Re: Study Question Help

From:
Lew <lew@lewscanon.com>
Newsgroups:
comp.lang.java.help
Date:
Thu, 3 Feb 2011 12:01:54 -0800 (PST)
Message-ID:
<40806552-af8a-44e7-ad45-f6034f5fbf0b@h19g2000prh.googlegroups.com>
On Feb 3, 2:59 pm, Lew <l...@lewscanon.com> wrote:

Steve wrote:

Thanks John. I had the idea in my head that access levels tightened
up 1 notch with inheritance. I think I am mixing up C++ rules for
Java.


An oft-overlooked access level is "default" (a.k.a. "package-
private"), which is the level when there is no other access keyword
('public', 'protected' or 'private'). This creates package-friendly
access, allowing access (and inheritance) for types in the same
package but not otherwise.

For real fun, consider an inner class that inherits from its
containing type:

 package eg;
 public class Foo
 {
    /* p-p */ void packageFriendly()
    {
        System.out.println( "Foo#packageFriendly()" );
    }

    private void hiddenInside()
    {
        System.out.println( "Foo#hiddenInside()" );
    }

    class InnerFoo extends Foo
    {
        void packageFriendly()
        {
            System.out.println( "InnerFoo#packageFriendly()" =

);

        }

        private void hiddenInside()
        {
            System.out.println( "InnerFoo#hiddenInside()" );
        }
    }

    public static void main( String[] args )
    {
        Foo foo = new Foo().new InnerFoo();
        foo.packageFriendly();
        foo.hiddenInside();
    }
 }
=========================

=====================

run:
InnerFoo#packageFriendly()
Foo#hiddenInside()


The output is the same if the inner class's 'hiddenInside()' is
declared package-private.

--
Lew

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