I am not clear about the procedure of extending a class. Given the following piece of code, why the output is 32?
class Rts {
public static void main(String[] args) {
System.out.println(zorg(new RtsC()));
}
static int zorg(RtsA x) {
return x.f()*10 + x.a;
}
}
abstract class RtsA {
int a = 2;
int f() { return a; }
}
class RtsB extends RtsA {
int a = 3;
int f() { return a; }
}
class RtsC extends RtsB {
int a = 4;
}
First off, fields aren't overridden, so all this is equivalent to
public class Rts {
public static void main(String[] args) {
System.out.println(zorg(new RtsC()));
}
static int zorg(RtsA x) {
return x.f()*10 + x.a;
}
}
abstract class RtsA {
int a = 2;
int f() { return a; }
}
class RtsB extends RtsA {
int b = 3;
int f() { return b; }
}
class RtsC extends RtsB {
int c = 4;
}
The implementation of f() for an object of type RtsC comes from RtsB, since that is the lowest-level class that overrides f(), so its implementation is used, and that returns b, which is 3. That's multiplied by 10, and then added to the a from RtsA, since zorg only knows that x is of type RtsA, so that field is used. That's 3 * 10 + 2 = 32.
(Note that the fact that RtsA is abstract didn't come into this at all; that mostly only matters when you have abstract methods to worry about.)
Related
Can someone explain why the function prints the variable from super and not from the subclass? Class variables cannot be overridden in Java?
class A {
int i = 1;
int fun() {
return i;
}
}
class B extends A {
int i = 2;
}
class Main {
public static void main(String[] args) {
System.out.println(new B().fun());
}
}
This prints out 1 instead of 2.
Because fields declared in the subclass never override fields of the super class.
Overriding is for methods.
If you want to use the i value of the current class, you could introduce getI() a method to provide the i value :
class A {
int i = 1;
int fun() {
return getI();
}
int getI(){
return i;
}
}
And override it in the subclass :
class B extends A {
int i = 2;
int getI(){
return i;
}
}
You are returning the value of i from fun() function . if you want to return the value of override variable from class B need to override that method, as fun method is a part of the super class it is referring i of super class only.
But always remember overriding of variable in java is always a bad idea it may give you unexpected result.
if you still want you can use this way.
class A {
int i = 1;
int fun() {
return i;
}
}
class B extends A {
int i = 2;
int fun() {
return i;
}
}
class Main {
public static void main(String[] args) {
System.out.println(new B().fun()); // this will refer the override i
}
}
I was playing around with classes since I'm learning java syntax and I came across this weird situation. So given a class A and B:
public class A {
public int x() {
return x;
}
}
public class B extends A {
int x = 5;
public int x() {
return x + 2;
}
public static void main(String[] args) {
B b = new B();
System.out.println(b.x());
}
When I execute the main method I get a compile time error that it doesn't find the x variable and it's calling the method from A because the error shows return x instead of return x + 2. Since b's static type is B, why is it looking in A for x?
error: cannot find symbol
return x;
symbol: variable x
location: class A
The class A doesn't know that it will be extended by B, where the x variable will exist.
In order to make this compile, make A.x() abstract and provide implementation within the subclass:
public abstract class A {
public abstract int x();
}
public class B extends A {
int x = 5;
#Override
public int x() {
return x + 2;
}
..
}
When you are creating the Object of class B, it automatically invoke the x() of class A, But x is not initialized in that class. As it is a local variable, it can not be used without initialization. so it is giving an error.
Try the below code, it is working correctly
class A
{
int x =10;
public int x()
{
return x;
}
}
class B extends A
{
int x = 5;
public int x()
{
return x + 2;
}
public static void main(String[] args)
{
B b = new B();
System.out.println(b.x());
}
}
I have output for this program that I don't understand how does it happens
here is the output
i from A is 40
i from A is 60
i from B is 60
I do understand the first line of output but nothing after that. Does this have to do with polymorphism ?
public class Test {
public static void main(String[] args) {
new A();
new B();
}
}
class A {
int i = 7;
public A() {
setI(20);
System.out.println("i from A is " + i);
}
public void setI(int i) {
this.i = 2 * i;
}
}
class B extends A {
public B() {
System.out.println("i from B is " + i);
}
public void setI(int i) {
this.i = 3 * i;
}
}
Yes. B extends A meaning it is a subclass of A.
Thus it shares i which is initialized only in A and then shared in B.
So B starts with what?
An i with "initial" value equal to 20.
It is java functionality
Whenever you create a child class object default constructor/argument-less constructor implicitly gets called.
And in your case first call is to A() constructor and then B() constuctor, and method call get to local method of B class
`ie public void setI(int i) {
this.i = 3 * i;
}
If you want to avoid this you can do this by making a call to super class constructor explicitly.
For example:
public class Test {
public static void main(String[] args) {
// new A();
new B();
}
}
class A {
int i = 7;
public A() {
setI(20);
System.out.println("i from A is " + i);
}
public A(int x)
{
System.out.println("This is a super call");
}
public void setI(int i) {
this.i = 2 * i;
}
}
class B extends A {
public B() {
super(10);
System.out.println("i from B is " + i);
}
public void setI(int i) {
this.i = 3 * i;
}
}
Output will be :
This is a super call
i from B is 7
I have the following scenario :
public class A {
private int x = 5;
public void print()
{
System.out.println(x);
}
}
public class B extends A {
private int x = 10;
/*public void print()
{
System.out.println(x);
}*/
public static void main(String[] args) {
B b = new B();
b.print();
}
}
On executing the code, the output is : 5.
How to access the child class(B's) variable(x) via the parent class method?
Could this be done without overriding the print() method (i.e. uncommenting it in B)?
[This is important because on overriding we will have to rewrite the whole code for the print() method again]
EDITED
More Clarification :-
The motive of the question is to use the value of a child class private variable from its parent class method. This doesn't require changing the value of the parent class private variable in order to achieve the desired result.
The answers posted here, though, led me to my desired answer, which I have posted below.
(Thanks all for your time and help )
class A {
private int x = 5;
protected int getX() {
return x;
}
protected void setX(int x) {
this.x = x;
}
public void print() {
// getX() is used such that
// subclass overriding getX() can be reflected in print();
System.out.println(getX());
}
}
class B extends A {
public B() {
// setX(10); // perhaps set the X to 10 in constructor or in main
}
public static void main(String[] args) {
B b = new B();
b.setX(10);
b.print();
}
}
EDITED
Below is a general answer using abstract class and method to solve similar scenario:
abstract class SuperA {
protected abstract Object getObj();
public void print() {
System.out.println(getObj());
}
}
class A extends SuperA {
#Override
protected Object getObj() {
// Your implementation
return null; // return what you want
}
}
class B extends A {
#Override
protected Object getObj() {
// Your implementation
return null; // return what you want
}
public static void main(String[] args) {
B b = new B();
b.print();
}
}
After reading all the answers posted here, I got what I was looking for. The following is what I feel is the best answer for my question :
public class A {
private int x = 5;
protected int getX(){
return x;
}
public void print(){
System.out.println(getX());
}
}
public class B extends A {
private int x = 10;
protected int getX(){
return x;
}
public static void main(String[] args) {
B b = new B();
b.print();
}
}
Setting up a protected getter and overriding it is better than overriding the print() method itself, as there could be any other huge method in place of the print method which might need to access the value of the child class variable(s).
To solve your question you have to define the fields in the parent class A like protected (so it will be inherited on the child class) and set the field value x inside the constructor in the child class B. The print method is also inherited from A class so you can invoke it directly from parent class.
I hope this can help you.
public class A
{
// fields declaration
protected int x = 5;
public void print()
{
System.out.println(x);
}
}
public class B extends A
{
public B()
{
// set child x value. The field have been defined in the parent class
x = 10;
}
public static void main(String[] args)
{
A a = new A();
a.print(); // print 5
B b = new B();
b.print(); // print 10
}
}
You can always add it to the constructor:
public class B extends A {
//this line is unnecessary: private int x = 10;
/*public void print()
{
System.out.println(x);
}*/
public B()
{
x=10;
}
public static void main(String[] args) {
B b = new B();
b.print();
}
}
The reason it won't work as you try it is that default values only get evaluated once. So when it's default 5 in A, it stays 5 even though you used default 10 in B.
You should expose a getter for the value you want and override that in the child class.
Like so:
public class A {
private int x = 5;
public void print()
{
System.out.println(getX());
}
protected void setX(int x)
{
this.x = x;
}
protected int getX()
{
return x;
}
}
public class B extends A {
/*public void print()
{
System.out.println(x);
}*/
public B()
{
setX(10);
}
public static void main(String[] args) {
B b = new B();
b.print();
}
}
Ex
class A () {
class A(int a, int b) {
}
}
class B extends A {
int m;
int n;
class B()
{
getInput(); // i wanna invoke this method first before calling super(). But it does not allow in Java. How to work around this ?
super(m,n);
}
public void getInput() {
Scanner scanner = new Scanner(System.in);
m = scanner.nextInt();
n = scanner.nextInt();
}
public static void main () {
B b = new B();
}
}
You can force your super class to run a method at the beginning of its constructor and then override that method in the subclass. Many frameworks have a "setup" type method that you can override to accomplish such things.
public class A {
protected int a; // 'protected' so subclass can see it
protected int b;
public A() {
setup(); // Runs whatever setup method is implemented, even in subclasses
}
protected void setup() { /* nothing */ } // 'protected' to be overridden by subclass
}
public class B extends A {
public B()
{
super();
}
/**
* When A's constructor calls setup(), this method will run.
*/
#Override
protected void setup() {
Scanner scanner = new Scanner(System.in);
a = scanner.nextInt(); // Stores value in A's protected variable.
b = scanner.nextInt();
}
}
Depending on the specifics of the classes you're writing, this is where you might have multiple constructors, public or protected methods for setting values, etc. This is where Java is fairly flexible. As the comments below indicate, this isn't a very good practice in constructors, but I'd need more context to figure out how to accomplish what you're asking.
You could chain multiple constructors together as jbrookover alluded to in such a manner. Sligtly convoluted though:
class A () {
public A(int a, int b) {
}
}
class B extends A {
int m;
int n;
public B()
{
this(new Scanner(System.in));
}
private B(Scanner scanner) {
this(scanner.nextInt(),scanner.nextInt())
}
private B(int m, int n) {
super(m,n)
this.m = m;
this.n = n;
}
public static void main (String ... args) {
B b = new B();
}
}