This question already has answers here:
Private methods in Inheritance
(9 answers)
Closed 7 years ago.
A friend of mine asked this question to me. Why the following code does not give error on invoking aa.x()?
I understand that aa is a reference to object of class B but is invoking private method of class A inside the method of class A where it is visible and hence accessible.
Is my understanding correct? Or is there any other reason behind this?
public class A {
public void xyz() {
System.out.println("A");
}
private void x() {
System.out.println("A:x");
}
public static void main(String[] args) {
B b = new B();
A aa = b;
aa.x();
aa.xyz();
B bb = (B) aa;
bb.xyz();
bb.xyz12();
}
}
class B extends A {
public void xyz() {
System.out.println("B");
}
public void xyz12() {
System.out.println("B-12");
}
}
I can't immediately find a duplicate using a subclass, but fundamentally it's the same answer as the answer to this question.
There are two things that govern access to x:
Where the code is that's doing the access. Since x is private to A, the code accessing it must be part of a method in A. It can't be in a subclass (B) or an unrelated class.
What kind of reference you're using. If you have an A reference, you can access x on it. If you have a B reference, you can't, even though your code is part of an A method. You could cast it to A and then access x, but you can't do it directly with a reference of type B.
It is only visible because the main method where it is invoked is contained in class A. Move it to class B and it will not work
As the private methods are not inherited, a superclass reference calls its own private method.
Your main method is the method of A, therefore it can can call x() private method.
private modifier—the field is accessible only within its own class.
Related
This question already has answers here:
Why java.lang.Object can not be cloned?
(10 answers)
Closed 4 months ago.
Object's clone method is protected, therefore it can be accessed in sub classes (class A), so why am I getting 'clone() has protected access in java.lang.Object' compiler error? I thought, that all Java classes are sub classes of Object. Thanks in advance.
The code below raises the compiler error:
public class A {
public static void main(String[] args) {
Object o = new Object();
o.clone();//error
}
}
But this one compiles perfectly, don't they have the same semantics tho?
public class A {
protected void foo() {
}
}
public class B extends A {
public static void main(String[] args) {
A a = new A();
a.foo();
}
}
No, they don't.
protected means 2 things:
It's like package, _and that explains why your second snippet can call foo(). It's not about the fact that B extends A, it's that A is in the same package as B.
Subclasses can invoke it.. on themselves only. Trivially (but this doesn't work if its final), you can simply override it, implement it as return super.clone(); and now you can call it just fine.
protected members can be accessed anywhere in the same package and outside package only in its child class and using the child class reference variable only, not on the reference variable of the parent class. We cant access protected members using the parent class reference.
This question already has answers here:
Can I get the instance of the calling object in Java?
(3 answers)
Closed 9 months ago.
Following example
public class C{
A myA;
public C(){
myA = new A();
}
}
public class A{
C myOrigin;
public A(){
// How to set myOrigin to the instance
// which invoke this.
}
}
Is there any way, to get the instance of class C, which creates the instance of A (inside this instance). With other words: Does an instance of A know the object from where it was initialised.
I know, I can use this as an parameter,
public class C{
A myA;
public C(){
myA = new A(this);
}
}
public class A{
C myOrigin;
public A(Object pObject){
myOrigin = pObject;
}
}
but I'm searching for a way without any parameter.
I very much doubt it's possible, but would be interested to hear otherwise. The reason I doubt it is because it's not an instance that calls a method, but another method. The calling method can be found easily through the stack trace (e.g. throw & catch an exception and examine its stack trace to find the caller). However, the instance is just a hidden argument to the method, and we have no way of knowing what arguments the calling method was called with. (At least, not to my knowledge.)
So, I think passing this to the other method is the only way.
This question already has answers here:
“overriding” private methods with upcasting call in java
(1 answer)
Overriding private methods in Java
(10 answers)
Closed 5 years ago.
Can anybody explain why output to below question is "A.test" ?
class A {
private void test(){
System.out.println("A.test");
}
public void mytest(){
this.test();
}
}
class B extends A{
protected void test(){
System.out.println("B.test");
}
}
public class Test{
public static void main(String[] args) {
A a = new B();
a.mytest();
}
}
The test() method of class A cannot be overridden by class B, since it is private. Therefore this.test(); invokes A's test() method even though it is executed on an instance of class B.
Super class can invoke the methods of the sub class without using typecasting, without using reflection and without using a reference other than this only if they are overridden.
In your case A a = new B(); , object of B is created which has both the behaviours private void test() as inherited by super class A as well as protected void test() as added by B. The reason for both the methods being there and not just one is that due to the acess modifier being private the method is not visible in subclass scope. Hence it can not overriden and adding the method with same name just simply adds another method.
As there is no run time polymorphism in this case hence compile time target method is resolved and method defined in A is invoked.
If in case you would have overriden the mytest in B and from the overriden mytest you would have made a call to test then method in B would have been invoked. This is easy to understand as any method of B can not see any private method of its super class.
This question already has answers here:
Protected member access from different packages in java - a curiosity [duplicate]
(9 answers)
Closed 7 years ago.
I am very confused with this . can any one clear me why it was not allowing the code that i have placed n comments
package pack1;
public class A {
protected void m1() {
System.out.println("This is very imp point");
}
package pack2;
import pack1.A;
class B extends A {
public static void main(String arg[]) {
// A a1 = new A();
//a1.m1();
B b1 = new B();
b1.m1();
//A a2 = new B();
//a2.m1(); }
}
}
Method m1 is protected, so it's accessible across packages to child classes.
Therefore, instances of B will be able to invoke or #Override m1.
Not so main static methods, even if pertaining to class B: the scope is different.
You can either make m1 public in A, or invoke it within your instance of B (e.g. in the constructor, etc.).
You can also override A's m1 in B and give it less access restrictions, thus making it public in this instance: then you could access it on an instance of B from your main method as you're trying to do.
You can access the protected members declared in A within B, but only for instances of B or subclasses of B. Check out this answer
This question already has answers here:
Understanding Java's protected modifier
(6 answers)
Closed 7 years ago.
I have a problem understanding protected members inheritance and visibility.
I know it is visible in the same package and subclasses.
But in the following code it is not visible in a subclass.
A.java
package a;
public class A {
public static void main(String[] args) {
}
protected void run() {
}
}
B.java
package b;
import a.A;
public class B extends A {
public static void main(String[] args) {
B b = new B();
b.run(); // this works fine
}
}
C.java
package b;
import a.A;
public class C extends A{ // it will not work also if extends B
public static void main(String[] args) {
B b = new B();
b.run(); // this is the problem; not visible
}
}
Why b.run() in the last class is invisible?
This is because class C can see A's protected methods from within its own inheritance tree. But it's not allowed to access A's protected methods for another class (B) from a different inheritance tree. C is not part of B's inheritance tree (by this, I mean that's it's not a parent of B), therefore the behavior is normal.
EDIT: Added documentation reference as requested
6.6.2.1. Access to a protected Member:
If the access is by a field access expression E.Id, or a method invocation expression E.Id(...), or a method reference expression E :: Id, where E is a Primary expression (§15.8), then the access is permitted if and only if the type of E is S or a subclass of S.
Applying the above to this case, because the variable b is not an instance of C or a subclass of C, access to the protected method b.run() is not allowed.
Also addressing Codebender's comment about the packages. Note that, if the C class would have been defined in the same package as the A class, where the protected run() method is defined, then the above rule wouldn't apply, and you would be able to access the method as shown in your code.