I'm a student, learning Java. I know, protected means access from children or the same package. Here we inherit and override a protected method. And after such an action, whenever the base class wants to call its own method it calls the new overridden one from the subclass. I've been debugging this for a while and marked the execution order with comments. But I can't understand why doesn't it call the base method when I clearly call that from inside the base class constructor?
public class Solution {
public static void main(String[] args) {
new B(); // first
}
public static class A {
public A() {
initialize(); // third
}
protected void initialize() {
System.out.println("class A"); // we never go here
}
}
public static class B extends A {
public B() {
super(); // second
initialize(); // fifth
}
protected void initialize() {
System.out.println("class B"); // fourth, sixth
}
}
}
That's a task from one website, so basically the solution is to change access modifier of the initialize method from protected to private. But I still fail to understand why is the problem happening.
What you're trying to do is defeat the purpose of polymorphism. You can, but you have to make the call specifically. Add a Boolean to your method and call the super.initialize(Boolean). Again, this defeats polymorphism and the extending class HAS to know about the super class. NOT VERY ELEGANT.
public class Solution {
public static void main(String[] args) {
new B(); // first
}
public static class A {
public static boolean USE_SUPER = true;
public A() {
initialize(USE_SUPER);
}
protected void initialize(boolean unusedHere) {
System.out.println("class A");
}
}
public static class B extends A {
public static boolean USE_EXTENDED = false;
public B() {
super();
initialize(USE_EXTENDED);
}
protected void initialize(boolean useSuper) {
if (useSuper)
super.initialize(useSuper);
else
System.out.println("class B");
}
}
}
As Dakoda answered, the root cause is polymorphism. That means we may create child objects, but refer to them as their parent type and when we call the methods of the parent layer we actually refer to the child's methods.
In my case, I create a child object (marked //first) B, which has its own body of the initialize method. One nuance of the inheritance is that it doesn't include constructors, so I can call the parent's constructor (marked //second). Inside the parent's constructor, I call the initialize method - that is the polymorphism because I call the method of the child from its parent abstraction layer.
Here is the answer to the question - this happens, because we only allocated memory for a B instance, that means, we took A as our base and started to extend it (while we can overwrite anything inside). The only two things we did are:
We created a constructor (it wasn't included in the base, as mentioned above)
We overwrote the initialize method code. The code for this method that is inside the base is now lost for this object.
This concept of polymorphism is designed that way and there is no way for us to access the base method unless we specifically create an object that is either A itself or a child that doesn't overwrite this method.
Related
public class Base {
public Base() {
foo();
}
public void foo() {
System.out.println("Base.foo()");
}
}
public class Derived extends Base {
public Derived () {}
public void foo() {
System.out.println("Derived.foo()");
}
}
And then, when i call those:
public class Running {
public static void main(String[] args) {
Base b = new Base();
Derived d = new Derived();
}
}
It outputs:
*Base.foo()*
*Derived.foo()*
So why, when it gets to derived constructor, it invokes the base constructor but uses the derived's method instead?
PS: If I mark those methods as private, it will print out:
*Base.foo()*
*Base.foo()*
This is how Java works read this page https://docs.oracle.com/javase/tutorial/java/IandI/super.html
And more specifically the Note here :
Note: If a constructor does not explicitly invoke a superclass
constructor, the Java compiler automatically inserts a call to the
no-argument constructor of the superclass. If the super class does not
have a no-argument constructor, you will get a compile-time error.
Object does have such a constructor, so if Object is the only
superclass, there is no problem.
So as you can see this is expected behavior. Even though you dot have a super call it is still automatically inserting it.
In regards of the second Question even though you are within the super constructor body still you Instance is of the Subtype. Also if you have some familiarity with C++ read this Can you write virtual functions / methods in Java?
The reason why it will write the base class when marking with private is because private methods are not Inherited. This is part of the Inheritance in Java topic.
To answer the question in your title. As I said, you cannot avoid the base class constructor being called (or one of the base class constructors if it has more than one). You can of course easily avoid the body of the constructor being executed. For example like this:
public class Base {
public Base(boolean executeConstructorBody) {
if (executeConstructorBody) {
foo();
}
}
public void foo() {
System.out.println("Base.foo()");
}
}
public class Derived extends Base {
public Derived() {
super(false);
}
public void foo() {
System.out.println("Derived.foo()");
}
}
public class Running {
public static void main(String[] args) {
Base b = new Base(true);
Derived d = new Derived();
}
}
Now the main method prints only:
Base.foo()
Because in the contructor of the Derived class it automatically gets injected a call to super(), if you do not add a call to super or to other constructor in the same class (using this).
I have a base class A, having a method "say" that calls from constructor of A. All the heritable classes uses the method "say" like it is. But one of the classes need to redefine this method. How is it possible?
For sure, I can denote base method "say" as abstract, but in that way, i have to copy the same method "say" in all the heritable classes.
If i just redefine method without denoting base one as abstract, it is not gonna be called.
public abstract class A(){
public A(){
say(); // <- wanna call this method from heritable class, if its redefined.
}
protected void say(){};
}
public class B extends A(){
public B(){
super();
}
private void say(){};
}
refactoring 1
public abstract class A(){
public A(){
// constructor methods
}
protected void say(){};
protected void executeSay(){
say();
}
}
public class B extends A(){
public B(){
super();
executeSay();
}
#Override
protected void say(){};
}
First of all one must be made clear: calling an overridable method from a constructor is a well-known antipattern. It will almost certainly break your code because the subclass method will be invoked before the subclass constructor is done and so will observe an uninitialized object. Thus I should better refrain from giving you detailed advice on Java technicalities involved in achieving this antipattern.
The only safe way to acomplish your requirement is to let the construction finish and only afterwards call an initialize-kind of method. If you want to ensure initialize is always invoked, make the constructors non-public and provide a factory method instead.
Unfortunately, Java requires quite a bit of work on your part to make this work properly.
You cannot instantiate a abstract class. That saying you have to link the abstract class reference to the concrete inherited class.
eg. A a = new B();
If that's the case, and B have redefined the say() method, then the say method in B will be called.
public class TestPad {
public static void main(String[] args) {
A a = new B();
}
}
abstract class A {
public A() {
say();
}
public void say(){
System.out.println("A");
};
}
class B extends A {
public B() {
super();
}
public void say() {
System.out.println("B");
}
}
The output will be B
public class B extends A {
public B() {
super();
}
#Override
protected void say() {
// your diffent say code
};
}
I'm not sure if you are allowed to reduce visibility to private.
Because of polymorphic method invocation, in your case the B.say() will be invoked if you override it.
But as #sanbhat commented, you need to change visibility of say() to protected.
Consider the following class
class A{
public void init(){
//do this first;
}
public void atEnd(){
//do this after init of base class ends
}
}
class B1 extends A{
#Override
public void init()
{
super.init();
//do new stuff.
//I do not want to call atEnd() method here...
}
}
I have several B1, B2,... Bn child classes which are already developed. All of them extend class A. If I want to add a new functionality in all of them, the best place to do so is define that in a method within class A. But the condition is that the method should always get called automatically just before the init() method of child class ends.
One basic way to do so is to again add atEnd() method call at end of init() method of child classes. But is there any other way to do this smartly ??
One way to do this is by making init() final and delegating its operation to a second, overridable, method:
abstract class A {
public final void init() {
// insert prologue here
initImpl();
// insert epilogue here
}
protected abstract void initImpl();
}
class B extends A {
protected void initImpl() {
// ...
}
}
Whenever anyone calls init(), the prologue and epilogue are executed automatically, and the derived classes don't have to do a thing.
Another thought would be to weave in an aspect. Add before and after advice to a pointcut.
Make init() final, and provide a separate method for people to override that init() calls in the middle:
class A{
public final void init(){
//do this first;
}
protected void initCore() { }
public void atEnd(){
//do this after init of base class ends
}
}
class B1 extends A{
#Override
protected void initCore()
{
//do new stuff.
}
}
The other answers are reasonable workarounds but to address the exact question: no, there is no way to do this automatically. You must explicitly call super.method().
class XYZ{
public static void show(){
System.out.println("inside XYZ");
}
}
public class StaticTest extends XYZ {
public static void show() {
System.out.println("inside statictest");
}
public static void main(String args[]){
StaticTest st =new StaticTest();
StaticTest.show();
}
}
though we know static methods cant be overridden. Then what actually is happening?
Static methods belong to the class. They can't be overridden. However, if a method of the same signature as a parent class static method is defined in a child class, it hides the parent class method. StaticTest.show() is hiding the XYZ.show() method and so StaticTest.show() is the method that gets executed in the main method in the code.
Its not overriding they are two different method in two different class with same signature. but method from XYZ isn't available in child class through inheritance .
It will call method from StaticTest
It's not overriden properly said... Static methods are 'tied' to the class so
StaticTest.show();
and
XYZ.show();
are two totally different things. Note you can't invoke super.show()
To see the difference you have to use more powerful example:
class Super {
public static void hidden(Super superObject) {
System.out.println("Super-hidden");
superObject.overriden();
}
public void overriden() {
System.out.println("Super-overriden");
}
}
class Sub extends Super {
public static void hidden(Super superObject) {
System.out.println("Sub-hidden");
superObject.overriden();
}
public void overriden() {
System.out.println("Sub-overriden");
}
}
public class Test {
public static void main(String[] args) {
Super superObject = new Sub();
superObject.hidden(superObject);
}
}
As Samit G. already have written static methods with same signature in both base and derived classes hide the implementation and this is no-overriding. You can play a bit with the example by changing the one or the another of the static methods to non-static or changing them both to non-static to see what are the compile-errors which the java compiler rises.
It's not an override, but a separate method that hides the method in XYZ.
So as I know, any static member (method or state) is an attribute of a class, and would not be associated with any instance of a class. So in your example, XYZ is a class, and so is StaticTest (as you know). So by calling the constructor two things first happen. An Object of type Class is created. It has a member on it call showed(). Class, XYZ.class, extends from Object so has all those Object methods on it plus show(). Same with the StaticClass, the class object has show() on it as well. They both extend java.lang.Object though. An instance of StaticClass would also be an instance of XYZ. However now the more interesting question would be what happens when you call show() on st?
StaticClass st = new StaticClass();
st.show();
XYZ xyz = st;
xyz.show();
What happens there? My guess is that it is StaticClass.show() the first time and XYZ.show() the second.
Static methods are tied to classes and not instances (objects).
Hence the invocations are always ClassName.staticMethod();
When such a case of same static method in a subclass appears, its called as refining (redefining) the static method and not overriding.
// Java allows a static method to be called from an Instance/Object reference
// which is not the case in other pure OOP languages like C# Dot net.
// which causes this confusion.
// Technically, A static method is always tied to a Class and not instance.
// In other words, the binding is at compile-time for static functions. - Early Binding
//
// eg.
class BaseClass
{
public static void f1()
{
System.out.println("BaseClass::f1()...");
} // End of f1().
}
public class SubClass extends BaseClass
{
public static void f1()
{
System.out.println("SubClass::f1()...");
// super.f1(); // non-static variable super cannot be referenced from a static context
} // End of f1().
public static void main(String[] args)
{
f1();
SubClass obj1 = new SubClass();
obj1.f1();
BaseClass b1 = obj1;
b1.f1();
} // End of main().
} // End of class.
// Output:
// SubClass::f1()...
// SubClass::f1()...
// BaseClass::f1()...
//
//
// So even though in this case, called with an instance b1 which is actually referring to
// an object of type SuperClass, it calls the BaseClass:f1 method.
//
I have the following classes:
class foo {
public void a() {
print("a");
}
public void b() {
a();
}
}
class bar extends foo {
public void a() {
print("overwritten a");
}
}
When I now call bar.b() I want it to call the overridden method a() in foo. It does, however, print "a".
Are your two classes in different packages? And is your foo class methods declared public, protected, or private or package local? Obviously if they are private, this won't work. Perhaps less obvious, is if they are package local (i.e. no public/protected/private scope) then you can only override them if you are in the same package as the original class.
For example:
package original;
public class Foo {
void a() { System.out.println("A"); }
public void b() { a(); }
}
package another;
public class Bar extends original.Foo {
void a() { System.out.println("Overwritten A"); }
}
package another;
public class Program {
public static void main(String[] args) {
Bar bar = new Bar();
bar.b();
}
}
In this case, you will still get 'A'. If you declare the original a() method in Foo public or protected, you will get the result you expected.
It may be that you are trying to use static methods, which won't work as they don't get overridden.
A good way of checking is to add the #Override annotation to bar.a() and see if the compiler gives you an error that a() isn't actually overidding anything
When I run the following:
public class Program {
public static void main(String[] args) {
bar b = new bar();
b.b();
}
}
class foo {
public void a() {
System.out.printf("a");
}
public void b() {
a();
}
}
class bar extends foo {
public void a() {
System.out.printf("overwritten a");
}
}
I get the following output:
overwritten a
which is what I would expect to see.
Are the methods defined as static? That's the only way I could see getting that result. I found a good explanation about that here: http://faq.javaranch.com/view?OverridingVsHiding
You may be confused if you are coming from C# or some other language where you have to explicitly declare virtual functions and/or overriding functions.
In Java, all instance functions are virtual, and can be overridden -- unless they are declared as private and/or final.
It is not necessary to specify the new #Override annotation to do so, adding the annotation just specifies that your intent is to override, and will cause a either a warning or error if it isn't an override. (If you accidentally misspelled the method name for example).
Andrew's example shows how this should work.
From the horse's mouth:
http://download.oracle.com/javase/tutorial/java/IandI/override.html
"The version of the overridden method that gets invoked is the one in the subclass. The version of the hidden method that gets invoked depends on whether it is invoked from the superclass or the subclass."
So if they are both static methods and you invoke the method from the super class, then the super class method is invoked, not the subclass method. So really, no overriding is taking place.
While I was working on an Android program, I had the same problem on my Java classes. It turned out that the problem was not in the classes but the way Android framework processes screen output.
for example if output was programmed in the onCreate() method in the parent class, Android failed to correctly fetch the output of overridden methods from child classes beyond first child. Honestly I don't understand whole method calling order.
To resolve the issue I simply programmed the output in the onResume() and now it seem to work fine.