Overriding methds

[tural ferhadov]

Why overriding method cannot have more restrictive access modifier.?

[Jeanne Boyarsky]

Suppose I have:

public class Fruit {
  public void eat() {}
}

And code in another class that calls it:

public void eatMultiple(Fruit f1, Fruit f2) {
f1.eat();
f2.eat();
}

All is well. The method is public. Now what happens if I have

public class Apple extends Fruit {
  protected void eat() {  // will not compile
  }
}

If this did compile, eatMultiple has a problem if I call it with Apples. It can no longer call Apple's method (assuming it is in another package) since it isn't public.

[Yin Stadfield]

tural ferhadov wrote:Why overriding method cannot have more restrictive access modifier.?

It can't because it will raise a compile error. It can however, declare a less restrictive access modifier. Here's the hierarchy of access modifier (1 being the most restrictive):
1 - private
2 - default
3 - protected
4 - public

If your superclass method is declared as default (no access modifier written), you can override it with the value of 2 - 4( just go down the hierarchy), but you can't go up to 1 (private) because you're making the method more restrictive.

Hope it helps. Thanks.