public methods in package-private classes

折月煮酒 提交于 2019-11-27 18:29:11

If the class is not going to be extended by another, more visible subclass*, the only difference is clarity of intent. Declaring all methods package private makes it more difficult for future readers to determine which of the methods are meant to be called by other classes in the same package.

*which would not make much sense as a design solution to me, but technically is possible nevertheless.

Example using inheritance:

A.java

package pkg1

class A {
  void foo();
  public void bar() {};
}

B.java

package pkg1

public class B extends A{

}

C.java

package pkg2

public class C {
  public void doSomething() {
   B b = new B();
   b.bar(); //ok
   b.foo(); //won't work, since foo() is not visible outside of package 'pkg1'

   A a = new A(); //won't work since A is not visible outside of package 'pkg1'
   a.bar(); //won't work, since a cannot be created
  }
}

Another case where the method has to be public is when you are creating a package private implementation of some public class or interface. Since you are not allowed to reduce the visibility of overridden methods, these have to be public.

Well... i had this doubt also (that's why i searched for this thread). This might be a good question.

But ...

After a second thought, things are really simpler than we thought.

A package-private method, is a package-private method.

Seems nonsense? But ...

A package-private method, even if its class is inherited, it is still a package-private method.

Does it make more sense now? For a more verbose explanation:

A package-private method, even if its class is inherited by a more visible subclass, it is still a package-private method.

If the subclass is of the same package, those package-private methods are also inherited, but they are still package-private.

If the subclass is of the different package (of here, we need the parent class to be public, with some package-private methods), those package-private methods are not inherited (because they are not visible at all).

A point to note, which may be the cause of this doubt, that a package-private method in a public class is not visible outside its package also.


The above explains about the package-private method. And the case of the public-method is just the same.

When a package-private class is inherited by a pubic subclass (of the same package, this is a must), its public methods are inherited as public methods, and thus they become public methods in a pubic class.

In the OP example, as foo() and bar() are both in a package-private class, their visibilities are limited to package-private at this moment, until further codes are added (e.g. inheritance).

Hope this is clear (and simple) enough.

P.S. the Java access control table

It makes very little difference, unless the class is extended by a public (or protected nested) class.

As a matter of consistency, I'd go for public. It should be rare for methods to be anything other than public or private.

There are a number of examples of public methods in the package private java.lang.AbstractStringBuilder class. For instance length() is public in AbstractStringBuilder and is exposed in the public class StringBuilder (overridden in StringBuffer to add synchronisation).

Yes. public methods in a package-private class can be useful (although they are rare). Illustrative example from the java library (and taken from the classic book "effective java"):

Have you ever seen the java.util.JumboEnumSet, java.util.RegularEnumSet?? Probably not, because they are private, and are not mentioned in the documentation.

Have you ever used them?? Probably yes. They are returned by the static method of java.util.EnumSet.noneOf(...). Although you can't construct them yourself, you can use their public methods when they are returned from this method.

The good thing about that is that you don't know which one you are using, or even that they exist. Java decides which one is more efficient depending on the arguments, and might use a different on in a newer version. You only access their functionality by referring to them through their common interface (which is a good practice anyway!!)

foo has default package visibility where as bar has public visibility

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