What I would like is to be able to create a variable with the superclass and then 'turn' it into the subclass later & be able to access the subClasses methods. It will always be defined to a subclass I just don't know which when defining.
class SuperClass {
//superclass methods
}
class SubClass1 extends SuperClass {
void subClass1method() {
//do something
}
}
class SubClass2 extends SuperClass {
void subClass2method() {
//do something different
}
}
class Main {
public static void main ( String [] args ) {
SuperClass a;
if (*some reson to be SubClass1*) {
a = new SubClass1();
a.subClass1Method(); //this is undefined it says
//a instanceof SubClass1 returns true which is what confused me
} else {
a = new SubClass2();
a.subClass2Method(); //this is undefined it says
}
}
}
I am not asking why it does this but the correct way to work around it & get the results I want.
I have looked for duplicates but could have easily missed one so please let me know.
When you don't have control of the source then there's not too much you can do besides casting (or possibly using adapters). If instanceof and casting sound familiar from your lessons then that's probably what your prof is expecting.
Here are some of your options:
// to execute those subclass specific methods without casting..
public static void main ( String [] args ) {
SuperClass superClass;
if (true /*some reson to be SubClass1*/) {
SubClass1 subClass1 = new SubClass1();
subClass1.subClass1Method();
superClass = subClass1;
} else {
SubClass2 subClass2 = new SubClass2();
subClass2.subClass2Method();
superClass = subClass2;
}
// ... ...
}
// using instanceof and casting
public static void main ( String [] args ) {
SuperClass superClass;
if (true /*some reson to be SubClass1*/) {
superClass = new SubClass1();
} else {
superClass = new SubClass2();
}
// ... ... ...
if(superClass instanceof SubClass1) {
((SubClass1) superClass).subClass1Method();
} else if(superClass instanceof SubClass2) {
((SubClass2) superClass).subClass2Method();
}
}
Below is my recommendation to achieve this goal when you have control of the source..
Rather than a new class for different behaviors you could just use composition to provide different behaviors to the one class. Note that this is especially convenient if you understand Functional Interfaces and Method References.
public class SuperClass {
private MyBehavior behavior;
public void setBehavior(MyBehavior behavior) { this.behavior = behavior; }
public void doBehavior() { this.behavior.doBehavior(); }
}
public interface MyBehavior { public void doBehavior(); }
public class MyGoodBehavior implements MyBehavior {
#Override public void doBehavior() { System.out.print("good behavior"); }
}
public class MyBadBehavior implements MyBehavior {
#Override public void doBehavior() { System.out.print("bad behavior"); }
}
public static void main(String[] args) {
SuperClass a = new SuperClass();
a.setBehavior(new MyBadBehavior());
a.doBehavior();
}
It's simple.
If you define "a" of class "SuperClass" you can only access "SuperClass" methods!
Eventually you could add abstract methods "subClass1method" and "subClass2method" to "SuperClass" and implement them in the subclasses.
When you do SuperClass a you tell the compiler that you intend to store objects of type SuperClass in the variable a. It means that you can store subclasses of SuperClass as well, but the compiler only sees a as an object of type SuperClass, regardless if you are storing a sub type.
At runtime there's no check for this, if you manage to write code that calls methods from a subclass on a, they will be called. If a is not of that particular sub class and don't implement the called method, an exception would be thrown.
But how do you force the compiler to accept that a is of the subclass type? Casting. You cast a to the subclass, then you can access methods from subclass on a. If it´s not even possible to cast an object to a subclass, you get an error at compile time.
Change your Main class to the following and it should do what you want:
class Main {
public static void main ( String [] args ) {
SuperClass a;
if (*some reson to be SubClass1*) {
a = new SubClass1();
((SubClass1)a).subClass1Method(); // Casting 'a' as SubClass1
} else {
a = new SubClass2();
((SubClass2)a).subClass2Method(); // Casting 'a' as SubClass2
}
}
}
A few examples of when casting work and don't work at compile time and runtime:
class SuperClass { }
class SubClass1 extends SuperClass { }
class SubClass2 extends SuperClass { }
class Main {
public static void main(String... args) {
SuperClass a;
SubClass1 b;
SubClass2 c;
b = new SubClass1();
a = b; // autocast to SuperClass
a = new SubClass1();
b = a; // Compile error
b = (SubClass1) a; // explicit cast
c = new SubClass2();
method(b); // works
method(c); // compiles, but will throw exception at runtime inside method
}
private static void method(SuperClass object) {
SubClass1 b = (SubClass1) object; // allowed by compiler but will throw exception at runtime in the second call, method(c)
}
}
How to clone a Java object with the clone() method
I have a question regarding properly implementing the clone() method for a class in java.
I know that this is bad practice, but I need to know this for an exam..
In the above discussion they say to call super.clone() - but I don't udnerstand what happens if the super function doesn't implement Clonable.
For example, say I have a class X that extends Y. X implements Clonable and Y doesnl't. Y's clone() method should throw an Exception. Then what do we do in this case?
All the explanations I could find somehow assume that all superclasses implement Clonable, or at least that's what I understood..
EDIT:
Check out this code please:
public class Employee implements Cloneable {
private String name;
public Employee(String name) {
this.name = name;
}
public String getName() {
return name;
}
public Object clone()throws CloneNotSupportedException{
return (Employee)super.clone();
}
public static void main(String[] args) {
Employee emp = new Employee("Abhi");
try {
Employee emp2 = (Employee) emp.clone();
System.out.println(emp2.getName());
} catch (CloneNotSupportedException e) {
e.printStackTrace();
}
}
}
It is taken from here: https://www.javacodegeeks.com/2018/03/understanding-cloneable-interface-in-java.html
Similar code can be found in many tutorials.
Why can they use super.clone() when the superclass (which in this case is Object) does not implement Clonable - that would result in an Exception.
If you have this structure:
class Y {}
class X extends Y implements Cloneable {
#Override
public X clone() {
try {
return (X) super.clone();
} catch (CloneNotSupportedException e) {
throw new InternalError(e);
}
}
}
Then clone on instances of X will work fine.
It won't work on direct instances of Y, because they are not declared cloneable. But the Cloneable interface on X is an indicator to the mechanisms of the default clone() implementation that they should be made to work.
Alternatively
You could also have a non-Cloneable class with a working clone() method, as long as you didn't rely on the default implementation of clone().
For instance:
class Y {
#Override
public Y clone() {
// Don't call super.clone() because it will error
return new Y(...); // whatever parameters
}
}
However, with this mechanism, if you called super.clone() from a subclass of Y, you would get an instance of Y, which is probably not what you would want.
As others have pointed out, the Cloneable mechanism is awkward and confusing, and usually copying mechanisms using new are easier to work with.
The Cloneable-interface is generally regarded as broken (and won't be fixed). At the core, the argument revolves around the fact that clone() is a method defined on Object, instead of being a method of the interface Cloneable.
I would not recommend using it at all. A better solution would be to provide copy-constructors. If one does not have the capability to fully recreate a parent-class object, then cloning is impossible.
Refactoring the code provided would lead to a result similar to this:
public class Employee implements Cloneable {
private String name;
public Employee(String name) {
this.name = name;
}
public Employee(Employee that) {
this.name = that.name;
}
public static void main(String[] args) {
Employee emp = new Employee("Abhi");
Employee emp2 = new Employee(emp);
System.out.println(emp2.getName());
}
public String getName() {
return name;
}
}
A remark on your code:
public class Employee {
public Object clone()throws CloneNotSupportedException{
return (Employee)super.clone();
}
}
The type cast is superfluous since the methode returns an Object.
I have this condition
public class A {
public action() {
System.out.println("Action done in A");
}
}
public class B extends A {
public action() {
System.out.println("Action done in B");
}
}
when I create an instance of B, the action will do just actions in B, as it overrides the action of the superclass.
the problem is that in my project, the super class A is already used too many times, and I am looking for a way that under certain conditions, when i create an instance of A it makes a check and if it is true, replace itself with B.
public class A {
public A() {
if ([condition]) {
this = new B();
}
}
public action() {
System.out.println("Action done in A");
}
}
A a = new A();
a.action();
// expect to see "Action done in B"...
is this possible in some way?
I would say that doing this:
this = new B();
within the constructor for A would violate OO design principles, even it were possible to do (and it isn't).
That being said, if I were faced with this scenario:
the problem is that in my project, the super class A is already used too many times
I would solve it in one of the two following ways:
I have assumed that your condition is that you do not want too many objects of type A, otherwise feel free to substitute in any other condition.
Option 1 : Use a factory design pattern.
public class AFactory
{
private static count = 0;
private static final MAX_COUNT = 100;
public A newObject() {
if (count < MAX_COUNT) {
count++;
return new A();
} else {
return new B();
}
}
}
And then somehwere else you generate the objects like so:
A obj1 = factory.newObject();
A obj2 = factory.newObject();
Option 2 : Static counter + try&catch
Use a static counter within your A class that keeps track of the number of times A has been instantiated, by incrementing the static variable by one in the constructor. If it hits a limit for the max number of object of type A, throw an InstantiationError in A's constructor.
This would mean that whenever you instantiate A, you have to a try..catch block to intercept the InstantionError, and then create a new object of type B instead.
public class A {
private static count = 0;
private static final MAX_COUNT = 100;
public A() {
if (count > 100) {
throw new InstationError();
}
}
}
And when generating your objects:
A obj1, obj2;
try {
obj1 = new A();
} catch (InstantiationError ie) {
obj1 = new B();
}
try {
obj2 = new A();
} catch (InstantiationError ie) {
obj2 = new B();
}
Option 2 is closest to what you ask directly in the question. However, I would personally choose to use the factory design pattern, because it is much more elegant solution, and it allows you to achieve what you want to do anyway.
Not directly, no. Calling new A() will always create an instance of A. However, you could make the constructor of A protected, and then have a static method:
public static A newInstance() {
// Either create A or B here.
}
Then convert all current calls to the constructor to calls to the factory method.
It is possible to choose whether or not to use a superclass' constructor?
It is not possible to conditionally control whether or not to use a superclass' constructor, as one of the superclass' constructors must be called before constructing one's own object.
From the above, there is a requirement in Java, that the first line of the constructor must call on of the superclass' constructor -- in fact, even if there is no explicit call to a superclass' constructor, there will be an implicit call to super():
public class X {
public X() {
// ...
}
public X(int i) {
// ...
}
}
public class Y extends X {
public Y() {
// Even if not written, there is actually a call to super() here.
// ...
}
}
It should be stressed that it is not possible to call the superclass' constructor after performing something else:
public class Y extends X {
public Y() {
doSomething(); // Not allowed! A compiler error will occur.
super(); // This *must* be the first line in this constructor.
}
}
The alternative
That said, a way to achieve what is desired here could be to use the factory method pattern, which can select the kind of implementation depending on some kind of condition:
public A getInstance() {
if (condition) {
return new B();
} else {
return new C();
}
}
In the above code, depending on condition, the method can return either an instance of B or C (assuming both are a subclass of class A).
Example
The following is a concrete example, using an interface rather than a class.
Let there be the following interfaces and classes:
interface ActionPerformable {
public void action();
}
class ActionPerformerA implements ActionPerformable {
public void action() {
// do something...
}
}
class ActionPerformerB implements ActionPerformable {
public void action() {
// do something else...
}
}
Then, there would be a class which will return one of the above classes depending on a condition which is passed in through a method:
class ActionPeformerFactory {
// Returns a class which implements the ActionPerformable interface.
public ActionPeformable getInstance(boolean condition) {
if (condition) {
return new ActionPerformerA();
} else {
return new ActionPerformerB();
}
}
}
Then, a class which uses the above factory method which returns the appropriate implementation depending on a condition:
class Main {
public static void main(String[] args) {
// Factory implementation will return ActionPerformerA
ActionPerformable ap = ActionPerformerFactory.getInstance(true);
// Invokes the action() method of ActionPerformable obtained from Factory.
ap.action();
}
}
This would be possible if you used a factory method. When you use a constructor: nope, completely impossible.
Assuming you aren't willing to move it to an interface or factory its kind of ugly but you could keep a delegeate copy of B in A and rewrite your methods ot invoke the delegate:
public class A{
B delegate;
public A(){
if([condition]){
delegate = new B()
return;
}
...//normal init
}
public void foo(){
if(delegate != null){
delegate.foo();
return;
}
...//normal A.foo()
}
public boolean bar(Object wuzzle){
if(delegate != null){
return delegate.bar(wuzzle);
}
...//normal A.bar()
}
...//other methods in A
}
I read this question and thought that would easily be solved (not that it isn't solvable without) if one could write:
#Override
public String toString() {
return super.super.toString();
}
I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.
What do you guys think?
EDIT:
To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it.
It violates encapsulation. You shouldn't be able to bypass the parent class's behaviour. It makes sense to sometimes be able to bypass your own class's behaviour (particularly from within the same method) but not your parent's. For example, suppose we have a base "collection of items", a subclass representing "a collection of red items" and a subclass of that representing "a collection of big red items". It makes sense to have:
public class Items
{
public void add(Item item) { ... }
}
public class RedItems extends Items
{
#Override
public void add(Item item)
{
if (!item.isRed())
{
throw new NotRedItemException();
}
super.add(item);
}
}
public class BigRedItems extends RedItems
{
#Override
public void add(Item item)
{
if (!item.isBig())
{
throw new NotBigItemException();
}
super.add(item);
}
}
That's fine - RedItems can always be confident that the items it contains are all red. Now suppose we were able to call super.super.add():
public class NaughtyItems extends RedItems
{
#Override
public void add(Item item)
{
// I don't care if it's red or not. Take that, RedItems!
super.super.add(item);
}
}
Now we could add whatever we like, and the invariant in RedItems is broken.
Does that make sense?
I think Jon Skeet has the correct answer. I'd just like to add that you can access shadowed variables from superclasses of superclasses by casting this:
interface I { int x = 0; }
class T1 implements I { int x = 1; }
class T2 extends T1 { int x = 2; }
class T3 extends T2 {
int x = 3;
void test() {
System.out.println("x=\t\t" + x);
System.out.println("super.x=\t\t" + super.x);
System.out.println("((T2)this).x=\t" + ((T2)this).x);
System.out.println("((T1)this).x=\t" + ((T1)this).x);
System.out.println("((I)this).x=\t" + ((I)this).x);
}
}
class Test {
public static void main(String[] args) {
new T3().test();
}
}
which produces the output:
x= 3
super.x= 2
((T2)this).x= 2
((T1)this).x= 1
((I)this).x= 0
(example from the JLS)
However, this doesn't work for method calls because method calls are determined based on the runtime type of the object.
I think the following code allow to use super.super...super.method() in most case.
(even if it's uggly to do that)
In short
create temporary instance of ancestor type
copy values of fields from original object to temporary one
invoke target method on temporary object
copy modified values back to original object
Usage :
public class A {
public void doThat() { ... }
}
public class B extends A {
public void doThat() { /* don't call super.doThat() */ }
}
public class C extends B {
public void doThat() {
Magic.exec(A.class, this, "doThat");
}
}
public class Magic {
public static <Type, ChieldType extends Type> void exec(Class<Type> oneSuperType, ChieldType instance,
String methodOfParentToExec) {
try {
Type type = oneSuperType.newInstance();
shareVars(oneSuperType, instance, type);
oneSuperType.getMethod(methodOfParentToExec).invoke(type);
shareVars(oneSuperType, type, instance);
} catch (Exception e) {
throw new RuntimeException(e);
}
}
private static <Type, SourceType extends Type, TargetType extends Type> void shareVars(Class<Type> clazz,
SourceType source, TargetType target) throws IllegalArgumentException, IllegalAccessException {
Class<?> loop = clazz;
do {
for (Field f : loop.getDeclaredFields()) {
if (!f.isAccessible()) {
f.setAccessible(true);
}
f.set(target, f.get(source));
}
loop = loop.getSuperclass();
} while (loop != Object.class);
}
}
I don't have enough reputation to comment so I will add this to the other answers.
Jon Skeet answers excellently, with a beautiful example. Matt B has a point: not all superclasses have supers. Your code would break if you called a super of a super that had no super.
Object oriented programming (which Java is) is all about objects, not functions. If you want task oriented programming, choose C++ or something else. If your object doesn't fit in it's super class, then you need to add it to the "grandparent class", create a new class, or find another super it does fit into.
Personally, I have found this limitation to be one of Java's greatest strengths. Code is somewhat rigid compared to other languages I've used, but I always know what to expect. This helps with the "simple and familiar" goal of Java. In my mind, calling super.super is not simple or familiar. Perhaps the developers felt the same?
There's some good reasons to do this. You might have a subclass which has a method which is implemented incorrectly, but the parent method is implemented correctly. Because it belongs to a third party library, you might be unable/unwilling to change the source. In this case, you want to create a subclass but override one method to call the super.super method.
As shown by some other posters, it is possible to do this through reflection, but it should be possible to do something like
(SuperSuperClass this).theMethod();
I'm dealing with this problem right now - the quick fix is to copy and paste the superclass method into the subsubclass method :)
In addition to the very good points that others have made, I think there's another reason: what if the superclass does not have a superclass?
Since every class naturally extends (at least) Object, super.whatever() will always refer to a method in the superclass. But what if your class only extends Object - what would super.super refer to then? How should that behavior be handled - a compiler error, a NullPointer, etc?
I think the primary reason why this is not allowed is that it violates encapsulation, but this might be a small reason too.
I think if you overwrite a method and want to all the super-class version of it (like, say for equals), then you virtually always want to call the direct superclass version first, which one will call its superclass version in turn if it wants.
I think it only makes rarely sense (if at all. i can't think of a case where it does) to call some arbitrary superclass' version of a method. I don't know if that is possible at all in Java. It can be done in C++:
this->ReallyTheBase::foo();
At a guess, because it's not used that often. The only reason I could see using it is if your direct parent has overridden some functionality and you're trying to restore it back to the original.
Which seems to me to be against OO principles, since the class's direct parent should be more closely related to your class than the grandparent is.
Calling of super.super.method() make sense when you can't change code of base class. This often happens when you are extending an existing library.
Ask yourself first, why are you extending that class? If answer is "because I can't change it" then you can create exact package and class in your application, and rewrite naughty method or create delegate:
package com.company.application;
public class OneYouWantExtend extends OneThatContainsDesiredMethod {
// one way is to rewrite method() to call super.method() only or
// to doStuff() and then call super.method()
public void method() {
if (isDoStuff()) {
// do stuff
}
super.method();
}
protected abstract boolean isDoStuff();
// second way is to define methodDelegate() that will call hidden super.method()
public void methodDelegate() {
super.method();
}
...
}
public class OneThatContainsDesiredMethod {
public void method() {...}
...
}
For instance, you can create org.springframework.test.context.junit4.SpringJUnit4ClassRunner class in your application so this class should be loaded before the real one from jar. Then rewrite methods or constructors.
Attention: This is absolute hack, and it is highly NOT recommended to use but it's WORKING! Using of this approach is dangerous because of possible issues with class loaders. Also this may cause issues each time you will update library that contains overwritten class.
#Jon Skeet Nice explanation.
IMO if some one wants to call super.super method then one must be want to ignore the behavior of immediate parent, but want to access the grand parent behavior.
This can be achieved through instance Of. As below code
public class A {
protected void printClass() {
System.out.println("In A Class");
}
}
public class B extends A {
#Override
protected void printClass() {
if (!(this instanceof C)) {
System.out.println("In B Class");
}
super.printClass();
}
}
public class C extends B {
#Override
protected void printClass() {
System.out.println("In C Class");
super.printClass();
}
}
Here is driver class,
public class Driver {
public static void main(String[] args) {
C c = new C();
c.printClass();
}
}
Output of this will be
In C Class
In A Class
Class B printClass behavior will be ignored in this case.
I am not sure about is this a ideal or good practice to achieve super.super, but still it is working.
Look at this Github project, especially the objectHandle variable. This project shows how to actually and accurately call the grandparent method on a grandchild.
Just in case the link gets broken, here is the code:
import lombok.val;
import org.junit.Assert;
import org.junit.Test;
import java.lang.invoke.*;
/*
Your scientists were so preoccupied with whether or not they could, they didn’t stop to think if they should.
Please don't actually do this... :P
*/
public class ImplLookupTest {
private MethodHandles.Lookup getImplLookup() throws NoSuchFieldException, IllegalAccessException {
val field = MethodHandles.Lookup.class.getDeclaredField("IMPL_LOOKUP");
field.setAccessible(true);
return (MethodHandles.Lookup) field.get(null);
}
#Test
public void test() throws Throwable {
val lookup = getImplLookup();
val baseHandle = lookup.findSpecial(Base.class, "toString",
MethodType.methodType(String.class),
Sub.class);
val objectHandle = lookup.findSpecial(Object.class, "toString",
MethodType.methodType(String.class),
// Must use Base.class here for this reference to call Object's toString
Base.class);
val sub = new Sub();
Assert.assertEquals("Sub", sub.toString());
Assert.assertEquals("Base", baseHandle.invoke(sub));
Assert.assertEquals(toString(sub), objectHandle.invoke(sub));
}
private static String toString(Object o) {
return o.getClass().getName() + "#" + Integer.toHexString(o.hashCode());
}
public class Sub extends Base {
#Override
public String toString() {
return "Sub";
}
}
public class Base {
#Override
public String toString() {
return "Base";
}
}
}
Happy Coding!!!!
I would put the super.super method body in another method, if possible
class SuperSuperClass {
public String toString() {
return DescribeMe();
}
protected String DescribeMe() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return "I am super";
}
}
class ChildClass extends SuperClass {
public String toString() {
return DescribeMe();
}
}
Or if you cannot change the super-super class, you can try this:
class SuperSuperClass {
public String toString() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return DescribeMe(super.toString());
}
protected String DescribeMe(string fromSuper) {
return "I am super";
}
}
class ChildClass extends SuperClass {
protected String DescribeMe(string fromSuper) {
return fromSuper;
}
}
In both cases, the
new ChildClass().toString();
results to "I am super super"
It would seem to be possible to at least get the class of the superclass's superclass, though not necessarily the instance of it, using reflection; if this might be useful, please consider the Javadoc at http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Class.html#getSuperclass()
public class A {
#Override
public String toString() {
return "A";
}
}
public class B extends A {
#Override
public String toString() {
return "B";
}
}
public class C extends B {
#Override
public String toString() {
return "C";
}
}
public class D extends C {
#Override
public String toString() {
String result = "";
try {
result = this.getClass().getSuperclass().getSuperclass().getSuperclass().newInstance().toString();
} catch (InstantiationException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
} catch (IllegalAccessException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
}
return result;
}
}
public class Main {
public static void main(String... args) {
D d = new D();
System.out.println(d);
}
}
run:
A
BUILD SUCCESSFUL (total time: 0 seconds)
I have had situations like these when the architecture is to build common functionality in a common CustomBaseClass which implements on behalf of several derived classes.
However, we need to circumvent common logic for specific method for a specific derived class. In such cases, we must use a super.super.methodX implementation.
We achieve this by introducing a boolean member in the CustomBaseClass, which can be used to selectively defer custom implementation and yield to default framework implementation where desirable.
...
FrameworkBaseClass (....) extends...
{
methodA(...){...}
methodB(...){...}
...
methodX(...)
...
methodN(...){...}
}
/* CustomBaseClass overrides default framework functionality for benefit of several derived classes.*/
CustomBaseClass(...) extends FrameworkBaseClass
{
private boolean skipMethodX=false;
/* implement accessors isSkipMethodX() and setSkipMethodX(boolean)*/
methodA(...){...}
methodB(...){...}
...
methodN(...){...}
methodX(...){
if (isSkipMethodX()) {
setSKipMethodX(false);
super.methodX(...);
return;
}
... //common method logic
}
}
DerivedClass1(...) extends CustomBaseClass
DerivedClass2(...) extends CustomBaseClass
...
DerivedClassN(...) extends CustomBaseClass...
DerivedClassX(...) extends CustomBaseClass...
{
methodX(...){
super.setSKipMethodX(true);
super.methodX(...);
}
}
However, with good architecture principles followed in framework as well as app, we could avoid such situations easily, by using hasA approach, instead of isA approach. But at all times it is not very practical to expect well designed architecture in place, and hence the need to get away from solid design principles and introduce hacks like this.
Just my 2 cents...
IMO, it's a clean way to achieve super.super.sayYourName() behavior in Java.
public class GrandMa {
public void sayYourName(){
System.out.println("Grandma Fedora");
}
}
public class Mama extends GrandMa {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName();
}else {
System.out.println("Mama Stephanida");
}
}
}
public class Daughter extends Mama {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName(lie);
}else {
System.out.println("Little girl Masha");
}
}
}
public class TestDaughter {
public static void main(String[] args){
Daughter d = new Daughter();
System.out.print("Request to lie: d.sayYourName(true) returns ");
d.sayYourName(true);
System.out.print("Request not to lie: d.sayYourName(false) returns ");
d.sayYourName(false);
}
}
Output:
Request to lie: d.sayYourName(true) returns Grandma Fedora
Request not to lie: d.sayYourName(false) returns Little girl Masha
I think this is a problem that breaks the inheritance agreement.
By extending a class you obey / agree its behavior, features
Whilst when calling super.super.method(), you want to break your own obedience agreement.
You just cannot cherry pick from the super class.
However, there may happen situations when you feel the need to call super.super.method() - usually a bad design sign, in your code or in the code you inherit !
If the super and super super classes cannot be refactored (some legacy code), then opt for composition over inheritance.
Encapsulation breaking is when you #Override some methods by breaking the encapsulated code.
The methods designed not to be overridden are marked
final.
In C# you can call a method of any ancestor like this:
public class A
internal virtual void foo()
...
public class B : A
public new void foo()
...
public class C : B
public new void foo() {
(this as A).foo();
}
Also you can do this in Delphi:
type
A=class
procedure foo;
...
B=class(A)
procedure foo; override;
...
C=class(B)
procedure foo; override;
...
A(objC).foo();
But in Java you can do such focus only by some gear. One possible way is:
class A {
int y=10;
void foo(Class X) throws Exception {
if(X!=A.class)
throw new Exception("Incorrect parameter of "+this.getClass().getName()+".foo("+X.getName()+")");
y++;
System.out.printf("A.foo(%s): y=%d\n",X.getName(),y);
}
void foo() throws Exception {
System.out.printf("A.foo()\n");
this.foo(this.getClass());
}
}
class B extends A {
int y=20;
#Override
void foo(Class X) throws Exception {
if(X==B.class) {
y++;
System.out.printf("B.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("B.foo(%s) calls B.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
}
class C extends B {
int y=30;
#Override
void foo(Class X) throws Exception {
if(X==C.class) {
y++;
System.out.printf("C.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("C.foo(%s) calls C.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
void DoIt() {
try {
System.out.printf("DoIt: foo():\n");
foo();
Show();
System.out.printf("DoIt: foo(B):\n");
foo(B.class);
Show();
System.out.printf("DoIt: foo(A):\n");
foo(A.class);
Show();
} catch(Exception e) {
//...
}
}
void Show() {
System.out.printf("Show: A.y=%d, B.y=%d, C.y=%d\n\n", ((A)this).y, ((B)this).y, ((C)this).y);
}
}
objC.DoIt() result output:
DoIt: foo():
A.foo()
C.foo(C): y=31
Show: A.y=10, B.y=20, C.y=31
DoIt: foo(B):
C.foo(B) calls C.super.foo(B)
B.foo(B): y=21
Show: A.y=10, B.y=21, C.y=31
DoIt: foo(A):
C.foo(A) calls C.super.foo(A)
B.foo(A) calls B.super.foo(A)
A.foo(A): y=11
Show: A.y=11, B.y=21, C.y=31
It is simply easy to do. For instance:
C subclass of B and B subclass of A. Both of three have method methodName() for example.
public abstract class A {
public void methodName() {
System.out.println("Class A");
}
}
public class B extends A {
public void methodName() {
super.methodName();
System.out.println("Class B");
}
// Will call the super methodName
public void hackSuper() {
super.methodName();
}
}
public class C extends B {
public static void main(String[] args) {
A a = new C();
a.methodName();
}
#Override
public void methodName() {
/*super.methodName();*/
hackSuper();
System.out.println("Class C");
}
}
Run class C Output will be:
Class A
Class C
Instead of output:
Class A
Class B
Class C
If you think you are going to be needing the superclass, you could reference it in a variable for that class. For example:
public class Foo
{
public int getNumber()
{
return 0;
}
}
public class SuperFoo extends Foo
{
public static Foo superClass = new Foo();
public int getNumber()
{
return 1;
}
}
public class UltraFoo extends Foo
{
public static void main(String[] args)
{
System.out.println(new UltraFoo.getNumber());
System.out.println(new SuperFoo().getNumber());
System.out.println(new SuperFoo().superClass.getNumber());
}
public int getNumber()
{
return 2;
}
}
Should print out:
2
1
0
public class SubSubClass extends SubClass {
#Override
public void print() {
super.superPrint();
}
public static void main(String[] args) {
new SubSubClass().print();
}
}
class SuperClass {
public void print() {
System.out.println("Printed in the GrandDad");
}
}
class SubClass extends SuperClass {
public void superPrint() {
super.print();
}
}
Output: Printed in the GrandDad
The keyword super is just a way to invoke the method in the superclass.
In the Java tutorial:https://docs.oracle.com/javase/tutorial/java/IandI/super.html
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super.
Don't believe that it's a reference of the super object!!! No, it's just a keyword to invoke methods in the superclass.
Here is an example:
class Animal {
public void doSth() {
System.out.println(this); // It's a Cat! Not an animal!
System.out.println("Animal do sth.");
}
}
class Cat extends Animal {
public void doSth() {
System.out.println(this);
System.out.println("Cat do sth.");
super.doSth();
}
}
When you call cat.doSth(), the method doSth() in class Animal will print this and it is a cat.