I have this code:
class A{
class B{
public void foo(){
class C{
public void bar(){
baz();
}
}
}
public void baz(){}
}
public void baz(){}
}
If i call baz() in C.bar(), it will call B.baz(). If i make a reference in A from this, (A aref = this;) then i can call A.baz() through it. Is there any other way to do this?
You may use A.this.baz(). A.this is called a qualified this expression.
Related
Here's something I quite understand:
abstract class A {
public void foo() {
System.out.println("a");
}
}
abstract class B extends A {
#Override
public abstract void foo();
public void bar() {
super.foo();
foo();
}
}
class C extends B {
#Override
public void foo() {
System.out.println("c");
}
}
public static void main(String[] args) {
new C().foo();
new C().bar();
}
new C().foo() prints c to the console, while new C().bar() prints a then c.
Calling super.foo() is illegal in the #foo() implementation of the C class.
I don't have a clear question, but if anyone could give a complete explanation of what is going on with the foo method, it may be interesting I think.
A is super class for B, so calling super.foo() inside B calls method defined in A, and calling foo() inside the same class will invoke its own implementation that should be delivered by any subclass.
You cannot use super.foo() within C class because it is defined as abstract in B and cannot be invoked directly.
I have a class A that extends a class B.
A is defined like this, it also overrides a method of B:
class A extends B
{
public A() {
super();
}
#Override
public void doSomething(){
//does something
}
}
B is defined like this:
public class B
{
public B(){
doSomething();
}
public void doSomething(){
//does something
}
}
So if I initialize an object of A, the constructor calls the one of the superclass that calls the method doSomething(). But which one will be executed? B's implementation or the overriden one in A?
That is a common bug, only call final methods in constructor, the method from A will be called.
Btw Sonar(if you have it) will trigger a rule here saying that you should not call polymorphic methods inside a constructor.
If the class Overrides a method, then the overriden method will be called. Try the example below:
public class A {
void doSomething() {
System.out.println("a");
}
}
public class B extends A {
#Override
void doSomething() {
System.out.println("b");
}
}
A a = new B();
a.doSomething(); // will print "b"
(newbie in Java) I couldn't find exactly this question on SO. I have project, with two files (phseudo-code):
First Java File (class)
public class A {
public void xyz() { System.out.println("hello");}
}
Second Java File (class)
public class B Extends ZZZZZ {
public void callme() {
xyz(); // <----------------- I want to call in this way, but It cant be done like this.
}
}
How to make xyz() to call successfully (like as if was defined inside b() class natively !!).
p.s. again, I don't want to call it with classname in front, like this:
a.xyz();
The whole idea of instance methods, like xyz is in this, is that you are using the state of an instance of A in the method, without having to pass that instance as an argument like this:
... String xyz(A thisInstance, ...) {...}
Instead you use:
A thisInstance = ...;
thisInstance.xyz(...);
That's why you need an instance of A, because it is practically an argument to the function.
However, if you don't need an instance of A, you can make the method static:
static String xyz(...) {...}
Then you can call it without passing an instance of A:
A.xyz(...);
You can use a static import so that you don't have to write A:
import static A.xyz;
...
xyz(...);
Okay several possibilities:
Instantiate A:
A a=new A();
a.xyz();
(you do not want this)
Heredity:
public class B extends A {...}
and
public class A extends ZZZZZ{...}
so you can still extend ZZZZZ;
Interface:
public interface A{...}
public class B extends ZZZZZ implements A{...}
Static Method:
public class A{
public static void xyz()
{
System.out.println("hello");
}
}
public class B{
public void callme()
{
A.xyz());
}
}
This will help you.
class A {
public void xyz() {
System.out.println("hello");
}
}
class ZZZZZ extends A{
}
class B extends ZZZZZ {
public void callme() {
xyz();// <----------------- calling this shows error
}
}
what will be the flow of execution in case of override? What i believe is , when we call a constructor/object of any class, during execution first it call parent constructor and than child. but what will happen in case of over ridding?
lets suppose:
class A {
public A(){
printStatus();
}
public void printStatus(){
System.out.println("In Class A");
}
}
class B extends A{
public B(){
printStatus();
}
#Override
public void printStatus(){
System.out.println("In Class b");
}
}
public class Test2 {
public static void main(String[] args){
B b = new B();
}
}
Out put of this code is:
In Class b
In Class b
what i don't understand is, why it's printing "In Class be" only, it should be "In class A and, In Class b",
when i remove override method from class b. it give me desired output.
All java methods are virtual. It means the method is called with using actual type of this. So inside of constructor A() {} this is the instance of B, so that is why you've got its method call.
Calling like this printStatus() will call the method from the same class. If you call with super.printStatus() it will envoke method from the super class (class which you have extended).
When you over-ride a method you over-ride it completely. The existence of the original implementation is completely invisible to other classes (except via reflection but that's a big topic of its own and not really relevant). Only your own class can access the original method and that is by calling super.methodName().
Note that your class can call super.methodName() anywhere, not just in the overriding function, although the most usual use for it is in the overriding function if you want the super implementation to run as well as your own.
Constructors are a slightly special case as there are rules about how and why constructors are called in order to make sure that your super-class is fully initialized when you try and use it in the inheriting class.
super is always called whether you write super(); or not.
In the example printStatus() method of Class A will never be called. Since you are creating an instance of class B and there will be method overriding. You can use the following to call the Class A printStatus() method.
public B()
{
super.printStatus();
}
When you override a method, it will override the one that you expect from class A.
Should use super keyword for calling super class method.
class A {
public A(){
printStatus();
}
public void printStatus(){
System.out.println("In Class A");
}
}
class B extends A{
public B(){
super.printStatus();
}
#Override
public void printStatus(){
System.out.println("In Class b");
}
}
Constructor public B(){ super.printStatus(); } calls Class A print method and constructor public A(){ printStatus(); } calls Class B print method since you've overridden.
But its wrong with overridable method calls in constructors.
Try with like this :
class A {
public A(){
printStatus();
}
public void printStatus(){
System.out.println("In Class A");
}
}
class B extends A{
public B(){
super.printStatus();
printStatus();
}
#Override
public void printStatus(){
System.out.println("In Class b");
}
}
public class Test2 {
public static void main(String[] args){
B b = new B();
}
}
For better understanding the concepts of Overloading and Overriding just go through this links:
http://en.wikibooks.org/wiki/Java_Programming/Overloading_Methods_and_Constructors
I have
public class D extends B
{
public void method() {}
}
public class B
{
public void method() {}
public void anotherMethod() { method(); }
}
In the above, if you hold an instance of D, say d, d.anotherMethod() results in calling D.method.
Is there a syntax in Java to call B.method() from inside anotherMethod()?
No, there isn't. The derived class would have to contain a call to super.method().
If B wants to prevent subclasses from overriding method(), it should declare method() as final.
You still can call the super method by explicitly using super like this this:
public class D extends B{
public void method() {}
public void anotherMethod() { super.method(); }
}
The only thing that is required is for you to override anotherMethod().
Another way is thinking like this. You want anotherMethod to call B method() so:
public class D extends B{
public void methodInternal() {}
}
public class B{
public final void method() {
//...
methodInternal();
}
public void methodInternal() {}
public void anotherMethod() { method(); }
}
Here the user can create his own version of method() by overriding methodInternal(), but still the behavior of the original method() is intact.
You can make method() static in B and then call it as B.method() from the instance methods where it is needed (you may need to rename the static method if you want to use the name for the instance method).
At least that's what I do in similar situations.
Maybe you should also reconsider your design.