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Holding Child Object to a Reference of Parent [duplicate]

I'm preparing for the SCJP (recently rebranded as OCPJP by Oracle) and one particular question that I got wrong on a mock exam has confused me, the answer description doesn't explain things clear enough.
This is the question :
class A
{
int x = 5;
}
class B extends A
{
int x = 6;
}
public class CovariantTest
{
public A getObject()
{
return new A();
}
public static void main(String[]args)
{
CovariantTest c1 = new SubCovariantTest();
System.out.println(c1.getObject().x);
}
}
class SubCovariantTest extends CovariantTest
{
public B getObject()
{
return new B();
}
}
The answer is 5, but I chose 6.
I understand that overriding applies to methods at runtime, and not variables, but the way my mind interpreted that println was :
call getObject on c1
c1 is actually a SubCovariantTest object, and has a valid override
for getObject(), so use the overridden method
The override returns B, so grab x from B which is 6
Is it a case of the JVM ignoring the getObject() part, and always taking x from c1 as variables are associated at compile time?

Although the override is done properly for SubCovariantTest the answer is 5 because of how the variable c1 is declared. It is declared as a CovariantTest and not as a SubCovariantTest.
When c1.getObject().x is run, it does not know that it is a SubCovariantTest (no casting was used). This is why 5 is returned from CovariantTest and not 6 from SubCovariantTest.
If you change
System.out.println(c1.getObject().x);
to
System.out.println(((SubCovariantTest) c1).getObject().x);
you will get 6 as you expected.
Edit: As pointed out in the comments
"fields are not polymorphic in Java. Only methods are. The x in the subclass hides the x in the base class. It doesn't override it." (Thanks to JB Nizet)

Okay I know this is a bit late to reply to this question but I and my friend had the same problem and the answers already here didn't quite clear it for us. So I'll just state what problem I had and how it makes sense now :)
Now I do understand that fields don't get overrided but instead they get hidden as miller.bartek pointed out and I also understand that overriding is for methods and not fields as Scott points out.
The problem I had however was this. According to me,
c1.getObject().x
This must transform into:
new B().x // and not newA().x since getObject() gets overrided
And that evaluates to 6.
And I couldn't get why the variable of class A (super-class) is being called by an object of class B (sub-class) without having explicitly asked for such a behaviour.
And guessing from the wording of the question, I feel the OP had the same question/doubt in mind.
My Answer:
You get a hint from Elbek's answer. Put the following lines in the main method and try to compile the code:
A a = c1.getObject(); //line 1
B b = c1.getObject(); //line 2
You'll notice that line 1 is completely legal while line 2 gives compilation error.
So when the function getObject() is being called, the CovariantTest (super) function IS getting overrided by SubCovariantTest (sub) function since that is valid overriding in the code and c1.getObject() WILL return new B().
However, since the super-function returns a reference of class-type A, even after getting overrided, it must return a reference of class-type A unless ofcourse we type-cast it. And here, class B is a class A (due to inheritance).
So practically, what we're getting from c1.getObject() is not
new B()
but this:
(A) new B()
That is why the output comes out to be 5 even though an object of class B is returned and class B has value of x as 6.

The technical term for what is happening here is "hiding". Variables names in Java are resolved by the reference type, not the object they are referencing.
A object has a A.x variable.
B object has both A.x and B.x variables.
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.
Note that hiding also applies to static methods with the same signature.
Your mock question in a simplified form (without overriding):
class A {
int x = 5;
}
class B extends A {
int x = 6;
}
public class CovariantTest {
public static void main(String[] args) {
A a = new B();
B b = new B();
System.out.println(a.x); // prints 5
System.out.println(b.x); // prints 6
}
}

You are calling method from c1: System.out.println(c1.getObject().x);
c1 reference type is:
public class CovariantTest
{
public A getObject()
{
return new A();
}
public static void main(String[]args)
{
CovariantTest c1 = new SubCovariantTest();
System.out.println(c1.getObject().x);
}
}
so for this: c1.getObject() return type is A. from A you getting directly attribute not method, as you mention java does not override attributes, so it is grabbing x from A

When methods are overridden, subclass methods are called, and when variables are overridden the superclass variables are used

When the child and parent class both have a variable with same name child class's variable hides parent class's variable and this is called variable hiding.
While variable hiding looks like overriding a variable similar to method overriding but it is not, Overriding is applicable only to methods while hiding is applicable variables.
In the case of method overriding, overridden methods completely replaces the inherited methods so when we try to access the method from parent's reference by holding child's object, the method from child class gets called.
But in variable hiding child class hides the inherited variables instead of replacing, so when we try to access the variable from parent's reference by holding child's object, it will be accessed from the parent class.
When an instance variable in a subclass has the same name as an instance variable in a superclass, then the instance variable is chosen from the reference type.
You can read more on my article What is Variable Shadowing and Hiding in Java.

Casting in Java - what does (Parent) this mean?

class A {
public int a = 100;
}
class B extends A {
public int a = 80;
}
class C extends B {
public int a = 10;
public void show() {
int a = 0;
System.out.println(a);
System.out.println(super.a);
System.out.println(((A) this).a);
}
}
What does ((A) this).a in the line System.out.println(((A) this).a); do?
Is it upcasting/downcasting thisor is something else happening here?
I also tried System.out.println(this); and System.out.println((A)this); and they both have the same output. What exactly is happening here?

In the java programming language, we have classes. When we write java code, we create instances of those classes, for example:
Object o = new Object();
Object is a class. Writing new Object() creates an instance of that class. The above code declares a variable o and assigns it [a reference to] an instance of class Object.
In the terminology of the java programming language, we say that variable o has type Object.
In the code in your question, a variable that is assigned an instance of class C, really has three types.
It has type C.
It has type B since B is the superclass of C.
It has type A because it indirectly extends class A also.
In the context of the code in your question, this is a special variable whose type is C. Writing (A) this is telling java to relate to the variable this as if its type is A.
Class A cannot access its subclasses. Hence it is only aware of its class member a. Hence when you write this line of code...
((A) this).a
You are accessing the member of class A only.

System.out.println(a);a is the one from the show method of your C class → a = 0
System.out.println(super.a);a is the one from the super-class of C, which is B → a = 80
System.out.println(((A) this).a);First, you cast your C instance (this) into A, then you call a which is a member of the A class → a = 100

There is also something to consider : method will always take the more specialized one (except if super is used), where field will be taken directly from the type referenced (even if there is an extending class).
For example, if I add getA() in each classes :
class A {
public int a = 100;
public int getA(){
return a;
}
}
class B extends A {
public int a = 80;
public int getA(){
return a;
}
}
class C extends B {
public int a = 10;
public int getA(){
return a;
}
public void show() {
int a = 0;
System.out.println(a);
System.out.println(super.a);
System.out.println(((A) this).a);
System.out.println(getA());
System.out.println(super.getA());
System.out.println(((A) this).getA());
}
}
class Scratch {
public static void main(String[] args) {
new C().show();
}
}
I get the following output :
0
80
100
10
80
10
Which means that in the case of the method, except in the case of super.getA() which explicitly goes to the superclass, casting your C into a A doesn't change much for methods, as it impacts the field.

If you write something like obj.a, obj.getA() or someMethod(obj), Java somehow has to find the actual field or method to be used, based on the type or class of obj. There are two distinct dispatch mechanisms involved (plus the special construct super).
Dynamic dispatch (polymorphism, overriding): This is used when calling an instance method on the object, as in obj.getA(). Then the runtime class of the obj is examined, and if this class contains a getA() method, this is used. Otherwise, the direct parent class is examined for a getA() method, and so on up to the Object class.
Static dispatch: In cases like obj.a or someMethod(obj), the runtime class of obj doesn't matter. Involved is only the compiler, and from his knowledge of obj's type, he decides which field or method to use.
super dispatch: If you write super.getA() or super.a, your getA() method or a field is ignored, and instead the next-higher class in the hierarchy is used that contains such a method or field.
In your case you have 3 fields plus one local variable, all with the same name a. (By the way, it's a very bad idea to have such name conflicts in professional code.) We are inside a method show() declared in the C class. Let's have a look at some different expressions and what they mean here:
a references the local variable a. There's no dispatch needed, it's just that local definitions take precedence over fields.
this.a is a static-dispatch expression, so it's important what the compiler thinks about the type of this. And that's always the class where this code has been written. In your case, it's class C, so the field a from class C is used, the one being 10.
super.a is a super-dispatch expression, meaning that the a field from this class C is ignored and the next higher one taken (the one from B, in our case).
((A) this).a is static dispatch, but the (A) casting has a significant effect. The expression before the dot originally comes from this, being of type C, but the (A) cast tells the compiler to believe it were of type A. This is okay, as every C also is an A, by inheritance. But now, static dispatch sees something of type A in front of the dot, and dispatches to the a field from the A class, and no longer from C.
getA(), this.getA() and ((A) this).getA() are all dynamic-dispatch examples, all giving the same result. The method called will be the one based on the runtime class of this object. This will typically be one defined in the C class. But if show() was called on an object of a subclass of C, e.g. D, and D had its own getA() method, that one would be used.
super.getA() is a case of super-dispatch, it will call the getA() method next higher up in the class hierarchy from the current class, e.g. B.

System.out.println(this);
And
System.out.println((A)this)
These two prints the object reference to class C with toString() method.
System.out.println(((A)this).a);
This is upcasting, child object to parent object.

How does accessing public fields work with inheritance? [duplicate]

I am studying overriding member functions in JAVA and thought about experimenting with overriding member variables.
So, I defined classes
public class A{
public int intVal = 1;
public void identifyClass()
{
System.out.println("I am class A");
}
}
public class B extends A
{
public int intVal = 2;
public void identifyClass()
{
System.out.println("I am class B");
}
}
public class mainClass
{
public static void main(String [] args)
{
A a = new A();
B b = new B();
A aRef;
aRef = a;
System.out.println(aRef.intVal);
aRef.identifyClass();
aRef = b;
System.out.println(aRef.intVal);
aRef.identifyClass();
}
}
The output is:
1
I am class A
1
I am class B
I am not able to understand why when aRef is set to b intVal is still of class A?

When you make a variable of the same name in a subclass, that's called hiding. The resulting subclass will now have both properties. You can access the one from the superclass with super.var or ((SuperClass)this).var. The variables don't even have to be of the same type; they are just two variables sharing a name, much like two overloaded methods.

Variables are not polymorphic in Java; they do not override one another.

There is no polymorphism for fields in Java.
Variables decision happens at a compile time so always Base Class variables (not child’s inherited variables) will be accessed.
So whenever upcasting happens always remember
1) Base Class variables will be accessed.
2) Sub Class methods(overridden methods if overriding happened else inherited methods as it is from parent) will be called.

Variables are resolved compile-time, methods run-time. The aRef is of type A, therefore aRef.Intvalue is compile-time resolved to 1.

OverRiding Concept in Java
Functions will override depends on object type and variables will accessed on reference type.
Override Function: In this case suppose a parent and child class both have same name of function with own definition. But which function will execute it depends on object type not on reference type on run time.
For e.g.:
Parent parent=new Child();
parent.behaviour();
Here parent is a reference of Parent class but holds an object of Child Class so that's why Child class function will be called in that case.
Child child=new Child();
child.behaviour();
Here child holds an object of Child Class, so the Child class function will be called.
Parent parent=new Parent();
parent.behaviour();
Here parent holds the object of Parent Class, so the Parent class function will be called.
Override Variable: Java supports overloaded variables. But actually these are two different variables with same name, one in the parent class and one in the child class. And both variables can be either of the same datatype or different.
When you trying to access the variable, it depends on the reference type object, not the object type.
For e.g.:
Parent parent=new Child();
System.out.println(parent.state);
The reference type is Parent so the Parent class variable is accessed, not the Child class variable.
Child child=new Child();
System.out.println(child.state);
Here the reference type is Child, so the Child class variable is accessed not the Parent class variable.
Parent parent=new Parent();
System.out.println(parent.state);
Here the reference type is Parent, so Parent class variable is accessed.

From JLS Java SE 7 Edition §15.11.1:
This lack of dynamic lookup for field accesses allows programs to be run efficiently with straightforward implementations. The power of late binding and overriding is available, but only when instance methods are used.
Answers from Oliver Charlesworth and Marko Topolnik are correct, I would like to elaborate a little bit more on the why part of the question:
In Java class members are accessed according the type of the reference and not the type of the actual object. For the same reason, if you had a someOtherMethodInB() in class B, you wouldn't be able to access it from aRef after aRef = b is run. Identifiers (ie class, variable, etc names) are resolved at compile time and thus the compiler relies on the reference type to do this.
Now in your example, when running System.out.println(aRef.intVal); it prints the value of intVal defined in A because this is the type of the reference you use to access it. The compiler sees that aRef is of type A and that's the intVal it will access. Don't forget that you have both fields in the instances of B. JLS also has an example similar to yours, "15.11.1-1. Static Binding for Field Access" if you want to take a look.
But why do methods behave differently? The answer is that for methods, Java uses late binding. That means that at compile time, it finds the most suitable method to search for during the runtime. The search involves the case of the method being overridden in some class.

I hope this can help:
public class B extends A {
// public int intVal = 2;
public B() {
super();
super.intVal = 2;
}
public void identifyClass() {
System.out.println("I am class B");
}
}
So overriding variable of base class is not possible, but base class variable value can be set (changed) from constructor of inherited class.

This is called variable hiding. When you assign aRef = b; , aRef has two intVal, 1 is named just intVal another is hidden under A.intVal (see debugger screenshot), Because your variable is of type class A , even when you print just intVal java intelligently picks up A.intVal.
Answer 1: One way of accessing child class's intVal is System.out.println((B)aRef.intVal);
Answer 2: Another way of doing it is Java Reflection because when you use reflection java cant intelligently pickup hidden A.intVal based on Class type, it has to pick up the variable name given as string -
import java.lang.reflect.Field;
class A{
public int intVal = 1;
public void identifyClass()
{
System.out.println("I am class A");
}
}
class B extends A
{
public int intVal = 2;
public void identifyClass()
{
System.out.println("I am class B");
}
}
public class Main
{
public static void main(String [] args) throws Exception
{
A a = new A();
B b = new B();
A aRef;
aRef = a;
System.out.println(aRef.intVal);
aRef.identifyClass();
aRef = b;
Field xField = aRef.getClass().getField("intVal");
System.out.println(xField.get(aRef));
aRef.identifyClass();
}
}
Output -
1
I am class A
2
I am class B

Well, I hope u got the answer. If not, you can try seeing in the debug mode. the subclass B has access to both the intVal. They are not polymorphic hence they are not overriden.
If you use B's reference you will get B's intVal. If you use A's reference , you will get A's intVal. It's that simple.

As per the Java specifications, the instance variables are not overridden from a super class by a sub class when it is extended.
Hence the variable in the sub class only can be seen as one sharing the same name.
Also when the constructor of A is called during the instance creation of B the variable (intVal) is initialized and hence the output.

It is because when you assign b to aRef, it is resolved, leading aRef to just be of class A. This means that aRef does not have access to any of class B's fields or methods. If you call for intVal instead by using b.intVal, you will get 2.

Java has a feather of encapsulation means it tightly binds the property and the behavior of an object. so only via a class reference we can call it's behavior to change it's property.
and in inheritance only method overrides so that it can affects only it's property.

As Many users have already pointed out, this is not polymorphism. Polymorphism only applies to methods(functions).
Now as to why the value of the intVal of class A is printed, this happens because as you can see the reference aRef is of type A.
I can see why you are confused by it. By the same procedure you have accessed the overridden methods for ex. the method identifyClass() but the not the variables which directly proves the first line that I have written .
Now in order to access the variable you can do ((Superclass)c).var
Note here that the Superclass can be many levels up for example
A<-B<-C. That is C extends B and B extends A. If you wanted the value of var of A then you could have done ((A)c).var .
EDIT: as one of the users have pointed out this 'trick' does not apply to static methods, because they are static.

Java subclass passing itself through a superclass constructor [duplicate]

I am studying overriding member functions in Java and thought about experimenting with overriding member variables.
So, I defined classes
public class A{
public int intVal = 1;
public void identifyClass()
{
System.out.println("I am class A");
}
}
public class B extends A
{
public int intVal = 2;
public void identifyClass()
{
System.out.println("I am class B");
}
}
public class mainClass
{
public static void main(String [] args)
{
A a = new A();
B b = new B();
A aRef;
aRef = a;
System.out.println(aRef.intVal);
aRef.identifyClass();
aRef = b;
System.out.println(aRef.intVal);
aRef.identifyClass();
}
}
The output is:
1
I am class A
1
I am class B
I am not able to understand why when aRef is set to b intVal is still of class A?

When you make a variable of the same name in a subclass, that's called hiding. The resulting subclass will now have both properties. You can access the one from the superclass with super.var or ((SuperClass)this).var. The variables don't even have to be of the same type; they are just two variables sharing a name, much like two overloaded methods.

Variables are not polymorphic in Java; they do not override one another.

There is no polymorphism for fields in Java.
Variables decision happens at a compile time so always Base Class variables (not child’s inherited variables) will be accessed.
So whenever upcasting happens always remember
1) Base Class variables will be accessed.
2) Sub Class methods(overridden methods if overriding happened else inherited methods as it is from parent) will be called.

Variables are resolved compile-time, methods run-time. The aRef is of type A, therefore aRef.Intvalue is compile-time resolved to 1.

OverRiding Concept in Java
Functions will override depends on object type and variables will accessed on reference type.
Override Function: In this case suppose a parent and child class both have same name of function with own definition. But which function will execute it depends on object type not on reference type on run time.
For e.g.:
Parent parent=new Child();
parent.behaviour();
Here parent is a reference of Parent class but holds an object of Child Class so that's why Child class function will be called in that case.
Child child=new Child();
child.behaviour();
Here child holds an object of Child Class, so the Child class function will be called.
Parent parent=new Parent();
parent.behaviour();
Here parent holds the object of Parent Class, so the Parent class function will be called.
Override Variable: Java supports overloaded variables. But actually these are two different variables with same name, one in the parent class and one in the child class. And both variables can be either of the same datatype or different.
When you trying to access the variable, it depends on the reference type object, not the object type.
For e.g.:
Parent parent=new Child();
System.out.println(parent.state);
The reference type is Parent so the Parent class variable is accessed, not the Child class variable.
Child child=new Child();
System.out.println(child.state);
Here the reference type is Child, so the Child class variable is accessed not the Parent class variable.
Parent parent=new Parent();
System.out.println(parent.state);
Here the reference type is Parent, so Parent class variable is accessed.

From JLS Java SE 7 Edition §15.11.1:
This lack of dynamic lookup for field accesses allows programs to be run efficiently with straightforward implementations. The power of late binding and overriding is available, but only when instance methods are used.
Answers from Oliver Charlesworth and Marko Topolnik are correct, I would like to elaborate a little bit more on the why part of the question:
In Java class members are accessed according the type of the reference and not the type of the actual object. For the same reason, if you had a someOtherMethodInB() in class B, you wouldn't be able to access it from aRef after aRef = b is run. Identifiers (ie class, variable, etc names) are resolved at compile time and thus the compiler relies on the reference type to do this.
Now in your example, when running System.out.println(aRef.intVal); it prints the value of intVal defined in A because this is the type of the reference you use to access it. The compiler sees that aRef is of type A and that's the intVal it will access. Don't forget that you have both fields in the instances of B. JLS also has an example similar to yours, "15.11.1-1. Static Binding for Field Access" if you want to take a look.
But why do methods behave differently? The answer is that for methods, Java uses late binding. That means that at compile time, it finds the most suitable method to search for during the runtime. The search involves the case of the method being overridden in some class.

I hope this can help:
public class B extends A {
// public int intVal = 2;
public B() {
super();
super.intVal = 2;
}
public void identifyClass() {
System.out.println("I am class B");
}
}
So overriding variable of base class is not possible, but base class variable value can be set (changed) from constructor of inherited class.

This is called variable hiding. When you assign aRef = b; , aRef has two intVal, 1 is named just intVal another is hidden under A.intVal (see debugger screenshot), Because your variable is of type class A , even when you print just intVal java intelligently picks up A.intVal.
Answer 1: One way of accessing child class's intVal is System.out.println((B)aRef.intVal);
Answer 2: Another way of doing it is Java Reflection because when you use reflection java cant intelligently pickup hidden A.intVal based on Class type, it has to pick up the variable name given as string -
import java.lang.reflect.Field;
class A{
public int intVal = 1;
public void identifyClass()
{
System.out.println("I am class A");
}
}
class B extends A
{
public int intVal = 2;
public void identifyClass()
{
System.out.println("I am class B");
}
}
public class Main
{
public static void main(String [] args) throws Exception
{
A a = new A();
B b = new B();
A aRef;
aRef = a;
System.out.println(aRef.intVal);
aRef.identifyClass();
aRef = b;
Field xField = aRef.getClass().getField("intVal");
System.out.println(xField.get(aRef));
aRef.identifyClass();
}
}
Output -
1
I am class A
2
I am class B

Well, I hope u got the answer. If not, you can try seeing in the debug mode. the subclass B has access to both the intVal. They are not polymorphic hence they are not overriden.
If you use B's reference you will get B's intVal. If you use A's reference , you will get A's intVal. It's that simple.

As per the Java specifications, the instance variables are not overridden from a super class by a sub class when it is extended.
Hence the variable in the sub class only can be seen as one sharing the same name.
Also when the constructor of A is called during the instance creation of B the variable (intVal) is initialized and hence the output.

It is because when you assign b to aRef, it is resolved, leading aRef to just be of class A. This means that aRef does not have access to any of class B's fields or methods. If you call for intVal instead by using b.intVal, you will get 2.

Java has a feather of encapsulation means it tightly binds the property and the behavior of an object. so only via a class reference we can call it's behavior to change it's property.
and in inheritance only method overrides so that it can affects only it's property.

As Many users have already pointed out, this is not polymorphism. Polymorphism only applies to methods(functions).
Now as to why the value of the intVal of class A is printed, this happens because as you can see the reference aRef is of type A.
I can see why you are confused by it. By the same procedure you have accessed the overridden methods for ex. the method identifyClass() but the not the variables which directly proves the first line that I have written .
Now in order to access the variable you can do ((Superclass)c).var
Note here that the Superclass can be many levels up for example
A<-B<-C. That is C extends B and B extends A. If you wanted the value of var of A then you could have done ((A)c).var .
EDIT: as one of the users have pointed out this 'trick' does not apply to static methods, because they are static.

Slight confusion regarding overriding where variables are concerned

I'm preparing for the SCJP (recently rebranded as OCPJP by Oracle) and one particular question that I got wrong on a mock exam has confused me, the answer description doesn't explain things clear enough.
This is the question :
class A
{
int x = 5;
}
class B extends A
{
int x = 6;
}
public class CovariantTest
{
public A getObject()
{
return new A();
}
public static void main(String[]args)
{
CovariantTest c1 = new SubCovariantTest();
System.out.println(c1.getObject().x);
}
}
class SubCovariantTest extends CovariantTest
{
public B getObject()
{
return new B();
}
}
The answer is 5, but I chose 6.
I understand that overriding applies to methods at runtime, and not variables, but the way my mind interpreted that println was :
call getObject on c1
c1 is actually a SubCovariantTest object, and has a valid override
for getObject(), so use the overridden method
The override returns B, so grab x from B which is 6
Is it a case of the JVM ignoring the getObject() part, and always taking x from c1 as variables are associated at compile time?

Although the override is done properly for SubCovariantTest the answer is 5 because of how the variable c1 is declared. It is declared as a CovariantTest and not as a SubCovariantTest.
When c1.getObject().x is run, it does not know that it is a SubCovariantTest (no casting was used). This is why 5 is returned from CovariantTest and not 6 from SubCovariantTest.
If you change
System.out.println(c1.getObject().x);
to
System.out.println(((SubCovariantTest) c1).getObject().x);
you will get 6 as you expected.
Edit: As pointed out in the comments
"fields are not polymorphic in Java. Only methods are. The x in the subclass hides the x in the base class. It doesn't override it." (Thanks to JB Nizet)

Okay I know this is a bit late to reply to this question but I and my friend had the same problem and the answers already here didn't quite clear it for us. So I'll just state what problem I had and how it makes sense now :)
Now I do understand that fields don't get overrided but instead they get hidden as miller.bartek pointed out and I also understand that overriding is for methods and not fields as Scott points out.
The problem I had however was this. According to me,
c1.getObject().x
This must transform into:
new B().x // and not newA().x since getObject() gets overrided
And that evaluates to 6.
And I couldn't get why the variable of class A (super-class) is being called by an object of class B (sub-class) without having explicitly asked for such a behaviour.
And guessing from the wording of the question, I feel the OP had the same question/doubt in mind.
My Answer:
You get a hint from Elbek's answer. Put the following lines in the main method and try to compile the code:
A a = c1.getObject(); //line 1
B b = c1.getObject(); //line 2
You'll notice that line 1 is completely legal while line 2 gives compilation error.
So when the function getObject() is being called, the CovariantTest (super) function IS getting overrided by SubCovariantTest (sub) function since that is valid overriding in the code and c1.getObject() WILL return new B().
However, since the super-function returns a reference of class-type A, even after getting overrided, it must return a reference of class-type A unless ofcourse we type-cast it. And here, class B is a class A (due to inheritance).
So practically, what we're getting from c1.getObject() is not
new B()
but this:
(A) new B()
That is why the output comes out to be 5 even though an object of class B is returned and class B has value of x as 6.

The technical term for what is happening here is "hiding". Variables names in Java are resolved by the reference type, not the object they are referencing.
A object has a A.x variable.
B object has both A.x and B.x variables.
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.
Note that hiding also applies to static methods with the same signature.
Your mock question in a simplified form (without overriding):
class A {
int x = 5;
}
class B extends A {
int x = 6;
}
public class CovariantTest {
public static void main(String[] args) {
A a = new B();
B b = new B();
System.out.println(a.x); // prints 5
System.out.println(b.x); // prints 6
}
}

You are calling method from c1: System.out.println(c1.getObject().x);
c1 reference type is:
public class CovariantTest
{
public A getObject()
{
return new A();
}
public static void main(String[]args)
{
CovariantTest c1 = new SubCovariantTest();
System.out.println(c1.getObject().x);
}
}
so for this: c1.getObject() return type is A. from A you getting directly attribute not method, as you mention java does not override attributes, so it is grabbing x from A

When methods are overridden, subclass methods are called, and when variables are overridden the superclass variables are used

When the child and parent class both have a variable with same name child class's variable hides parent class's variable and this is called variable hiding.
While variable hiding looks like overriding a variable similar to method overriding but it is not, Overriding is applicable only to methods while hiding is applicable variables.
In the case of method overriding, overridden methods completely replaces the inherited methods so when we try to access the method from parent's reference by holding child's object, the method from child class gets called.
But in variable hiding child class hides the inherited variables instead of replacing, so when we try to access the variable from parent's reference by holding child's object, it will be accessed from the parent class.
When an instance variable in a subclass has the same name as an instance variable in a superclass, then the instance variable is chosen from the reference type.
You can read more on my article What is Variable Shadowing and Hiding in Java.