方法覆盖如何工作？

Question

Look at the following code snippet:

class A
{
  void fun1()
   {

    System.out.println("fun1 of A");
    fun2();
   }

  void fun2()
  {
     System.out.println("fun2 of A");
  }
}
class B extends A
{
  void fun2()
  {
    System.out.println("fun2 of B");
   }
}

in main method:

A a=new B();
a.fun1()

The output is:

fun1 of A
fun2 of B

Can you please explain this output.

According to me a.fun1 is calling fun1 of A since fun1 is not overriden by B (otherwise it would have called fun1 of B ). And, fun2() in fun1 of A is calling fun2 of B since fun2 is overriden and object of B is created at runtime. Am I thinking in correct direction ?

Answer 1

It has been answered but I'm putting this as an answer anyway because I object to the simplification of the example code and I can't properly express that in a comment. Using names such as A and B and fun() really does not help anyone understand anything, including yourself. Try this:

class Animal {

  public void  makeSound(){
    System.out.println("<silence>");
  }
} 

class Cow extends Animal {
  public void  makeSound(){
    System.out.println("Moooooooo");
  } 
}


public class Test {

    public static void main(String[] args){

       Animal animal = new Cow();
       animal.makeSound(); // what sound is the animal going to make?
    }

}

If you use something "realistic" that is easy to envision, it all of a sudden becomes almost self-explanatory.

Note: I purposely left out any reference to the abstract keyword because that is not within the context of this question.

Answer 2

Your understanding is mostly correct. Just remember that all functions in Java are virtual and methods will be called depending on run-time type of the object you're working with. The trick is that when you do fun2(); there is an implicit this so it becomes this.fun2() . Since this in your exapmle this points to an object of class B , the overriden method will be called.

Answer 3

Yes. Not much more to say than that your interpretation is correct.

Answer 4

A a=new B();

That line says that get the implementations in the type B when they called. Keep it simple. You are right.