suppose I have an interface:
public interface abcd{
public int a();
public void b();
public void c();
public String d(String h, String j);
}
and I implement it in some class:
public class xyzw implements abcd{
}
but I want the method d() to be static, but I can't do this:
public class xyzw implements abcd{
public static void c(){
//some code
}
}
neither can I do this:
public interface abcd{
public int a();
public void b();
public static void c();
public String d(String h, String j);
}
I wonder if there is something or some workaround or some language construct which allows me to make a method defined by an interface a static method?
You can define a static method in interface, but only with implementation.
public interface A {
public static void b() {
System.out.println("Hi");
}
}
Overriding of static methods is not allowed in Java, because you call it on Class object, not on implementation object.
If you can implement a static method in an interface, but you cannot overwrite it, remember that a static method is referenced by the class itself and not by an instance of it.
To solve your problem maybe you could define an abstract class
No, its not possible and doesn't make any sense. An interface is meant to be implemented by subclasses you can only hava a non-abstract, implemented, static method in an interface. You could not statically call your interface method with
abcd.c()
when it has no implementation. Thats why static elements can not be overridden.
It's not possible to override static methods in java.
However, in the subclass, you can declare static method with the same name and "mask it as" the original method - which is as close as you can get.
interface a {
public static int f() {
return 0;
}
}
interface b extends a {
public static int f() {
return 1;
}
}
System.out.println(a.f());
>> 0
System.out.println(b.f());
>> 1
I'd like to access a static method on a class, but have that class passed in a generic.
I've done the following:
class Base{
public static String getStaticName(){
return "Base";
}
}
class Child extends Base{
public static String getStaticName(){
return "Child";
}
}
class StaticAccessor{
public static <T extends Base>String getName(Class<T> clazz){
return T.getStaticName();
}
}
StaticAccessor.getName() // --> "Base"
This will return "Base" but what i'd like is "Child" anybody a suggestion without reflections?
You can't do that without reflection, because the type T is erased at runtime (meaning it will be reduced to its lower bound, which is Base).
Since you do have access to a Class<T> you can do it with reflection, however:
return (String) clazz.getMethod("getStaticName").invoke(null);
Note that I'd consider such code to be code smell and that it is pretty fragile. Could you tell us why you need that?
If it is OK for you to pass an object instance rather than Class in your static accessor, then, there is a simple and elegant solution:
public class Test {
static class Base {
public static String getStaticName() { return "Base"; }
public String myOverridable() { return Base.getStaticName(); };
}
static class Child extends Base {
public static String getStaticName() { return "Child"; }
#Override
public String myOverridable() { return Child.getStaticName(); };
}
static class StaticAccessor {
public static <T extends Base>String getName(T instance) {
return instance.myOverridable();
}
}
public static void main(String[] args) {
Base b = new Base();
Child c = new Child();
System.out.println(StaticAccessor.getName(b));
System.out.println(StaticAccessor.getName(c));
}
}
The output is:
Base
Child
I don't believe you can do this without reflection.
It appears you should be doing is not using static methods. You are using inheritance but static methods do not follow inheritance rules.
I was told that static methods in java didn't have Inheritance but when I try the following test
package test1;
public class Main {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
TB.ttt();
TB.ttt2();
}
}
package test1;
public class TA {
static public Boolean ttt()
{
System.out.println("TestInheritenceA");
return true;
}
static public String test ="ClassA";
}
package test1;
public class TB extends TA{
static public void ttt2(){
System.out.println(test);
}
}
it printed :
TestInheritenceA
ClassA
so do java static methods (and fields) have inheritance (if you try to call a class method does it go along the inheritance chain looking for class methods). Was this ever not the case? And are there any inheritance OO languages that are messed up like that for class methods?
So apparently static methods are inherited but can't be overidden, so does c# share that problem? Do any other Languages?
In Java, fields and static methods are inherited, but cannot be overridden - I believe that is what whoever told you that "they are not inherited" meant.
Non-private, non-static methods are inherited and can be overridden.
This was always the case, but you cannot override class methods:
class A {
public static void a() { system.out.println("A"); }
}
class B {
public static void a() { system.out.println("B"); }
}
A a = new A();
a.a(); // "A"
B b = new B();
b.a() // "B"
a = b;
a.a(); // "A"
That's the meaning of static. It means per class. Static fields and methods are shared among instances. If you change a static value, it's reflected across instances.
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.