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javainheritanceinner-classesprotected
Kat.S
Kat.S

Reputation: 637

Java - Unable to access protected method of superclass extended by nested class

I'm new to java and getting the hang of protected visibility.

I have Class A which has protected method method1 as follows -

public abstract class A { 
    public A(...){}

    protected void method1(){
        // do something
    }

Now I have another Class B which extends a different Class C. Class B has a nested static class, Class D extending Class A

public class B extends C {
    public B(...){
        super(...);
    }

    private static class D extends A {
        D(...){super(...);}
    }

    public void method2() {
        D record = new D();
        record.method1(); // need to access the protected method1 of class A, which is the superclass of nested class D. Getting the error here
    }
}

I get this error - method1 has protected access in Class A Now my question is how can I access method1? I cannot change it to public in class A

Upvotes: 0

Views: 2109

Answers (3)

Yunus
Yunus

Reputation: 84

The problem was caused by the protected method1() called from method2(), which belongs to B, which is outside A and D, and B is not in the same package with A or D.

The method1() should only used/called inside A or D. That's what protected is used for, to ensure it is used only in the owner class, its inheriting classes or from the same package.

That's why when the wrapper method created inside D, the problem solved, since the protected method1() is called from inside D.

We still can override the protected method1() of A and make it public through D if necessary. (Modified from the wrapper example above)

class A {
    protected void protectedMethod() {}
}

class D extends A {
    @Override
    public void protectedMethod() {
        super.protectedMethod();
    }
}

final D record = new D();
record.protectedMethod();

Upvotes: 0

Chicky
Chicky

Reputation: 328

Since A.method1() is protected, it is only accessible from same class/same package/child of A.

In your code, your are calling A.method1() in B, that is not accessible if B not in same package as A

The solution is using a wrapper as @Alex Ander or change modifier of A.method1() to public

Java Access Modifiers

Upvotes: 0

CodeTheorist
CodeTheorist

Reputation: 1672

You need to create a proxy method in class D as outlined below. Class D can access Class A's protected methods, but an instance of Class D can't call protected methods directly, and must have a proxy method to call the protected method of the super class.

In class D, you need to create a public getter for the method in Class A, so:

public class A {
    protected void protectedMethod() {}
}

public class D extends A {
    public void callProtectedMethod() {
        super.protectedMethod();
    }
}

final D record = new D();

record.callProtectedMethod();

Upvotes: 3

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