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Java protected modifier not working as expected


By : Tdogiii
Date : October 17 2020, 11:12 PM
With these it helps The behavior you observe is because of the fact that printName is static. The protected modifier provides visibility of that method in the sub-class but if the method is non-static, invocation of that method is possible only through the instance of a sub-class (the concern you were raising "if I inherit from the class superClass as follows I do not get any error!"). The JLS section 6.6.2.1 defines this succinctly.
code :


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Java protected modifier


By : David
Date : March 29 2020, 07:55 AM
I hope this helps . I also had the impression what protected meant "accessible from the same package or from a subclass" but the Java Language Specification is of course more precise, and explains that in a subclass S of C, "If the access is by a qualified name Q.Id, where Q is an ExpressionName, then the access is permitted if and only if the type of the expression Q is S or a subclass of S."
So you can only access a protected method of the superclass via a reference to the subclass you are calling from, like this:
code :
public class C2 extends C1 {
    private C2 c2_other_instance;
    public void doItAgain() {
        c2_other_instance.doIt();
    }
}

Protected access modifier in Java


By : Alice
Date : March 29 2020, 07:55 AM
To fix this issue The webpage @MadProgrammer linked gives a decent explanation:
code :
package first; // Current package

First fst = new First(); // from package first and does not extend anything
fst.pro();
package second; // Current package

First fst = new First(); // from package first and does not extend anything
fst.pro();
package second; // Current package

Second sec = new Second(); // from package second and extends First from package first
sec.pro();
package first; // Current package

Second sec = new Second(); // from package second and extends First from package first
sec.pro();
package first; // Current package

Third third = new Third(); // from package third and extends Second from package second,
                           // which extends First from package first
third.pro();

Java - protected access modifier


By : elijk88
Date : March 29 2020, 07:55 AM
I hope this helps . I think a code illustration would help here to understand better.
Add a protected member in Class E
code :
public class E extends A {
    protected int protectedIntE;
    ...
e.protectedInt = 0; // ok, but why?
e.protectedIntE = 0; // error, exactly as you expected

Understanding java's protected modifier


By : John
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have a class called A in package1 and another class called C in package2. Class C extends class A. ,
What's going on here?
code :
C c = new C();
System.out.println(c.publicInt);
System.out.println(c.protectedInt);

Java - Protected Modifier with Interface


By : Renatto Mysael Mysae
Date : March 29 2020, 07:55 AM
hop of those help? You cannot add more restriction to the method you override in the sub class.In the same way, you cannot add more restriction to the method you implement from an interface.
Now, since methods defined in an interface is by default - public (Yes, you don't need to write it explicitly), so you cannot make your method protected in implementing class.
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