Related
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.
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.
I have one class and one interface:
public interface A {
public void getNum();
}
public class B {
public void getNum() {
System.out.println("4");
}
}
public class C extends B implements A {
protected void getNum() {
System.out.println("3");
}
}
Now my question is, why this code is giving compilation error and how can we avoid it. Is there any way in which we can override this method in class C?
From Java Language Specification:
jls-8.4.8.3
The access modifier (§6.6) of an overriding or hiding method must provide at least as much access as the overridden or hidden method, as follows:
If the overridden or hidden method is public, then the overriding or hiding method must be public; otherwise, a compile-time error occurs.
...
Notice that you are trying to override public method getNum() inherited from class B (and also from interface A) with new one that has protected access modifier. It means that you are trying to reduce visibility of this method which according to specification is incorrect.
To be able to override this method you need to use public access modifier with your new version of that method.
Why you cant reduce visibility? Take a look at below code which uses your classes but is placed inside some other package and ask yourself "how should this code behave?".
package my.pckage;
import your.pckage.A;
import your.pckage.C;
public class Test{
public static void main (String[] args){
C C = new C();
c.getNum();// ERROR: Test class doesn't have access to `c`s protected method.
// Why should it have, Test doesn't extend C.
A a = (A)c;// Lets try using other reference
a.getNum();// Should `a` have access to method that is protected in `C`?
// If yes, then what is the point of declaring this method
// protected if all I would have to do to get access to it is
// casting instance of C to A interface?
}
}
Fix the typos and try again ;)
public interface A {
public void getNum();
}
public class B {
protected void getNum() {
System.out.println("4");
}
}
public class C extends B implements A {
public void getNum() {
System.out.println("3");
}
}
First of all scope should be from lower to higher while you are overriding method in Java. scope of subclass method should be high then super class for e.g
Valid Overriding
class B {
protected void getNum() {
System.out.println("4");
}
class C extends B {
public void getNum() {
System.out.println("3");
}
InValid Overriding
class B {
public void getNum() {
System.out.println("4");
}
class C extends B {
protected void getNum() {
System.out.println("3");
}
Your second problem is you have created two public class which is not valid you can create only one public class in your java file.
When you implement an interface you need to compulsorily override it to provide concrete implementation of function(unless the class implementing the interface is abstract). In your case you are implementing an interface which make you implement getNum() function and due to overriding class you have another function with same signature which is not allowed. So you get compilation error.
Possible solution : You can make B as an interface.
The explanation by Pshemo is perfectly right that you can not reduce visibility of overridden or the interface functions.
Lets take an exapmle
class B
{
protected void getProtected1()
{
System.out.println("4");
}
protected void getProtected2()
{
System.out.println("4");
}
public void getPublic1()
{
System.out.println("4");
}
public void getPublic2()
{
System.out.println("4");
}
}
class C extends B
{
#Override
private void getPublic1() //COMPILATION ERROR : Cannot reduce the visibility of the inherited method from myzeromqApp.B
{
System.out.println("3");
}
#Override
protected void getPublic2() //COMPILATION ERROR :Cannot reduce the visibility of the inherited method from myzeromqApp.B
{
System.out.println("3");
}
#Override
private void getProtected1() //COMPILATION ERROR : Cannot reduce the visibility of the inherited method from myzeromqApp.B
{
System.out.println("3");
}
#Override
public void getProtected2() // NO ERROR IT MEANS YOU ARE ALLOWED TO INCREASE THE VISIBILITY
{
System.out.println("3");
}
}
From the above example it is clear that you are not allowed to decrease the visibility of function in any case.
In your question you are trying to implement the interface function and we know interface in Java has rules that,
Method: only public & abstract are permitted
Field: (Variables) only public, static & final are permitted
As thumb of rule, you can never decrease the visibility, of overridden or implemented methods or variables and for interface it is always public (if visibility is concerned) so those should always be public in implemented classes.
As stated, only one public class can be used per file. So, to have them all public, one must create three separate .java files. I will write the code up below, as well as detailing how to override the method to use the correct version of it in each case.
One may always have methods with the same name, but for overriding, they must have different argument lists. This is one of the compiler errors, you have three methods with the same argument lists, namely none. You may create and call the method with the correct argument list to achieve the desired result.
A.java:
package stackOverflow.tests; // Sample package for visibility
public Interface A {
public void getNum(int a); // Method takes a single integer argument
}
B.java:
package stackOverflow.tests;
public class B {
protected void getNum(int a, int b) { // Method takes two integer arguments, differing in the argument list but equal in name
System.out.println("4");
}
}
C.java:
package stackOverflow.tests;
import stackOverflow.tests.A; // Importing both classes to use their methods
import stackOverflow.tests.B;
public class C extends B implements A {
public void getNum(int a, String x) { // Takes an integer and a string argument
System.out.println("3");
}
public void getNum(int a) {
//Do nothing, as in A.java, this code is necessary to be able to override the method.
}
public static void main(String[] arguments) { // Sample main method for implementation
C c = new C(); // Instantiating class C
int test = 0; // Initializing two integer variables and one String variable
int test2 = 0;
String test3 = "";
c.getNum(test); // takes one integer, using getNum() from A.java
c.getNum(test, test2); // takes two integers, using getNum() from B.java
c.getNum(test, test3); // takes an integer and a String, using getNum() from C.java
}
}
Output:
4
3
As seen in the code above, the argument lists define which version of the method is used. As a side tip, the definition getNum(int a) is no different from getNum(int b), so this would result in it not compiling.
In order to get this working you can do something like this since there can be only one public class per file and the file name should be the same name as that of the class
public class HelloWorld{
public static void main(String []args){
C obj=new C();
obj.getNum();
}
}
//interface
interface A {
public void getNum();
}
class B {
protected void getNum() {
System.out.println("4");
}
}
class C extends B implements A {
public void getNum() {
System.out.println("3");
}
}
output:
3
A java class file can have only one public class or interface.
Change the visibility of the interface and the defined class to default level or declare it in separate files.
Only public and abstract modifiers can be applied to interface methods. The class implementing the interface cannot change the visibility of the method (we cannot change it from public to protected).
I have FinanceRequests and CommisionTransactions in my domain.
If I have a list of FinanceRequests each FinanceRequest could contain multiple CommisionTransactions that need to be clawed back. Dont worry how exactly that is done.
The class below (very bottom) makes me feel all fuzzy and warm since its succint and reuses existing code nicely. One problem Type erasure.
public void clawBack(Collection<FinanceRequest> financeRequestList)
public void clawBack(Collection<CommissionTrns> commissionTrnsList)
They both have the same signature after erasure, ie:
Collection<FinanceRequest> --> Collection<Object>
Collection<CommissionTrns> --> Collection<Object>
So eclipse complainst that:
Method clawBack(Collection) has the same erasure clawBack(Collection) as another method in type CommissionFacade
Any suggestions to restructure this so that it still an elegant solution that makes good code reuse?
public class CommissionFacade
{
/********FINANCE REQUESTS****************/
public void clawBack(FinanceRequest financeRequest)
{
Collection<CommissionTrns> commTrnsList = financeRequest.getCommissionTrnsList();
this.clawBack(commTrnsList);
}
public void clawBack(Collection<FinanceRequest> financeRequestList)
{
for(FinanceRequest finReq : financeRequestList)
{
this.clawBack(finReq);
}
}
/********COMMISSION TRANSACTIOS****************/
public void clawBack(CommissionTrns commissionTrns)
{
//Do clawback for single CommissionTrns
}
public void clawBack(Collection<CommissionTrns> commissionTrnsList)
{
for(CommissionTrns commTrn : commissionTrnsList)
{
this.clawBack(commTrn);
}
}
}
Either rename the methods, or use polymorphism: use an interface, and then either put the clawback code in the objects themselves, or use double-dispatch (depending on your design paradigm and taste).
With code in objects that would be:
public interface Clawbackable{
void clawBack()
}
public class CommissionFacade
{
public <T extends Clawbackable> void clawBack(Collection<T> objects)
{
for(T object: objects)
{
object.clawBack();
}
}
}
public class CommissionTrns implements Clawbackable {
public void clawback(){
// do clawback for commissions
}
}
public class FinanceRequest implements Clawbackable {
public void clawBack(){
// do clwaback for FinanceRequest
}
}
I prefer this approach, since I'm of the belief your domain should contain your logic; but I'm not fully aware of your exact wishes, so I'll leave it up to you.
With a double dispatch, you would pass the "ClawbackHandler" to the clawback method, and on the handler call the appropriate method depending on the type.
I think your best option is to simply name the method differently.
public void clawBackFinReqs(Collection<FinanceRequest> financeRequestList) {
}
public void clawBackComTrans(Collection<CommissionTrns> commissionTrnsList) {
}
In fact, it's not too bad, since you don't get anything extra out of having the same name on them.
Keep in mind, that the JVM will not decide which method to call at runtime. As opposed to virtual methods / method overriding resolution of overloaded methods are done at compile time. The Java Tutorials on method overloading even points out that "Overloaded methods should be used sparingly...".
Here is a trick with overloading by the second varargs parameter for the CommissionFacade class from the question:
public class CommissionFacade {
public void clawBack(Collection<FinanceRequest> financeRequestList, FinanceRequestType ...ignore) {
// code
}
public void clawBack(Collection<CommissionTrns> commissionTrnsList, CommissionTrnsType ...ignore) {
// code
}
/*******TYPES TO TRICK TYPE ERASURE*******/
private static class FinanceRequestType {}
private static class CommissionTrnsType {}
}
The code snippet to fast-check this trick works:
import java.util.ArrayList;
class HelloType {
public static void main(String[] args) {
method(new ArrayList<Integer>());
method(new ArrayList<Double>());
}
static void method(ArrayList<Integer> ints, IntegerType ...ignore) {
System.out.println("Hello, Integer!");
}
static void method(ArrayList<Double> dbs, DoubleType ...ignore) {
System.out.println("Hello, Double!");
}
static class IntegerType {}
static class DoubleType {}
}
In Java, I'd like to have something as:
class Clazz<T> {
static void doIt(T object) {
// ...
}
}
But I get
Cannot make a static reference to the non-static type T
I don't understand generics beyond the basic uses and thus can't make much sense of that. It doesn't help that I wasn't able to find much info on the internet about the subject.
Could someone clarify if such use is possible, by a similar manner? Also, why was my original attempt unsuccessful?
You can't use a class's generic type parameters in static methods or static fields. The class's type parameters are only in scope for instance methods and instance fields. For static fields and static methods, they are shared among all instances of the class, even instances of different type parameters, so obviously they cannot depend on a particular type parameter.
It doesn't seem like your problem should require using the class's type parameter. If you describe what you are trying to do in more detail, maybe we can help you find a better way to do it.
Java doesn't know what T is until you instantiate a type.
Maybe you can execute static methods by calling Clazz<T>.doit(something) but it sounds like you can't.
The other way to handle things is to put the type parameter in the method itself:
static <U> void doIt(U object)
which doesn't get you the right restriction on U, but it's better than nothing....
I ran into this same problem. I found my answer by downloading the source code for Collections.sort in the java framework. The answer I used was to put the <T> generic in the method, not in the class definition.
So this worked:
public class QuickSortArray {
public static <T extends Comparable> void quickSort(T[] array, int bottom, int top){
//do it
}
}
Of course, after reading the answers above I realized that this would be an acceptable alternative without using a generic class:
public static void quickSort(Comparable[] array, int bottom, int top){
//do it
}
I think this syntax has not been mentionned yet (in the case you want a method without arguments) :
class Clazz {
static <T> T doIt() {
// shake that booty
}
}
And the call :
String str = Clazz.<String>doIt();
Hope this help someone.
It is possible to do what you want by using the syntax for generic methods when declaring your doIt() method (notice the addition of <T> between static and void in the method signature of doIt()):
class Clazz<T> {
static <T> void doIt(T object) {
// shake that booty
}
}
I got Eclipse editor to accept the above code without the Cannot make a static reference to the non-static type T error and then expanded it to the following working program (complete with somewhat age-appropriate cultural reference):
public class Clazz<T> {
static <T> void doIt(T object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
private static class KC {
}
private static class SunshineBand {
}
public static void main(String args[]) {
KC kc = new KC();
SunshineBand sunshineBand = new SunshineBand();
Clazz.doIt(kc);
Clazz.doIt(sunshineBand);
}
}
Which prints these lines to the console when I run it:
shake that booty 'class com.eclipseoptions.datamanager.Clazz$KC' !!!
shake that booty 'class com.eclipseoptions.datamanager.Clazz$SunshineBand' !!!
It is correctly mentioned in the error: you cannot make a static reference to non-static type T. The reason is the type parameter T can be replaced by any of the type argument e.g. Clazz<String> or Clazz<integer> etc. But static fields/methods are shared by all non-static objects of the class.
The following excerpt is taken from the doc:
A class's static field is a class-level variable shared by all
non-static objects of the class. Hence, static fields of type
parameters are not allowed. Consider the following class:
public class MobileDevice<T> {
private static T os;
// ...
}
If static fields of type parameters were allowed, then the following code would be confused:
MobileDevice<Smartphone> phone = new MobileDevice<>();
MobileDevice<Pager> pager = new MobileDevice<>();
MobileDevice<TabletPC> pc = new MobileDevice<>();
Because the static field os is shared by phone, pager, and pc, what is the actual type of os? It cannot be Smartphone, Pager, and
TabletPC at the same time. You cannot, therefore, create static fields
of type parameters.
As rightly pointed out by chris in his answer you need to use type parameter with the method and not with the class in this case. You can write it like:
static <E> void doIt(E object)
Something like the following would get you closer
class Clazz
{
public static <U extends Clazz> void doIt(U thing)
{
}
}
EDIT: Updated example with more detail
public abstract class Thingo
{
public static <U extends Thingo> void doIt(U p_thingo)
{
p_thingo.thing();
}
protected abstract void thing();
}
class SubThingoOne extends Thingo
{
#Override
protected void thing()
{
System.out.println("SubThingoOne");
}
}
class SubThingoTwo extends Thingo
{
#Override
protected void thing()
{
System.out.println("SuThingoTwo");
}
}
public class ThingoTest
{
#Test
public void test()
{
Thingo t1 = new SubThingoOne();
Thingo t2 = new SubThingoTwo();
Thingo.doIt(t1);
Thingo.doIt(t2);
// compile error --> Thingo.doIt(new Object());
}
}
Since static variables are shared by all instances of the class. For example if you are having following code
class Class<T> {
static void doIt(T object) {
// using T here
}
}
T is available only after an instance is created. But static methods can be used even before instances are available. So, Generic type parameters cannot be referenced inside static methods and variables
When you specify a generic type for your class, JVM know about it only having an instance of your class, not definition. Each definition has only parametrized type.
Generics work like templates in C++, so you should first instantiate your class, then use the function with the type being specified.
Also to put it in simple terms, it happens because of the "Erasure" property of the generics.Which means that although we define ArrayList<Integer> and ArrayList<String> , at the compile time it stays as two different concrete types but at the runtime the JVM erases generic types and creates only one ArrayList class instead of two classes. So when we define a static type method or anything for a generic, it is shared by all instances of that generic, in my example it is shared by both ArrayList<Integer> and ArrayList<String> .That's why you get the error.A Generic Type Parameter of a Class Is Not Allowed in a Static Context!
#BD at Rivenhill: Since this old question has gotten renewed attention last year, let us go on a bit, just for the sake of discussion.
The body of your doIt method does not do anything T-specific at all. Here it is:
public class Clazz<T> {
static <T> void doIt(T object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
// ...
}
So you can entirely drop all type variables and just code
public class Clazz {
static void doIt(Object object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
// ...
}
Ok. But let's get back closer to the original problem. The first type variable on the class declaration is redundant. Only the second one on the method is needed. Here we go again, but it is not the final answer, yet:
public class Clazz {
static <T extends Saying> void doIt(T object) {
System.out.println("shake that booty "+ object.say());
}
public static void main(String args[]) {
Clazz.doIt(new KC());
Clazz.doIt(new SunshineBand());
}
}
// Output:
// KC
// Sunshine
interface Saying {
public String say();
}
class KC implements Saying {
public String say() {
return "KC";
}
}
class SunshineBand implements Saying {
public String say() {
return "Sunshine";
}
}
However, it's all too much fuss about nothing, since the following version works just the same way. All it needs is the interface type on the method parameter. No type variables in sight anywhere. Was that really the original problem?
public class Clazz {
static void doIt(Saying object) {
System.out.println("shake that booty "+ object.say());
}
public static void main(String args[]) {
Clazz.doIt(new KC());
Clazz.doIt(new SunshineBand());
}
}
interface Saying {
public String say();
}
class KC implements Saying {
public String say() {
return "KC";
}
}
class SunshineBand implements Saying {
public String say() {
return "Sunshine";
}
}
T is not in the scope of the static methods and so you can't use T in the static method. You would need to define a different type parameter for the static method. I would write it like this:
class Clazz<T> {
static <U> void doIt(U object) {
// ...
}
}
For example:
public class Tuple<T> {
private T[] elements;
public static <E> Tuple<E> of(E ...args){
if (args.length == 0)
return new Tuple<E>();
return new Tuple<E>(args);
}
//other methods
}