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.
Is it possible to call a constructor from another (within the same class, not from a subclass)? If yes how? And what could be the best way to call another constructor (if there are several ways to do it)?
Yes, it is possible:
public class Foo {
private int x;
public Foo() {
this(1);
}
public Foo(int x) {
this.x = x;
}
}
To chain to a particular superclass constructor instead of one in the same class, use super instead of this. Note that you can only chain to one constructor, and it has to be the first statement in your constructor body.
See also this related question, which is about C# but where the same principles apply.
Using this(args). The preferred pattern is to work from the smallest constructor to the largest.
public class Cons {
public Cons() {
// A no arguments constructor that sends default values to the largest
this(madeUpArg1Value,madeUpArg2Value,madeUpArg3Value);
}
public Cons(int arg1, int arg2) {
// An example of a partial constructor that uses the passed in arguments
// and sends a hidden default value to the largest
this(arg1,arg2, madeUpArg3Value);
}
// Largest constructor that does the work
public Cons(int arg1, int arg2, int arg3) {
this.arg1 = arg1;
this.arg2 = arg2;
this.arg3 = arg3;
}
}
You can also use a more recently advocated approach of valueOf or just "of":
public class Cons {
public static Cons newCons(int arg1,...) {
// This function is commonly called valueOf, like Integer.valueOf(..)
// More recently called "of", like EnumSet.of(..)
Cons c = new Cons(...);
c.setArg1(....);
return c;
}
}
To call a super class, use super(someValue). The call to super must be the first call in the constructor or you will get a compiler error.
[Note: I just want to add one aspect, which I did not see in the other answers: how to overcome limitations of the requirement that this() has to be on the first line).]
In Java another constructor of the same class can be called from a constructor via this(). Note however that this has to be on the first line.
public class MyClass {
public MyClass(double argument1, double argument2) {
this(argument1, argument2, 0.0);
}
public MyClass(double argument1, double argument2, double argument3) {
this.argument1 = argument1;
this.argument2 = argument2;
this.argument3 = argument3;
}
}
That this has to appear on the first line looks like a big limitation, but you can construct the arguments of other constructors via static methods. For example:
public class MyClass {
public MyClass(double argument1, double argument2) {
this(argument1, argument2, getDefaultArg3(argument1, argument2));
}
public MyClass(double argument1, double argument2, double argument3) {
this.argument1 = argument1;
this.argument2 = argument2;
this.argument3 = argument3;
}
private static double getDefaultArg3(double argument1, double argument2) {
double argument3 = 0;
// Calculate argument3 here if you like.
return argument3;
}
}
When I need to call another constructor from inside the code (not on the first line), I usually use a helper method like this:
class MyClass {
int field;
MyClass() {
init(0);
}
MyClass(int value) {
if (value<0) {
init(0);
}
else {
init(value);
}
}
void init(int x) {
field = x;
}
}
But most often I try to do it the other way around by calling the more complex constructors from the simpler ones on the first line, to the extent possible. For the above example
class MyClass {
int field;
MyClass(int value) {
if (value<0)
field = 0;
else
field = value;
}
MyClass() {
this(0);
}
}
Within a constructor, you can use the this keyword to invoke another constructor in the same class. Doing so is called an explicit constructor invocation.
Here's another Rectangle class, with a different implementation from the one in the Objects section.
public class Rectangle {
private int x, y;
private int width, height;
public Rectangle() {
this(1, 1);
}
public Rectangle(int width, int height) {
this( 0,0,width, height);
}
public Rectangle(int x, int y, int width, int height) {
this.x = x;
this.y = y;
this.width = width;
this.height = height;
}
}
This class contains a set of constructors. Each constructor initializes some or all of the rectangle's member variables.
As everybody already have said, you use this(…), which is called an explicit constructor invocation.
However, keep in mind that within such an explicit constructor invocation statement you may not refer to
any instance variables or
any instance methods or
any inner classes declared in this class or any superclass, or
this or
super.
As stated in JLS (§8.8.7.1).
Yes, any number of constructors can be present in a class and they can be called by another constructor using this() [Please do not confuse this() constructor call with this keyword]. this() or this(args) should be the first line in the constructor.
Example:
Class Test {
Test() {
this(10); // calls the constructor with integer args, Test(int a)
}
Test(int a) {
this(10.5); // call the constructor with double arg, Test(double a)
}
Test(double a) {
System.out.println("I am a double arg constructor");
}
}
This is known as constructor overloading.
Please note that for constructor, only overloading concept is applicable and not inheritance or overriding.
Using this keyword we can call one constructor in another constructor within same class.
Example :-
public class Example {
private String name;
public Example() {
this("Mahesh");
}
public Example(String name) {
this.name = name;
}
}
Yes it is possible to call one constructor from another. But there is a rule to it. If a call is made from one constructor to another, then
that new constructor call must be the first statement in the current constructor
public class Product {
private int productId;
private String productName;
private double productPrice;
private String category;
public Product(int id, String name) {
this(id,name,1.0);
}
public Product(int id, String name, double price) {
this(id,name,price,"DEFAULT");
}
public Product(int id,String name,double price, String category){
this.productId=id;
this.productName=name;
this.productPrice=price;
this.category=category;
}
}
So, something like below will not work.
public Product(int id, String name, double price) {
System.out.println("Calling constructor with price");
this(id,name,price,"DEFAULT");
}
Also, in the case of inheritance, when sub-class's object is created, the super class constructor is first called.
public class SuperClass {
public SuperClass() {
System.out.println("Inside super class constructor");
}
}
public class SubClass extends SuperClass {
public SubClass () {
//Even if we do not add, Java adds the call to super class's constructor like
// super();
System.out.println("Inside sub class constructor");
}
}
Thus, in this case also another constructor call is first declared before any other statements.
I will tell you an easy way
There are two types of constructors:
Default constructor
Parameterized constructor
I will explain in one Example
class ConstructorDemo
{
ConstructorDemo()//Default Constructor
{
System.out.println("D.constructor ");
}
ConstructorDemo(int k)//Parameterized constructor
{
this();//-------------(1)
System.out.println("P.Constructor ="+k);
}
public static void main(String[] args)
{
//this(); error because "must be first statement in constructor
new ConstructorDemo();//-------(2)
ConstructorDemo g=new ConstructorDemo(3);---(3)
}
}
In the above example I showed 3 types of calling
this() call to this must be first statement in constructor
This is Name less Object. this automatically calls the default constructor.
3.This calls the Parameterized constructor.
Note:
this must be the first statement in the constructor.
You can a constructor from another constructor of same class by using "this" keyword.
Example -
class This1
{
This1()
{
this("Hello");
System.out.println("Default constructor..");
}
This1(int a)
{
this();
System.out.println("int as arg constructor..");
}
This1(String s)
{
System.out.println("string as arg constructor..");
}
public static void main(String args[])
{
new This1(100);
}
}
Output -
string as arg constructor..
Default constructor..
int as arg constructor..
Calling constructor from another constructor
class MyConstructorDemo extends ConstructorDemo
{
MyConstructorDemo()
{
this("calling another constructor");
}
MyConstructorDemo(String arg)
{
System.out.print("This is passed String by another constructor :"+arg);
}
}
Also you can call parent constructor by using super() call
There are design patterns that cover the need for complex construction - if it can't be done succinctly, create a factory method or a factory class.
With the latest java and the addition of lambdas, it is easy to create a constructor which can accept any initialization code you desire.
class LambdaInitedClass {
public LamdaInitedClass(Consumer<LambdaInitedClass> init) {
init.accept(this);
}
}
Call it with...
new LambdaInitedClass(l -> { // init l any way you want });
Pretty simple
public class SomeClass{
private int number;
private String someString;
public SomeClass(){
number = 0;
someString = new String();
}
public SomeClass(int number){
this(); //set the class to 0
this.setNumber(number);
}
public SomeClass(int number, String someString){
this(number); //call public SomeClass( int number )
this.setString(someString);
}
public void setNumber(int number){
this.number = number;
}
public void setString(String someString){
this.someString = someString;
}
//.... add some accessors
}
now here is some small extra credit:
public SomeOtherClass extends SomeClass {
public SomeOtherClass(int number, String someString){
super(number, someString); //calls public SomeClass(int number, String someString)
}
//.... Some other code.
}
Hope this helps.
Yes it is possible to call one constructor from another with use of this()
class Example{
private int a = 1;
Example(){
this(5); //here another constructor called based on constructor argument
System.out.println("number a is "+a);
}
Example(int b){
System.out.println("number b is "+b);
}
You can call another constructor via the this(...) keyword (when you need to call a constructor from the same class) or the super(...) keyword
(when you need to call a constructor from a superclass).
However, such a call must be the first statement of your constructor. To overcome this limitation, use this answer.
The keyword this can be used to call a constructor from a constructor, when writing several constructor for a class, there are times when you'd like to call one constructor from another to avoid duplicate code.
Bellow is a link that I explain other topic about constructor and getters() and setters() and I used a class with two constructors. I hope the explanations and examples help you.
Setter methods or constructors
I know there are so many examples of this question but what I found I am putting here to share my Idea. there are two ways to chain constructor. In Same class you can use this keyword. in Inheritance, you need to use super keyword.
import java.util.*;
import java.lang.*;
class Test
{
public static void main(String args[])
{
Dog d = new Dog(); // Both Calling Same Constructor of Parent Class i.e. 0 args Constructor.
Dog cs = new Dog("Bite"); // Both Calling Same Constructor of Parent Class i.e. 0 args Constructor.
// You need to Explicitly tell the java compiler to use Argument constructor so you need to use "super" key word
System.out.println("------------------------------");
Cat c = new Cat();
Cat caty = new Cat("10");
System.out.println("------------------------------");
// Self s = new Self();
Self ss = new Self("self");
}
}
class Animal
{
String i;
public Animal()
{
i = "10";
System.out.println("Animal Constructor :" +i);
}
public Animal(String h)
{
i = "20";
System.out.println("Animal Constructor Habit :"+ i);
}
}
class Dog extends Animal
{
public Dog()
{
System.out.println("Dog Constructor");
}
public Dog(String h)
{
System.out.println("Dog Constructor with habit");
}
}
class Cat extends Animal
{
public Cat()
{
System.out.println("Cat Constructor");
}
public Cat(String i)
{
super(i); // Calling Super Class Paremetrize Constructor.
System.out.println("Cat Constructor with habit");
}
}
class Self
{
public Self()
{
System.out.println("Self Constructor");
}
public Self(String h)
{
this(); // Explicitly calling 0 args constructor.
System.out.println("Slef Constructor with value");
}
}
It is called Telescoping Constructor anti-pattern or constructor chaining. Yes, you can definitely do. I see many examples above and I want to add by saying that if you know that you need only two or three constructor, it might be ok. But if you need more, please try to use different design pattern like Builder pattern. As for example:
public Omar(){};
public Omar(a){};
public Omar(a,b){};
public Omar(a,b,c){};
public Omar(a,b,c,d){};
...
You may need more. Builder pattern would be a great solution in this case. Here is an article, it might be helpful
https://medium.com/#modestofiguereo/design-patterns-2-the-builder-pattern-and-the-telescoping-constructor-anti-pattern-60a33de7522e
Yes, you can call constructors from another constructor. For example:
public class Animal {
private int animalType;
public Animal() {
this(1); //here this(1) internally make call to Animal(1);
}
public Animal(int animalType) {
this.animalType = animalType;
}
}
you can also read in details from
Constructor Chaining in Java
Originally from an anser by Mirko Klemm, slightly modified to address the question:
Just for completeness: There is also the Instance initialization block that gets executed always and before any other constructor is called. It consists simply of a block of statements "{ ... }" somewhere in the body of your class definition. You can even have more than one. You can't call them, but they're like "shared constructor" code if you want to reuse some code across constructors, similar to calling methods.
So in your case
{
System.out.println("this is shared constructor code executed before the constructor");
field1 = 3;
}
There is also a "static" version of this to initialize static members: "static { ... }"
I prefer this way:
class User {
private long id;
private String username;
private int imageRes;
public User() {
init(defaultID,defaultUsername,defaultRes);
}
public User(String username) {
init(defaultID,username, defaultRes());
}
public User(String username, int imageRes) {
init(defaultID,username, imageRes);
}
public User(long id, String username, int imageRes) {
init(id,username, imageRes);
}
private void init(long id, String username, int imageRes) {
this.id=id;
this.username = username;
this.imageRes = imageRes;
}
}
I am a little confused on how to set up the TestHomework method so that it prints properly when using the toString() method. Right now when I run the main method it prints "null - 0" but what I would like it to say is "Math - 6". This program is supposed to extend an abstract class. It is supposed to say how many pages there are for homework and for what subject.
public abstract class Homework {
private int pagesToRead;
private String typeHomework;
{
// initialise instance variables
pagesToRead = 0;
typeHomework = "none";
}
public Homework(int pages, String hw) {
this.pagesToRead = pages;
this.typeHomework = hw;
}
public abstract void createAssignment(int p);
public int getPages() {
return pagesToRead;
}
public void setPagesToRead(int p) {
pagesToRead = p;
}
public String getTypeHomework() {
return typeHomework;
}
public void setTypeHomework(String hw) {
typeHomework = hw;
}
}
public class MyMath extends Homework {
private int pagesRead;
private String typeHomework;
public MyMath(int pages, String hw) {
super(pages,hw);
}
public void createAssignment(int p) {
setTypeHomework("Math");
setPagesToRead(p);
}
public String toString() {
return typeHomework + " - " + pagesRead;
}
}
public class TestHomework {
public static void main(String[] args) {
MyMath one = new MyMath(6, "Math");
one.createAssignment(6);
System.out.println(one);
}
}
That's because you are defining the 2 properties (that one of them happen to have the same name as one of the abstract class's) but you are not initializing them, you are initializing those of the abstract class. (So their values is always set to their type's default)
You need to drop those from the MyMath class, & define the toString method in your abstract class: it's the one to be used by default by its inheriting classes.
public abstract class Homework {
private int pagesToRead;
private String typeHomework;
// Same code
// Define the toString here
#Override
public String toString() {
return typeHomework + " - " + pagesToRead;
}
}
public class MyMath extends Homework {
// You don't need to define any extra attributes
public MyMath(int pages, String hw) {
super(pages,hw);
}
public void createAssignment(int p) {
setTypeHomework("Math");
setPagesToRead(p);
}
}
public static void main(String[] args) {
// Calls the constructor of the MyMath class, which in turn
// invokes the constructor of its superclass, the 'Homework' class
MyMath one = new MyMath(6, "Math");
one.createAssignment(6);
// Invokes the toString of the MyMath class. Since it does not have one,
// The toString method of its superclass (Homework) is called.
System.out.println(one);
}
Your derived class has its own typeHomework and pagesRead fields, which are never set (even though the base class happens to have fields with the same names). Therefore, they stay null and 0.
You should delete those fields and use the data from the base class, via the public getter methods.
Why it doesn't work:
Be careful you redeclared the attribute typeHomework of you parent class. Attributes are automatically added to your extending class so you don't have to write them again.
By redeclaring it you confused the compiler, viewing your code in debug shows, that your one object contains your typeHomework twice:
typeHomework = null // The one from the super class
typeHomework = "Math" // The one from your child class
Your method now uses the typeHomework from your super-class therefor the output is null!
pagesRead is 0 because you are setting the pagesToRead of your super-class to 6(not pagesRead!) when calling setPagesToRead(p);.
Some style tips
Use the #Override annotation when overriding methods like this:
#Override
public void createAssignment(int p) {
setTypeHomework("Math");
setPagesToRead(p);
}
It's not really needed but it's good practice (readers of your code know that it overrides something).
When referring to attributes of your class it's also good practice to use the this statement so it's clear, that you're referring to an attribute and not a local variable:
#Override
public String toString() {
return this.typeHomework + " - " + this.pagesRead;
}
I ran into this block of code, and there is this one line I don't quit understand the meaning or what it is doing.
public Digraph(In in) {
this(in.readInt());
int E = in.readInt();
for (int i = 0; i < E; i++) {
int v = in.readInt();
int w = in.readInt();
addEdge(v, w);
}
}
I understand what this.method() or this.variable are, but what is this()?
This is constructor overloading:
public class Diagraph {
public Diagraph(int n) {
// Constructor code
}
public Digraph(In in) {
this(in.readInt()); // Calls the constructor above.
int E = in.readInt();
for (int i = 0; i < E; i++) {
int v = in.readInt();
int w = in.readInt();
addEdge(v, w);
}
}
}
You can tell this code is a constructor and not a method by the lack of a return type.
This is pretty similar to calling super() in the first line of the constructor in order to initialize the extended class. You should call this() (or any other overloading of this()) in the first line of your constructor and thus avoid constructor code duplications.
You can also have a look at this post: Constructor overloading in Java - best practice
Using this() as a function like that, essentially calls the Constructor of the class. This allows you to all the generic initializations in one constructor and have specializations in others. So in this piece of code for example, the call to this(in.readInt()) is calling the Digraph constructor that has one int argument.
This code snippet is a constructor.
This call to this calls another constructor of the same class
public App(int input) {
}
public App(String input) {
this(Integer.parseInt(input));
}
In the above example we have a constructor that takes an int and one that takes a String. The constructor that takes a String converts the String to an int and then delegates to the int constructor.
Note that a call to another constructor or a superclass constructor (super()) must be the first line in a constructor.
Maybe take a look at this for a more detailed explanation of constructor overloading.
It's nearly the same
public class Test {
public Test(int i) { /*construct*/ }
public Test(int i, String s){ this(i); /*construct*/ }
}
Calling this essentially calls the class Constructor.
For example, if you're extending something, than along with add(JComponent), you could do: this.add(JComponent).
An other constructor of the class Digraph with an int parameter.
Digraph(int param) { /* */ }
Constructor Overloading:
ex:
public class Test{
Test(){
this(10); // calling constructor with one parameter
System.out.println("This is Default Constructor");
}
Test(int number1){
this(10,20); // calling constructor with two parameter
System.out.println("This is Parametrized Constructor with one argument "+number1);
}
Test(int number1,int number2){
System.out.println("This is Parametrized Constructor with two argument"+number1+" , "+number2);
}
public static void main(String args[]){
Test t = new Test();
// first default constructor,then constructor with 1 parameter , then constructor with 2 parameters will be called
}
}
this(); is constructor which is used to call another constructor in a class,
for example:-
class A{
public A(int,int)
{ this(1.3,2.7);-->this will call default constructor
//code
}
public A()
{
//code
}
public A(float,float)
{ this();-->this will call default type constructor
//code
}
}
Note:
i did not use this() constructor in default constructor because it will lead to deadlock state.
Hope this will help you:)