superclass accesing methods of a sublass in oop

Question

I was wondering, is it possible that a superclass to access the methods of a inherited subclass, like for example in Java?

I know that a subclass can override and even implements, in case of abstract classes, the methods of the superclass, but the question mentioned above is possible?

Thanks

Answer 1

Example in c#.. in superclass make abstract method, which is implemented in derived class

public abstract class SuperCLass
{
    public void CallSubMethod()
    {
        Test(); // calls method in derived class
    }
    public abstract void Test();
}

public class SubClas : SuperCLass
{
    public override void Test()
    {
        // code here 
    }
}

Answer 2

Java, PHP, Ruby, Python, C# (and so on) methods are always virtual, so, no matter what, when you override a method in a subclass, this version will be called:

public class SuperClass {
    public void someMethod() {
        otherMethod();
    }

    public void otherMethod() {
        System.out.println("Super");
    }
}

public class SubClass extends SuperClass {
    public void otherMethod() {
        System.out.println("Sub");
    }
}

SubClass o1 = new SubClass();
o1.someMethod(); // Outputs: Sub 

SuperClass o2 = new SubClass();
o2.someMethod(); // Also outputs: Sub

So, you not just CAN access your subclass method, you HAVE TO.

Just for comparison, in C++, for example, things work different. If you don't declare a method as virtual, you can't override it.

Answer 3

I' ll try to explain as they explained to me at university.

You have a reference:

Object o = new Object()

• His static type(ST) is Object : this is his own type and never changes.

• His dynamic type(DT) is also Object(in this case): the reference point to an object of type Object, but it can change.

  for example if i write :

  Object o = new String("abc") // now his ST == Object but the DT == String
  

That being said:

• Upcasting is always permitted: consider two references s and r. the assignment s=r compile and execute always if ST(r) <= ST(s) (the static type of r is, in the hierarchi, less or equals to the static type of s)

  for example:

  class A { }
  class B extends A { }
  ...
  A a = new A(); B b = new B();
  a = b  // OK, upcast
  

• Downcasting: at compile-time it is always legal to downcast from a type X to a type Y if X and Y belong to hierarchy. Consider the reference s. I want to cast s to a type C, so if C <= TS(s) it will always compile if I do the cast as : C r = (C)s

  for example:

  class A { }
  class B extends A { }
  class C extends A { }
  ...
  A a = new A(); B b = new B();
  C c = new C();
  ...
  b = c // ILLEGAL
  b = (B)a // OK at compile-time but maybe at run-time it is not!
  

  When we run our application if the downcast fails, Java raise an Exception. Otherwise it success.

  To downcast correctly: consider a reference ref and we want to cast to a type C. So a downcast will success if DT(ref) <= C <= ST(ref) .
  And the downcast will be obtained as: C ref2 = (C)ref

  for example:

  // I suggest to write the hierarchy in a piece of paper and 
  // try the rules before coding.
  
  class A { }
  class B extends A { }
  class C extends A { }
  class D extends B { }
  ...
  
  A a = new A(); B b = new B();
  C c = new C(); D d = new D();
  A r = new B();
  A s = new D();
  a = b; // OK, upcast
  a = d; // OK, upcast
  /* b = c; */ // ILLEGAL
  b = (B)r; // OK, downcast
  d = (D)r; // downcast: it compiles, but fails at run-time
  d = (D)s; // OK, downcast
  /* b = s; */ // ILLEGAL
  /* d = (D)c; */ // ILLEGAL
  b = (B)s; // OK, downcast
  b = (D)s; // OK, downcast
  

PS: please forgive if I made some mistake but I wrote a bit in a hurry.

Answer 4

In Java, It's not possible, and I think what you are asking would go against OOP.