public class Test3 {
public static void main(String[] args) {
Derived04 b = new Derived04();
Base04 a = (Base04)b;
System.out.println(a.i);
System.out.println(a.f());
}
}
class Base04 {
int i=1;
public int f() {return i;}
}
class Derived04 extends Base04 {
int i = 2;
public int f(){return -i;}
}
Both classes include the f() method so when I cast d onto a which method does it go to when I call a.f()? I thinking since a is declared as Base04 type when you do a.i it gives 1 but then why does it use the method from the subclass?
The whole point is about implementing and casting. so when you create a Derived04 then you have implemented a Derived04 class, not its parents Base04 so even if you cast it like that it will run the implemented method not the casting method unless there is a different between f() in both classes. and that's not our case.
why does it use the method from the subclass?
You create an object of Derived04 class you just cast it to the type Base04. It is still Derived04 implementation.
Related
Is there a way to get the methods of an Class object, without getting the methods defined by the Object class?
Right now I am using getDeclaredMethods() to look for a specific method with a list of parameters.
My problem is that this also returns functions like
"equals", "hashCode", etc...
and thus it could be ambiguous between these functions and the one I'm looking for.
Looking at the documentation, it says that it only returns the public methods defined by this class (or in my case an interface), and my objects never override these methods
Is there any workaround for this?
Example:
class Test implements ITest {
void myMethod() {...}
}
and in my code I have something like
Object o = new Test();
for (Method m : o.getClass().getDeclaredMethods()) {
System.out.println(m.getName()...);
}
and this prints me methods defined in the Object class
getMethods()returns all methods declared by a class, plus methods declared in the object class. You can use getDeclaredMethods() to access methods that are class specific or implemented from super class. Example:
interface IChild {
int child1(int a, int b);
}
static class ChildImpl implements IChild {
#Override
public int child1(int a, int b) {
return a*b;
}
public int child2(int x, int y){
return x+x*y;
}
}
public static void main(String[] args) {
IChild i=new ChildImpl();
System.out.println("===Non Object Methods===");
for(Method m:i.getClass().getDeclaredMethods()){
System.out.println("Declared Method"+m.getName());
}
System.out.println("===All Methods===");
for(Method m:i.getClass().getMethods()){
System.out.println("Method: "+m.getName());
}
}
}
Sample output:
===Non Object Methods===
Declared Method: child2
Declared Method: child1
===All Methods===
Method: child2
Method: child1
Method: wait
Method: wait
Method: wait
Method: equals
Method: toString
Method: hashCode
Method: getClass
Method: notify
Method: notifyAll
I have a super class named TestSuper
public class TestSuper {
int a = 0;
}
and I have 2 sub classes named TestSub and TestSub2 that extend TestSuper
public class TestSub extends TestSuper{
int a=1;
}
public class TestSub2 extends TestSuper{
int a=2;
}
in my main class i created a method that takes in a type TestSuper and returns the a value of it and in the main i display it on the console
public class Main {
public static void main(String[] args){
System.out.println(test(new TestSub())+" "+test(new TestSub2()));
}
public static int test(TestSuper b){
return b.a;
}
}
but the output is "0 0" instead of "1 2", what do I do?
You need to cast the reference so say which one you want.
public static int test(TestSuper b){
return b instanceof TestSub ? ((TestSub) b).a :
b instanceof TestSub2 ? ((TestSub2) b).a :
b.a;
}
If this seems needlessly complicated, it is. You should use polymorphism instead.
public class TestSuper {
int a = 0;
public int getA() { return a; }
}
public class TestSub extends TestSuper {
int a = 1;
public int getA() { return a; }
}
public class TestSub2 extends TestSuper {
int a = 2;
public int getA() { return a; }
}
public static int test(TestSuper b) {
return b.getA();
}
First understand the difference between hiding and overriding: https://docs.oracle.com/javase/tutorial/java/IandI/override.html
Then create a getter method in the base-class which you can override in the subclass.
You can look into the theory behind this, and then do the only reasonable thing -forget about writing such kind of code.
In good OOP you consider your fields to be part of your "secret" internal implementation. You don't use fields of sub classes in the super class context. Period.
You are even very conservative about making a field protected in the superclass and to use that in subclasses.
When you call test method like this:
test(new TestSub())+" "+test(new TestSub2())
You use upcasting. Upcasting seperates interface and implementation for an object. But for seperating interface and implementation and achieving true implementation in polymorphism, you must use polymorphic structures. The instance variables aren't polymorphic. Because of this, actually you call a variable which is in TestSuper class.
Only instance methods are polymorphic.
I'm new to java. Recently I saw some code which was similiar to this:
class A {
protected int myInt;
public static void main(String[] args) {
B b = new B();
b.myFunction();
}
}
class B extends A {
public void myFunction() {
this.myInt = 10;
}
}
As far as I know, when creating a subclass instance, an instance of its parent is created as well. All protected and public members of base class are accessible from the subclass.
If I override myInt there will be a difference between this.myInt to super.myInt because each class will have its own myInt (B will have access to both).
So, my question is: if I don't override myInt, which form is preferable, this.myInt or super.myInt?
You only need to use this or super when need to specify which scope are you using/referring to. In your case, I'll prefer to omit the this to simplify the readability.
super is used to represents the current instante of a parent class while this is used to represents the current class. You only need to used this or super if some variable or method overlaps (Have the same name) with one in a wide scope.
eg. If you have define a method parameter with the same name as class attribute, you need to use this to indicate that you are using the class attribute and not the method parameter.
public class A {
public int myInt = 1;
public static void main(String[] args) {
B b = new B();
b.myFunction(3);
}
}
class B extends A {
public int myInt = 2;
public void myFunction(int myInt){
System.out.println(myInt); // The parameter
System.out.println(this.myInt); // myInt from the current class (B)
System.out.println(super.myInt); // myInt from the parent class (A)
}
}
This example will print:
3
2
1
If you don't have this kind of collission, the use of this is optional:
public void myFunction2(){
System.out.println(myInt); // Both refers to the same
System.out.println(this.myInt); // variable myInt from class B
}
It's a matter of taste and the project's standards/guidelines more than anything else.
Personally, I wouldn't use either, and would just write myInt = 10.
Only one instance is created. If you instantiate a derived object, the parents constructor is called, but only one object is created. Also, the term this is more so used when there are different variables with the same name being referenced in a class.
For example a simple constructor:
class SupClass{
public int a = 1;
int incA(){
return ++a;
}
}
class MyClass extends SupClass {
public int a = 10;
public int b = 20;
MyClass() {};
MyClass(int a, int b){
this.a = a;
this.b = b;
}
int incA(){
return ++a;
}
public static void main(String args[])
{
SupClass d = new MyClass();
System.out.println(d.a); //1, members known of type SupClass at compile-time,
System.out.println(d.incA()); //11, methods are virtual, decided at run-time
}
}
Only use the super method when you want to explicitly use the value that is in the super class. To answer your question, only methods can be overwritten, member variables can not.
I know that we cannot override static methods in Java, but can someone explain the following code?
class A {
public static void a() {
System.out.println("A.a()");
}
}
class B extends A {
public static void a() {
System.out.println("B.a()");
}
}
How was I able to override method a() in class B?
You didn't override anything here. To see for yourself, Try putting #Override annotation before public static void a() in class B and Java will throw an error.
You just defined a function in class B called a(), which is distinct (no relation whatsoever) from the function a() in class A.
But Because B.a() has the same name as a function in the parent class, it hides A.a() [As pointed by Eng. Fouad]. At runtime, the compiler uses the actual class of the declared reference to determine which method to run. For example,
B b = new B();
b.a() //prints B.a()
A a = (A)b;
a.a() //print A.a(). Uses the declared reference's class to find the method.
You cannot override static methods in Java. Remember static methods and fields are associated with the class, not with the objects. (Although, in some languages like Smalltalk, this is possible).
I found some good answers here: Why doesn't Java allow overriding of static methods?
That's called hiding a method, as stated in the Java tutorial Overriding and Hiding Methods:
If a subclass defines a class method with the same signature as a
class method in the superclass, the method in the subclass hides the
one in the superclass.
static methods are not inherited so its B's separate copy of method
static are related to class not the state of Object
You didn't override the method a(), because static methods are not inherited. If you had put #Override, you would have seen an error.
A.java:10: error: method does not override or implement a method from a supertype
#Override
^
1 error
But that doesn't stop you from defining static methods with the same signature in both classes.
Also, the choice of method to call depends on the declared type of the variable.
B b = null;
b.a(); // (1) prints B.a()
A a = new B();
a.a(); // (2) prints a.a()
At (1), if the system cared about the identity of b, it would throw a NPE. and at (2), the value of a is ignored. Since a is declared as an A, A.a() is called.
Your method is not overridden method. you just try to put #Override annotation before your method in derived class. it will give you a compile time error. so java will not allow you to override static method.
While goblinjuice answer was accepted, I thought the example code could improved:
public class StaticTest {
public static void main(String[] args) {
A.print();
B.print();
System.out.println("-");
A a = new A();
B b = new B();
a.print();
b.print();
System.out.println("-");
A c = b;
c.print();
}
}
class A {
public static void print() {
System.out.println("A");
}
}
class B extends A {
public static void print() {
System.out.println("B");
}
}
Produces:
A
B
-
A
B
-
A
If B had overridden print() it would have write B on the final line.
Static methods will called by its Class name so we don't need to create class object we just cal it with class name so we can't override static
for example
class AClass{
public static void test(){
}
}
class BClass extends AClass{
public static void test(){}
}
class CClass extends BClass{
public static void main(String args[]){
AClass aclass=new AClass();
aclass.test(); // its wrong because static method is called
// by its class name it can't accept object
}
}
we just call it
AClass.test();
means static class can't be overridden
if it's overridden then how to cal it .
Static members belong to class not to any objects. Therefore static methods cannot be overriden. Also overiding happens at run time therefore compiler will not complain.
Howeve, you can add #Override annotation to method. This will flag compiler error.
Why can't I cast a base class instance to a derived class?
For example, if I have a class B which extends a class C, why can't I do this?
B b=(B)(new C());
or this?
C c=new C();
B b=(B)c;
Alright let me be more specific as to what I'm trying to do. Here's what I have:
public class Base(){
protected BaseNode n;
public void foo(BaseNode x){
n.foo(x);
}
}
public class BaseNode(){
public void foo(BaseNode x){...}
}
Now I want to create a new set of classes which extend Base and Basenode, like this:
public class Derived extends Base(){
public void bar(DerivedNode x){
n.bar(x);//problem is here - n doesn't have bar
}
}
public class DerivedNode extends BaseNode(){
public void bar(BaseNode){
...
}
}
So essentially I want to add new functionality to Base and BaseNode by extending them both, and adding a function to both of them. Furthermore, Base and BaseNode should be able to be used on their own.
I'd really like to do this without generics if possible.
Alright so I ended up figuring it out, partly thanks to Maruice Perry's answer.
In my constructor for Base, n is instantiated as a BaseNode. All I had to do was re-instantiate n as a DerivedNode in my derived class in the constructor, and it works perfectly.
because if B extends C, it means B is a C and not C is a B.
rethink what you are trying to do.
The existing answers are fine in terms of an abstract argument, but I'd like to make a more concrete one. Suppose you could do that. Then this code would have to compile and run:
// Hypothetical code
Object object = new Object();
InputStream stream = (InputStream) object; // No exception allowed?
int firstByte = stream.read();
Where exactly would the implementation of the read method come from? It's abstract in InputStream. Where would it get the data from? It simply isn't appropriate to treat a bare java.lang.Object as an InputStream. It's much better for the cast to throw an exception.
In my experience it's tricky to get "parallel class hierarchies" like the one you're describing to work. You may find that generics help, but it can get hairy very quickly.
You need to use the instanceof keyword to check the type of object referenced by n and typecast the object and call the bar() method. Checkout Derived.bar() method bellow
public class Test{
public static void main(String[] args){
DerivedNode dn = new DerivedNode();
Derived d = new Derived(dn);
d.bar( dn );
}
}
class Base{
protected BaseNode n;
public Base(BaseNode _n){
this.n = _n;
}
public void foo(BaseNode x){
n.foo(x);
}
}
class BaseNode{
public void foo(BaseNode x){
System.out.println( "BaseNode foo" );
}
}
class Derived extends Base{
public Derived(BaseNode n){
super(n);
}
public void bar(DerivedNode x){
if( n instanceof DerivedNode ){
// Type cast to DerivedNode to access bar
((DerivedNode)n).bar(x);
}
else {
// Throw exception or what ever
throw new RuntimeException("Invalid Object Type");
}
}
}
class DerivedNode extends BaseNode{
public void bar(BaseNode b){
System.out.println( "DerivedNode bar" );
}
}
You can create a constructor for B that takes C as a parameter.
See this post for ideas to do what you're trying to do.
Base classes shouldn't know anything about classes derived from them, otherwise the problems highlighted above will arise. Downcasting is a 'code smell', and downcasting in the base class to a derived class is particularly 'smelly'. Such designs can lead to difficult to resolve circular dependencies too.
If you want a base class to make use of derived class implementations use the Template method pattern i.e add a virtual or abstract method in your base class and override and implement it in the derived class. You can then safely call this from the base class.
You can't do that because C does not necessarily implement the behaviours you created when you extended it in B.
So, say C has a method foo(). Then you know that you can call foo() on a B, as B extends C, so you can cast accordingly a treat a B as if it was a C with (C)(new B()).
However - if B has a method bar(), nothing in the subclass relationship says that you can call bar() on C too. Thus you cannot treat a C as if it were a B, and so you cannot cast.
In your exemple, you can cast n into a DerivedNode if you are certain that n is an instance of DerivedNode, or you can use generics:
public class Base<N extends BaseNode> {
protected N n;
public void foo(BaseNode x){
n.foo(x);
}
}
public class BaseNode {
public void foo(BaseNode x){...}
}
public class Derived extends Base<DerivedNode> {
public void bar(DerivedNode x){
n.bar(x); // no problem here - n DOES have bar
}
}
public class DerivedNode extends BaseNode {
public void bar(BaseNode){
...
}
}
Because if B extends C, then B might have stuff that isn't in C (like instance variables you initialize in the constructor that are not in new C())