance variable a? What happens when both base class and derived class have the same variable?
The output of the above program is
b 3
How does this happen?
There are actually two distinct public instance variables called a.
A Foo object has a Foo.a variable.
A Bar object has both Foo.a and Bar.a variables.
When you run this:
Foo f = new Bar();
f.addFive();
System.out.println(f.a);
the addFive method is updating the Bar.a variable, and then reading the Foo.a variable. To read the Bar.a variable, you would need to do this:
System.out.println(((Bar) f).a);
The technical term for what is happening here is "hiding". Refer to the JLS section 8.3, and section 8.3.3.2 for an example.
Note that hiding also applies to static methods with the same signature.
However instance methods with the same signature are "overridden" not "hidden", and you cannot access the version of a method that is overridden from the outside. (Within the class that overrides a method, the overridden method can be called using super. However, that's the only situation where this is allowed. The reason that accessing overridden methods is generally forbidden is that it would break data abstraction.)
The recommended way to avoid the confusion of (accidental) hiding is to declare your instance variables as private and access them via getter and setter methods. There are lots of other good reasons for using getters and setters too.
It should also be noted that: 1) Exposing public variables (like a) is generally a bad idea, because it leads to weak abstraction, unwanted coupling, and other problems. 2) Intentionally declaring a 2nd public a variable in the child class is a truly awful idea.
From JLS
8.3.3.2 Example: Hiding of Instance Variables This example is similar to
that in the previous section, but uses
instance variables rather than static
variables. The code:
class Point {
int x = 2;
}
class Test extends Point {
double x = 4.7;
void printBoth() {
System.out.println(x + " " + super.x);
}
public static void main(String[] args) {
Test sample = new Test();
sample.printBoth();
System.out.println(sample.x + " " +
((Point)sample).x);
}
}
produces the output:
4.7 2
4.7 2
because the declaration of x in class
Test hides the definition of x in
class Point, so class Test does not
inherit the field x from its
superclass Point. It must be noted,
however, that while the field x of
class Point is not inherited by class
Test, it is nevertheless implemented
by instances of class Test. In other
words, every instance of class Test
contains two fields, one of type int
and one of type double. Both fields
bear the name x, but within the
declaration of class Test, the simple
name x always refers to the field
declared within class Test. Code in
instance methods of class Test may
refer to the instance variable x of
class Point as super.x.
Code that uses a field access
expression to access field x will
access the field named x in the class
indicated by the type of reference
expression. Thus, the expression
sample.x accesses a double value, the
instance variable declared in class
Test, because the type of the variable
sample is Test, but the expression
((Point)sample).x accesses an int
value, the instance variable declared
in class Point, because of the cast to
type Point.
In inheritance, a Base class object can refer to an instance of Derived class.
So this is how Foo f = new Bar(); works okay.
Now when f.addFive(); statement gets invoked it actually calls the 'addFive() method of the Derived class instance using the reference variable of the Base class. So ultimately the method of 'Bar' class gets invoked. But as you see the addFive() method of 'Bar' class just prints 'b ' and not the value of 'a'.
The next statement i.e. System.out.println(f.a) is the one that actually prints the value of a which ultimately gets appended to the previous output and so you see the final output as 'b 3'. Here the value of a used is that of 'Foo' class.
Hope this trick execution & coding is clear and you understood how you got the output as 'b 3'.
Here F is of type Foo and f variable is holding Bar object but java runtime gets the f.a from the class Foo.This is because in Java variable names are resolved using the reference type and not the object which it is referring.