I've inherited some code, and I'm trying to refactor it in an attempt to clean it up.
This is a simplified case of how I received the code:
public class Foo
{
private void doSomething()
{
someOtherAction(Foo.this);
}
protected class FooChildClass
{
protected void anotherAction()
{
createDialogBox(Foo.this);
}
}
}
//totally not related to class Foo even though most code is duplicated
public class Bar
{
private void doSomething()
{
someOtherAction(Bar.this);
}
protected class BarChildClass
{
protected void anotherAction()
{
createDialogBox(Bar.this);
}
}
}
Now, here's the catch - after further examination of the code, Bar is almost a carbon-copy of Foo with some slight changes here and there.
I'd like to condense my code into something like this:
//superclass
public class Foo
{
protected void doSomething()
{
someOtherAction(Foo.this);
}
protected class ChildClass
{
protected void anotherAction()
{
createDialogBox(Foo.this);
}
}
}
//inherits from Foo and only overrides when necessary
public class Bar extends Foo
{
//rely on base implementation of doSomething
//override other functions only when necessary
}
Will my condensed code run successfully? I have a C# background, so I understand a good bit of Java syntax, but C# doesn't have anything similar to ClassName.this.
I'm figuring that ClassName.this is meant to resolve scoping issues by defining what 'this' is intended - and that with my modified code, Bar is indeed a Foo object too - but I figure I would ask to be sure!
Edited original question for clarification
Foo.this is equivalent to this. Therefore your condensed code is Okay with no requirement of being changed.
The following test code is to test your updated program:
Foo f = new Foo();
Foo.ChildClass fc = f.new ChildClass();
f.doSomething();
fc.anotherAction();
Bar b = new Bar();
Bar.ChildClass bc = b.new ChildClass();
b.doSomething();
bc.anotherAction();
The output is:
Foo#52e922
Foo#52e922
Bar#25154f
Bar#25154f
Referring to your most recent edit I composed the following example:
public class Foo {
public void printMe() {
System.out.println("I'm foo");
}
public class Child {
public void printMe() {
Foo.this.printMe();
}
}
}
public class Bar extends Foo {
#Override
public void printMe() {
System.out.println("I'm bar");
}
}
public class Main {
public static void main(String[] args) {
new Bar().new Child().printMe();
}
}
The result was:
I'm bar
Which concludes, that Foo.this.printMe() does not statically refer to the outer class Foo, but to the outer instance of the Child instance, which in this case is an instance of Bar.
For further reading:
The Classname.this syntax is used for inner classes.
You can use the following example to try a bit around and see its effects:
public class Foo {
private int value = 1;
public class Bar {
private int value = 2;
public void printValue() {
System.out.println(Foo.this.value);
}
}
}
Regarding derivation, there's no way to explicitly refer to a certain super type in java, like in C. However, you can refer to your super class using the super keyword.
With java 1.8 and default interface methods, the Classname.super syntax was introduced:
public interface Foo {
default int getValue() {
return 1;
}
}
public interface Bar {
default int getValue() {
return 2;
}
}
public class Impl implements Foo, Bar {
#Override
public int getValue() {
return Bar.super.getValue();
}
}
However, you can still only refer to direct super types. Everything other will be denied by the compiler.
I am not sure how am I suppose to go about my question. It is about Android can Instantiate Interface. I am trying to do in C#. Now I am pretty sure that the rules for both Java and C# is you can't create an Instance of abstract and Interface as being said.
But I would really like to know how Android does this practice.
In Android you can do this.
public interface Checkme{
void Test();
void Test2();
}
public void myFunc(Checkme my){
//do something
}
// Now this is the actual usage.
public void Start(){
myFunc(new Checkme(){
#Override
public void Test()
{
}
#Override
public void Test2()
{
}
});
}
Actually once you press Enter on new Checkme() You will automatically get the Override methods of the Interface. Like auto Implement method of an Interface in C#.
I hope my question make sense.
C# doesn't support anonymously auto-implemented interfaces because it has delegates:
public void Foo(Func<string> func, Action action) {}
// call it somewhere:
instance.Foo(() => "hello world", () => Console.WriteLine("hello world"));
With delegates you can fill the gap and it can be even more powerful than implementing interfaces with anonymous classes.
Learn more about delegates.
This is an Anonymous Class:
public void Start(){
myFunc(new Checkme() {
#Override
public void Test() {
}
#Override
public void Test2() {
}
});
}
An anonymous class is an unnamed class implemented inline.
You could also have done it using a Local Class, but those are rarely seen in the wild.
public void Start(){
class LocalCheckme implements Checkme {
#Override
public void Test() {
}
#Override
public void Test2() {
}
}
myFunc(new LocalCheckme());
}
These both have the advantage that they can use method parameters and variables directly, as long as they are (effectively) final.
As a third option, you could do it with an Inner Class.
private class InnerCheckme implements Checkme {
#Override
public void Test() {
}
#Override
public void Test2() {
}
}
public void Start(){
myFunc(new InnerCheckme());
}
An inner class cannot access method variables (obviously because it's outside the method), but can be used by multiple methods.
Any local values from the method can however be passed into the constructor and stored as fields of the inner class, to get the same behavior. Just requires a bit more code.
If the inner class doesn't need access to fields of the outer class, it can be declared static, making it a Static Nested Class.
So, all 3 ways above a very similar. The first two are just Java shorthands for the third, i.e. syntactic sugar implemented by the compiler.
C# can do the third one, so just do it that way for C#.
Of course, if the interface only has one method, using a Java lambda or C# delegate is much easier than Anonymous / Local / Inner classes.
If I understand correcly, you're defining a class that implements an interface, and when you specify that the class implements an interface, you want it to automatically add the interface's methods and properties.
If you've declared this:
public interface ISomeInterface
{
void DoSomething();
}
And then you add a class:
public class MyClass : ISomeInterface // <-- right-click
{
}
Right-click on the interface and Visual Studio will give you an option to implement the interface, and it will add all the interface's members to the class.
you mean something like this?
pulic interface Foo{
void DoSomething();
}
public class Bar : Foo {
public void DoSomething () {
//logic here
}
}
myFunc(new Checkme(){
#Override
public void Test()
{
}
#Override
public void Test2()
{
}
});
You're passing into myFunc() something that is called an anonymous class. When it says "new Checkme() { .... }", it is defining an anonymous implementation of the Checkme interface. So, it's not an instance of the interface itself, just an instance of a type that implements it.
In C# anonymously implemented classes for Interface are not auto generated just like in java, you need to follow the below procedure to workout.
public class MyClass {
public void someMethod (string id, IMyInterface _iMyInterface) {
string someResponse = "RESPONSE FOR " + id;
_iMyInterface.InterfaceResponse (someResponse);
}
}
public interface IMyInterface {
void InterfaceResponse (object data);
void InterfaceResponse2 (object data, string x);
}
public class MyInterfaceImplementor : IMyInterface {
private readonly Action<object> actionname;
private readonly Action<object, string> actionInterfaceResponse2;
public MyInterfaceImplementor (Action<object> InterfaceResponse) {
this.actionname = InterfaceResponse;
}
public MyInterfaceImplementor(Action<object> interfaceResponseMethod, Action<object, string> interfaceResponseMethod1) {
this.actionname = interfaceResponseMethod ?? throw new ArgumentNullException(nameof(interfaceResponseMethod));
this.actionInterfaceResponse2 = interfaceResponseMethod1 ?? throw new ArgumentNullException(nameof(interfaceResponseMethod1));
}
public void InterfaceResponse (object data) {
this.actionname (data);
}
public void InterfaceResponse2(object data, string x) {
this.actionInterfaceResponse2(data, x);
}
}
Gist Source : https://gist.github.com/pishangujeniya/4398db8b9374b081b0670ce746f34cbc
Reference :
I've inherited an abstract parent class with only a single child class, both in
separate files.
They're big and ugly. And they're no-longer structured correctly.
I'm going to refactor them to be much nicer - but as part of this I'd
first like to collapse them into a single class. I.e. to go from something like this:
public abstract class ParentClass implements SomeInterface {
ParenClass(SomeOtherClass v) {
aMember = v;
}
#NonNull final SomeOtherClass aMember;
abstract void doSomething();
// Several hundred lines of other functions and members.
}
public class ChildClass extends ParentClass {
ChildClass(AnotherClass v2) {
super(v2.getSomething());
this.childMember = v2.somethingElse();
}
#NonNull final SomeOtherClass aMember;
#Override void doSomething() {
//...
}
// Several hundred lines of other functions and members
}
To a merged version of both classes (Could actually be named like either
of the original classes though...)
public class MergedParentAndChild implements SomeInterface {
JustOneClass(AnotherClass v2) {
this.aMember = v2.getSomething();
this.childMember = v2.somethingElse();
}
void doSomething() {
//...
}
// Lots of other functions etc.
}
I've done this by hand using the "pull up" refactor on all members but it's time consuming as
it gets constructors wrong and has issues with final members and abstract functions etc.
Maybe eclipse probably has a quick way to do this directly in its lists of refactorings. I just can't seem to find it.
Is there a quick and easy way to do this?
Using the "pull up" refactor isn't as repetitive/painful as I'd initially thought. I'm going to borrow JClassic's example and modify it a little (to make it worse) and show what happens
Here's the original code.
public class Temp {
public static interface I1 {
public void interfaceFn1();
public void interfaceFn2();
}
public static abstract class A implements I1 {
int a;
public void foo(){
System.out.printf("%s In A a=%d\n", this, a);
}
#Override public void interfaceFn1() {
System.out.println("In A.interfaceFn1");
}
}
public static class AA extends A {
int aa;
#Override
public void foo(){
super.foo();
System.out.printf("%s In AA a=%d, aa=%d\n", this, a , aa);
}
#Override public void interfaceFn2() {
System.out.println("In A.interfaceFn1");
}
}
}
If you then put the cursor on one of the functions in AA and select the "
"pull up" refactor. (Alt+Shift+T, U) - you get the following dialog.
Then click select all on the right, and next.
You get warnings about duplicate functions, which will cause compile errors, in a dialog that looks like this. You can just ignore these, and fix them later.
Then going to finish you get code that looks like this:
public class Temp {
public static interface I1 {
public void interfaceFn1();
public void interfaceFn2();
}
public static abstract class A implements I1 {
int a;
int aa;
public void foo(){
System.out.printf("%s In A a=%d\n", this, a);
}
#Override public void interfaceFn1() {
System.out.println("In A.interfaceFn1");
}
#Override public void foo() {
super.foo();
System.out.printf("%s In AA a=%d, aa=%d\n", this, a , aa);
}
#Override public void interfaceFn2() {
System.out.println("In A.interfaceFn1");
}
}
public static class AA extends A {
}
}
You can now remove the class AA and fix the errors in A (duplication of foo() being the main issue) . Resulting in code like this:
public class Temp {
public static interface I1 {
public void interfaceFn1();
public void interfaceFn2();
}
public static abstract class A implements I1 {
int a;
int aa;
private void fooBase(){
System.out.printf("%s In A.fooBase a=%d\n", this, a);
}
#Override public void interfaceFn1() {
System.out.println("In A.interfaceFn1");
}
public void foo() {
fooBase();
System.out.printf("%s In AA a=%d, aa=%d\n", this, a , aa);
}
#Override public void interfaceFn2() {
System.out.println("In A.interfaceFn1");
}
}
}
The biggest remaining problem is to switch all creations/references of AA to A. But if you selected the use destination type where possible that will reduce the number of places you will need to fix.
I suspect there are still cases where constructors cause some issues, but this seems OK for most cases.
It is really hard (if not impossible) to automate the combination of two hierarchy classes: consider the following: you have two classes
public static class A {
int a;
public void foo(){
//do stuff
}
}
public static class AA extends A {
int aa;
#Override
public void foo(){
//do stuff
}
}
When you merge them, you run into the problem:
public static class AMerged{
int a;
int aa; //this is fine
public void foo(){ //but use who's implementation?
//do what?
}
}
With this, you run into the problem of method names being the same name, same with variables being the same, who's implementation will I use? The child or parent? As you can see, there are lots of problems in merging two classes you will run into, so it is nearly impossible to automate.
Even with abstract classes there is no way to combine them other than to do it manually
Actually, this can be done by class inline.
First, you need to pull up subclass method to parent class;
Then, you can inline the subclass. (Intellij Idea, place cursor on subclass, refactor->inline. Intellij Idea will remove subclass, and replace subclass with parent class)
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.