I want an implementation where I need to invoke a method from one of two classes based on a condition. To illustrate, lets say I have two simplistic classes:
public class A implements IObject {
#Override
public void doIt() {
System.out.println("OBJECTA");
}
}
public class B implements IObject {
#Override
public void doIt() {
System.out.println("OBJECTB");
}
}
and an interface
public interface IObject {
void doIt();
}
My method to dynamically call a class function is implemented as:
void call(String s, A a, B b, Consumer<IObject> o) {
if(s.equalsIgnoreCase("CONDITION")) {
o.accept(a);
} else {
o.accept(b);
}
}
I can call the method as
A objectA = new A();
B objectB = new B();
call("CONDITION", objectA, objectB, IObject::doIt);
call("OTHER", objectA, objectB, IObject::doIt);
This will essentially invoke doIt on class A or B depending upon the condition parameter
Is there a cleaner way to achieve this by perhaps reducing the number of parameters an hence the function call signature?
Thanks
For my understanding of your problem:
You have a condition
You have a context to use (object a and b)
You would not ask such a question if there was not an intent to generalize your code for more than one condition or object (a, b). Otherwise, a ternary operator or a if would be far more easier to maintain (eg: ((condition) a:b).doIt() vs the code below).
Why not using a pattern like this:
A a = new A();
B b = new B();
String hint = ...;
List<Executor> list = new ArrayList<>();
list.add(new Executor("CONDITION"::equalsIgnoreCase, a::doIt));
list.add(new Executor(s -> true, b::doIt));
for (Executor executor : list) {
if (executor.process(hint)) {
break;
}
}
With Executor class:
class Executor {
private final Predicate<String> predicate;
private final Runnable runnable;
...
public boolean process(String s) {
if (!predicate.test(s)) {
return false;
}
runnable.run();
return true;
}
}
The loop will evaluate the condition, then run the code if true, otherwise continue onto the next element.
The Executor class is technically not bound to a or b; only the initial setup is.
Here is a complete different approach, utilizing java.lang.reflect.Proxy:
import java.lang.reflect.InvocationHandler;
import java.lang.reflect.Method;
import java.lang.reflect.Proxy;
/**
*
* #author ben
*/
public class Test {
public interface IObject {
void doIt();
}
public static class A implements IObject {
#Override
public void doIt() {
System.out.println("OBJECTA");
}
}
public static class B implements IObject {
#Override
public void doIt() {
System.out.println("OBJECTB");
}
}
public static void main(String[] args) {
final boolean condition = true;
IObject proxyObj = (IObject)Proxy.newProxyInstance(Test.class.getClassLoader(), new Class[]{IObject.class}, new InvocationHandler() {
#Override
public Object invoke(Object proxy, Method method, Object[] args) throws Throwable {
if (condition)
new A().doIt();
else
new B().doIt();
return null;
}
});
proxyObj.doIt();
}
}
Here, you are creating a Proxy-Object for your Interface.
When you call .doIt() on this object, the invocation-handler will call the appropriate implementation, based on a condition.
You could then pass the proxy around and work with the invocation handler.
(It should be clear that this code should only give an idea and is only an example of how proxied interfaces/objects could be used to solve this problem.)
The above requirement can be implemented easily by using the 'Factory design pattern'
The factory design pattern involves these 3 steps:
Define the common interface for both the object types
Define the different classes that implement the common interface
Create a Factory class with a static method that returns one of the object types based on the input condition
Here is a working demo:
// File name: Demo.java
interface IObject {
public void doIT();
}
class A implements IObject {
public void doIT() {
System.out.println("DoIT - Class A");
}
}
class B implements IObject {
public void doIT() {
System.out.println("DoIT - Class B");
}
}
class IObjectFactory {
static IObject getObject(String CONDITION) {
if(CONDITION.equalsIgnoreCase("CONDITION")) {
return (new A());
}
return (new B());
}
}
public class Demo {
public static void main(String[] args) {
IObject io1 = IObjectFactory.getObject("condition");
IObject io2 = IObjectFactory.getObject("no condition");
io1.doIT();
io2.doIT();
}
}
Output:
> javac Demo.java
> java Demo
DoIT - Class A
DoIT - Class B
It really depends on your use cases. If the use cases are as simple as what you've described, you could remove the Consumer and just call doIt directly in the call function. However, I would recommend generalizing the function even further:
static <T> void ifElse(boolean decider, T optionA, T optionB, Consumer<T> callback) {
if (decider) {
callback.accept(optionA);
} else {
callback.accept(optionB);
}
}
Then call it with:
A a = new A();
B b = new B();
ifElse(condition.equals("CONDITION"), a, b, IObject::doIt);
Doing it this way is cleaner and more expandable. It also takes the same amount of code. (Untested, may have errors, but you get the idea). This approach is what you might see in purely ("purely") functional programming languages, where an if statement is just a function that takes a conditional and two functions (one for true, one for false).
I have 2 classes, and I have made one class (Class A) instantiate a Class B object.
I have a method in Class B that I want to call a method in Class A.
I'm working on a larger project for practicing Java, so I am simplifying things here.
// Class A:
public class ClassA {
private int number;
private ClassB instanceOfB = new ClassB();
public ClassA {
number = 0;
}
public void incrementNumber {
number++;
}
public void incrementNumberLongWay {
instanceOfB.incrementNumberInA()
}
}
// Class B:
public class ClassB {
public void incrementNumberInA() {
// My desire: Call Class A's incrementNumber method
// What should I put here?
}
}
How do I make sure incrementNumberLongWay works? Class A has been instantiated, and it's method incrementNumberLongWay is called, so this should call ClassB's method incrementNumberInA
I know this seems extremely convoluted, but the reason I'm doing this, is because in my program I'm not incrementing numbers, but instead doing some logic in Class B, and only wanting to affect Class A in certain cases.
You can't do this with the code provided. Relationships are by default one way. B doesn't know about A so cannot access it.
What you can do is pass a reference of A to B in it's construction process and then access A via that reference.
One solution would be to pass a method of A as a callback.
For example:
public class ClassA {
private int number;
private ClassB instanceOfB = new ClassB();
public ClassA {
number = 0;
}
public void incrementNumber {
number++;
}
public void incrementNumberLongWay {
instanceOfB.incrementNumberInA(this::increment);
// alternatively
// instanceOfB.incrementNumberInA(() -> incrementNumber());
}
}
public class ClassB {
public void incrementNumberInA(Runnable callbackMethod) {
callbackMethod.run();
}
}
This removes B's dependency on A, and instead allows a general callback mechanism.
However, for such a simple scenario this approach isn't advised.
It's probably a bad idea in general to have a circular dependency in this way. One approach to break the cycle would be to have a third class (classC?) that implements the increment logic (or whatever your real-world equivalent is), and have classA and classB instances each reference classC. That way there's no case where two classes know about each other.
ClassB doesn't know anything about ClassA. So, you couldn't do it.
The ugly decision is
public void incrementNumberLongWay() {
instanceOfB.incrementNumberInA(this);
}
and in
public class ClassB {
public void incrementNumberInA(ClassA cl) {
cl.incrementNumber();
}
}
You can't call methods from class A from class B as class B has no reference to an object of class a. You could, however, pass class A's current number state to class B as parameter, and return a value from class B which class A can then get and use.
For example:
public class A {
private int number;
public A(int number) {
this.number = number;
}
public void incrementNumber(boolean largeIncrement) {
if(largeIncrement) {
B bInstance = new this.B();
number = bInstance.incrementNumberLongWay(number);
}
else {
number++;
}
}
private class B {
private B() {
// if some initialization is needed...
}
public int incrementNumberLongWay(int num) {
num += 1000;
return num;
}
}
}
Hope this is what you wanted.
I ended up the following scenario in code earlier today (which I admit is kinda weird and I have since refactored). When I ran my unit test I found that a field initialization was not set by the time that the superclass constructor has run. I realized that I do not fully understand the order of constructor / field initialization, so I am posting in the hopes that someone explain to me the order in which these occur.
class Foo extends FooBase {
String foo = "foobar";
#Override
public void setup() {
if (foo == null) {
throw new RuntimeException("foo is null");
}
super.setup();
}
}
class FooBase {
public FooBase() {
setup();
}
public void setup() {
}
}
#Test
public void testFoo() {
new Foo();
}
The abbreviated backtrace from JUnit is as follows, I guess I expected $Foo.<init> to set foo.
$Foo.setup
$FooBase.<init>
$Foo.<init>
.testFoo
Yes, in Java (unlike C#, for example) field initializers are called after the superclass constructor. Which means that any overridden method calls from the constructor will be called before the field initializers are executed.
The ordering is:
Initialize superclass (recursively invoke these steps)
Execute field initializers
Execute constructor body (after any constructor chaining, which has already taken place in step 1)
Basically, it's a bad idea to call non-final methods in constructors. If you're going to do so, document it very clearly so that anyone overriding the method knows that the method will be called before the field initializers (or constructor body) are executed.
See JLS section 12.5 for more details.
A constructor's first operation is always the invocation of the superclass constructor. Having no constructor explicitely defined in a class is equivalent to having
public Foo() {
super();
}
The constructor of the base class is thus called before any field of the subclass has been initialized. And your base class does something which should be avoided: call an overridable method.
Since this method is overridden in the subclass, it's invoked on an object that is not fully constructed yet, and thus sees the subclass field as null.
Here's an example of polymorphism in pseudo-C#/Java:
class Animal
{
abstract string MakeNoise ();
}
class Cat : Animal {
string MakeNoise () {
return "Meow";
}
}
class Dog : Animal {
string MakeNoise () {
return "Bark";
}
}
Main () {
Animal animal = Zoo.GetAnimal ();
Console.WriteLine (animal.MakeNoise ());
}
The Main function doesn't know the type of the animal and depends on a particular implementation's behavior of the MakeNoise() method.
class A
{
A(int number)
{
System.out.println("A's" + " "+ number);
}
}
class B
{
A aObject = new A(1);
B(int number)
{
System.out.println("B's" + " "+ number);
}
A aObject2 = new A(2);
}
public class myFirstProject {
public static void main(String[] args) {
B bObj = new B(5);
}
}
out:
A's 1
A's 2
B's 5
My rules:
1. Don't initialize with the default values in declaration (null, false, 0, 0.0...).
2. Prefer initialization in declaration if you don't have a constructor parameter that changes the value of the field.
3. If the value of the field changes because of a constructor parameter put the initialization in the constructors.
4. Be consistent in your practice. (the most important rule)
public class Dice
{
private int topFace = 1;
private Random myRand = new Random();
public void Roll()
{
// ......
}
}
or
public class Dice
{
private int topFace;
private Random myRand;
public Dice()
{
topFace = 1;
myRand = new Random();
}
public void Roll()
{
// .....
}
}
I ended up the following scenario in code earlier today (which I admit is kinda weird and I have since refactored). When I ran my unit test I found that a field initialization was not set by the time that the superclass constructor has run. I realized that I do not fully understand the order of constructor / field initialization, so I am posting in the hopes that someone explain to me the order in which these occur.
class Foo extends FooBase {
String foo = "foobar";
#Override
public void setup() {
if (foo == null) {
throw new RuntimeException("foo is null");
}
super.setup();
}
}
class FooBase {
public FooBase() {
setup();
}
public void setup() {
}
}
#Test
public void testFoo() {
new Foo();
}
The abbreviated backtrace from JUnit is as follows, I guess I expected $Foo.<init> to set foo.
$Foo.setup
$FooBase.<init>
$Foo.<init>
.testFoo
Yes, in Java (unlike C#, for example) field initializers are called after the superclass constructor. Which means that any overridden method calls from the constructor will be called before the field initializers are executed.
The ordering is:
Initialize superclass (recursively invoke these steps)
Execute field initializers
Execute constructor body (after any constructor chaining, which has already taken place in step 1)
Basically, it's a bad idea to call non-final methods in constructors. If you're going to do so, document it very clearly so that anyone overriding the method knows that the method will be called before the field initializers (or constructor body) are executed.
See JLS section 12.5 for more details.
A constructor's first operation is always the invocation of the superclass constructor. Having no constructor explicitely defined in a class is equivalent to having
public Foo() {
super();
}
The constructor of the base class is thus called before any field of the subclass has been initialized. And your base class does something which should be avoided: call an overridable method.
Since this method is overridden in the subclass, it's invoked on an object that is not fully constructed yet, and thus sees the subclass field as null.
Here's an example of polymorphism in pseudo-C#/Java:
class Animal
{
abstract string MakeNoise ();
}
class Cat : Animal {
string MakeNoise () {
return "Meow";
}
}
class Dog : Animal {
string MakeNoise () {
return "Bark";
}
}
Main () {
Animal animal = Zoo.GetAnimal ();
Console.WriteLine (animal.MakeNoise ());
}
The Main function doesn't know the type of the animal and depends on a particular implementation's behavior of the MakeNoise() method.
class A
{
A(int number)
{
System.out.println("A's" + " "+ number);
}
}
class B
{
A aObject = new A(1);
B(int number)
{
System.out.println("B's" + " "+ number);
}
A aObject2 = new A(2);
}
public class myFirstProject {
public static void main(String[] args) {
B bObj = new B(5);
}
}
out:
A's 1
A's 2
B's 5
My rules:
1. Don't initialize with the default values in declaration (null, false, 0, 0.0...).
2. Prefer initialization in declaration if you don't have a constructor parameter that changes the value of the field.
3. If the value of the field changes because of a constructor parameter put the initialization in the constructors.
4. Be consistent in your practice. (the most important rule)
public class Dice
{
private int topFace = 1;
private Random myRand = new Random();
public void Roll()
{
// ......
}
}
or
public class Dice
{
private int topFace;
private Random myRand;
public Dice()
{
topFace = 1;
myRand = new Random();
}
public void Roll()
{
// .....
}
}
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.