package packageOne;
public class Base
{
protected void display(){
System.out.println("in Base");
}
}
package packageTwo;
public class Derived extends packageOne.Base {
public void show(){
new Base().display(); //this is not working throws compilation error that display() from the type Base is not visible
new Derived().display(); //is working
display(); //is working
}
}
The two packages are in two different files. But why this behaviour?
The protected access modifier is accessible within the package. However, it can also accessible outside the package but through inheritance only. We can't assign protected to outer class and interface. If you make any constructor protected, you cannot create the instance of that class from outside the package.
The protected members are inherited by the child classes and can access them as its own members. But we can't access these members using the reference of the parent class. We can access protected members only by using child class reference.
Protected Access Modifier - Protected Variables, methods, and constructors, which are declared protected in a superclass can be accessed only by the subclasses in other package or any class within the package of the protected members' class. The protected access modifier cannot be applied to class and interfaces.
Protected members in a class are similar to private members as they cannot be accessed from outside the class. But they can be accessed by derived classes or child classes while private members cannot.
protected
allows access from subclasses and from other classes in the same package. That's why any Derived
class instance can access the protected method in Base
.
The other line creates a Base
instance (not a Derived
instance!!). And access to protected methods of that instance is only allowed from objects of the same package.
display();
-> allowed, because the caller, an instance of Derived
has access to protected members and fields of its subclasses, even if they're in different packages
new Derived().display();
-> allowed, because you call the method on an instance of Derived
and that instance has access to the protected methods of its subclasses
new Base().display();
-> not allowed because the caller's (the this
instance) class is not defined in the same package like the Base
class, so this
can't access the protected method. And it doesn't matter - as we see - that the current subclasses a class from that package. That backdoor is closed ;)
http://java.sun.com/docs/books/jls/third_edition/html/names.html#6.6
class C
protected member;
// in a different package
class S extends C
obj.member; // only allowed if type of obj is S or subclass of S
The motivation is probably as following. If obj
is an S
, class S
has sufficient knowlege of its internals, it has the right to manipulate its members, and it can do this safely.
If obj
is not an S
, it's probably another subclass S2
of C
, which S
has no idea of. S2
may have not even been born when S
is written. For S
to manipulate S2
's protected internals is quite dangerous. If this is allowed, from S2
's point of view, it doesn't know who will tamper with its protected internals and how, this makes S2
job very hard to reason about its own state.
Now if obj
is D
, and D extends S
, is it dangerous for S
to access obj.member
? Not really. How S
uses member
is a shared knowlege of S
and all its subclasses, including D
. S
as the superclass has the right to define behaviors, and D
as the subclass has the obligation to accept and conform.
For easier understanding, the rule should really be simplified to require obj
's (static) type to be exactly S
. After all, it's very unusual and inappropriate for subclass D
to appear in S
. And even if it happens, that the static type of obj
is D
, our simplified rule can deal with it easily by upcasting: ((S)obj).member
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