Protected inner-class is NOT accessible from subclass of another package

Question

I have the following code in Java:

package a; public classs ClassInOtherPackage{ protected void protectedMethod(){} protected class protectedInnerClass {} } package b; import a.*; public class MyClass extends ClassInOtherPackage{ public MyClass(){ MyClass x = new MyClass(); x.protectedMethod(); //<-- This one ok //UPDATED: Declaration is ok MyClass.protectedInnerClass y; //<-- This one ok //UPDATED: Only when instantiated is not allowed, why? y = x.new protectedInnerClass(); //<-- Not allow when instantiated. } }

Refer to the Java documentation:

"The protected modifier specifies that the member can only be accessed within its own package (as with package-private) and, in addition, by a subclass of its class in another package."

Why can't I instantiate the inner protected class as shown above?

possible duplicate of Can't access protected inner class while inheriting — Narendra Pathai
– Narendra Pathai, Commented Jul 12, 2013 at 8:25
It is because the protected class is visible, but the constructor is also protected and so the constructor is only visible to its direct subclasses, which MyClass is not. So you will have to make constructor public. — Narendra Pathai
– Narendra Pathai, Commented Jul 12, 2013 at 8:27
First you should write public class ClassInOtherPackage{} instead of public ClassInOtherPackage{}. — Shreyos Adikari
– Shreyos Adikari, Commented Jul 12, 2013 at 8:27

Community · Accepted Answer · 2020-06-20 09:12:55Z

In JLS 8.8.9

8.8.9. Default Constructor

... if the class is declared protected, then the default constructor is implicitly given the access modifier protected (§6.6); ...

So the implicitly declared constructor is:

protected class protectedInnerClass { protected protectedInnerClass(){ super(); } }

You code won't compile because the constructer is inaccessible.

I just realized that this statement is ok --> MyClass.protectedInnerClass y; but not --> y = x.new protectedInnerClass(); Not sure why....
The class protectedInnerClass is accessible, but its constructor isn't.

AllTooSir · Accepted Answer · 2013-07-12 08:31:38Z

The default constructor of protectedInnerClass is protected , not the class . You need to define a public constructor to your inner class :

protected class protectedInnerClass { public protectedInnerClass () {} }

MyClass can access the protected members of ClassInOtherPackage , but it cannot access the protected members of protectedInnerClass , since its constructor is protected hence you get such a compilation error.

Thank you, in addition to johnchen902's answer, you answered with a solution that makes the concept clearer. In other words, if I have a public inner-class with a protected constructor, it won't work as well. Thank you for sharing the knowledge.

René Link · Accepted Answer · 2013-07-12 08:38:22Z

As johnchen902 said the default constructor of a protected class is implicitly protected. Protected means that only subclasses and classes within the same package can access it.

The protectedInnerClass doesn't declare a constructor thus it is implicitly protected.
MyClass is not a subclass of protectedInnerClass, thus it can not access it.

So you have to make it public or you can modify your code like this:

public class MyClass extends ClassInOtherPackage { public MyClass() { MyClass x = new MyClass(); x.protectedMethod(); // <-- This one ok MyClass.protectedInnerClass y = x.new MyProtectedInnerClass(); } protected class MyProtectedInnerClass extends protectedInnerClass { } }

My name doesn't follow the naming convention... it begins with lower case j.
Thank you so much. That makes me understand better now. Cheers!

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Protected inner-class is NOT accessible from subclass of another package

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8.8.9. Default Constructor

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