Question

Here are two examples:

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


public class B extends A {      
    public void foo(B b) { // (1)
        System.out.println("in B");
    }
    public void f(Object o) { // (2)
        System.out.println("in B");
    }
}

I don’t understand how come (1) and (2) are considered to be an overrided method for A’s foo(). method number 1 accepts a lower class than the original foo(), I mean that I can’t send her any class of A. as I see it, (1) does not extend foo(), but it is still counts as an override- why?( Similar argument for (2), but opposite).

The reason that I made me think that it is overrided is this:

when I’m trying to run

  B b = new B();

  b.foo(b);

It checks if A has a foo that accepts B. since every B is an A, it does have one so it ought to print “in A” and not “in B” because B does not overrides it. but it prints “in B”..

Answer 1

Neither of them override the super class A’a method.

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


class B extends A {

  @Override
  public void foo(B b) {
    System.out.println("in B");
  }
  @Override
  public void foo(Object o) {
    System.out.println("in B");
  }
}

When I compile the above I get errors:

$ javac -g O.java 
O.java:10: method does not override or implement a method from a supertype
  @Override
  ^
O.java:14: method does not override or implement a method from a supertype
  @Override
  ^
2 errors

But note that it is ok to return a subclass from an overriding method. The following compiles without error.

class A {
  public A foo() {
    System.out.println("in A");
    return this;
  }
}

class B extends A {
  @Override
  public B foo() {
    System.out.println("in B");
    return this;
  }
}