Java: calling a super class' protected method from a subclass - not visible?

Question

I am calling a super class' protected method from a subclass. Why is this method "not visible"?

I've been reading some posts such as this one , that seem to contradict the following:

Super class:

package com.first;

public class Base
{
    protected void sayHello()
    {
        System.out.println("hi!");
    }
}

Subclass:

package com.second;

import com.first.Base;

public class BaseChild extends Base
{
    Base base = new Base();

    @Override
    protected void sayHello()
    {
        super.sayHello(); //OK :)
        base.sayHello(); //Hmmm... "The method sayHello() from the type Base is not visible" ?!?
    }   
}

Answer 1

base is a variable that is not special in any way: it's not part of a class hierarchy and protected access is not available through it. Even though sayHello has access to protected members of Base , it has that access only through inheritance (since it's not in the same package: the protected keyword allows access via both inheritance and package, see the table in this Oracle tutorial ).

Access through this and super is allowed because those are part of the inheritance hierarchy.

Answer 2

This is the correct behavior. In fact, Java Language Specification , section , 6.6.2-1, has an example very similar to yours, with a comment that it should not compile.

The specifics of accessing protected members are detailed in section 6.6.2.1:

6.6.2.1. Access to a protected Member

Let C be the class in which a protected member is declared. Access is permitted only within the body of a subclass S of C .

In addition, if Id denotes an instance field or instance method, then:

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 .

If the access is by a field access expression E.Id , where E is a Primary expression, or by a method invocation expression E.Id(. . .) , where E is a Primary expression, then the access is permitted if and only if the type of E is S or a subclass of S .

It's the last paragraph that describes why the access should be denied. In your example, C is Base , S is BaseChild , and E , the type of variable base , is also Base . Since Base is neither a BaseChild nor a subclass of BaseChild , the access is denied.

Answer 3

The keyboard protected is for visibility on the same package. If the child class is not of the same package, the protected methods of parent is not visible to the child class.