Question

This is a code sample of inheritance while the code functions fine.I am unable to understand the output of this code kindly explain it to me.

class R {
    private void S1() {
        System.out.println("R:S1");
    }

    protected void S2() {
        System.out.println("R:S2");
    }

    protected void S1S2() {
        S1();
        S2();
    }
}

class S extends R {
    private void S1() {
        System.out.println("S:S1");
    }

    protected void S2() {
        System.out.println("S:S2");
    }
}

public class Inheritance {
    public static void main(String srgs[]) {
        new S().S1S2();
    }
}

The output is:

R:S1
S:S2

Why is the first call made to,R class’ S1 while second to S class’ S2.

Answer 1

R.S1 is private, so it’s not called polymorphically, and S.S1 doesn’t override it.

R.S2 is protected, so S.S2 overrides it, and when you call S2 on an instance of S2 (even if it’s only statically known to be an instance of R), S.S2 will be called.

From section 8.4.8.1 of the JLS:

An instance method m1, declared in class C, overrides another instance method m2, declared in class A iff all of the following are true:

• C is a subclass of A.

• The signature of m1 is a subsignature (§8.4.2) of the signature of m2.

• Either:

  • m2 is public, protected, or declared with default access in the same package as C, or

  • m1 overrides a method m3 (m3 distinct from m1, m3 distinct from m2), such that m3 overrides m2.

Note how m2 can’t be private.