CODEWITHSUNDEEP
javavirtualabstract
Stevie Kideckel
Stevie Kideckel

Reputation: 2106

Abstract/Virtual Functions in java

I've heard that all Java functions are implicitly virtual, but I'm still not sure if this will run how I want.

Suppose I have a class A, with child B. both A and B have functions called foo(), so B's definition is overriding A's.

Suppose also that A has a function called that takes an instance of A as a parameter:

If I pass in an instance of B to the function, which definition of foo() will it call, A's or B's?

Upvotes: 1

Views: 4667

Answers (2)

hoaz
hoaz

Reputation: 10161

As I mentioned in my comment private functions are not virtual and I want to demonstrate it using following example:

class A {
    public void foo() {
        System.out.println("A#foo()");
    }

    public void bar() {
        System.out.println("A#bar()");
        qux();
    }

    private void qux() {
        System.out.println("A#qux()");
    }
}

class B extends A {
    public void foo() {
        System.out.println("B#foo()");
    }

    private void qux() {
        System.out.println("B#qux()");
    }
}

Now lets run following code:

A foobar = new B();
foobar.foo(); // outputs B#foo() because foobar is instance of B
foobar.bar(); // outputs A#bar() and A#qux() because B does not have method bar 
              // and qux is not virtual

Upvotes: 4

SLaks
SLaks

Reputation: 888087

B's implementation will be called.
That's exactly what virtual means.

Upvotes: 3

Related Questions