CODEWITHSUNDEEP
java
Nik
Nik

Reputation: 7283

Java Class Protection

I suspect the answer is no, but I want to check. If I have this structure in a Java application:

-package
    -ClassA
    -subpackage
        -ClassB
        -subsubpackage
            -ClassC

I want package.subpackage.ClassB to have access to package.subpackage.subsubpackage.ClassC, but I don't want package.ClassA to have access to package.subpackage.subsubpackage.ClassC. Is there any way to enforce this?

Upvotes: 2

Views: 266

Answers (5)

tskuzzy
tskuzzy

Reputation: 36476

No, the only access modifiers are:

  • public - global access
  • protected - package and subclass access
  • "package-private" (no modifier) - package access
  • private - class access only

protected and package-private doesn't recursively grant access to subpackages. In short, subpackages don't really have any relationship with their parent package except for the fact that they share the same name prefix.

Here is a Java Specification Request that (I believe) deals with this issue: http://jcp.org/en/jsr/detail?id=294

This was supposed to be implemented in the just recently released Java 7, but apparently has been forgotten.

Upvotes: 2

Thilo
Thilo

Reputation: 262824

No. The package hierarchy in Java has no effect on package visibility.

Things are either totally public or package-private.

There is no such thing as a "friend" package or a "parent" package here.

Upvotes: 1

cwallenpoole
cwallenpoole

Reputation: 82078

The only way that you can get this to work is through the use of inner classes. And, honestly, if that doesn't work then maybe you should be asking, "What is so important about C that A shouldn't be able to instantiate it but B and at least one other class can?"

Upvotes: 2

user784540
user784540

Reputation:

Making ClassC as an internal class for ClassB may solve your task.

Upvotes: 1

Andreas Dolk
Andreas Dolk

Reputation: 114817

No, there isn't. A package name is not an hierarchical construct If we have a class

foo.bar.Application

then bar is not a child-package of foo. The package is foo.bar and there is no relation between foo and foo.bar, they are totally different packages (namespaces).

Upvotes: 1

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