指向基类的动态多态性引用

Question

I need a clarification in Dynamic polymorphism of Java.

class Foo {
  int a=3;
  public void display() {
    System.out.println(" in foo "+a);
  }
}

class Bar extends Foo {
  int a=8;
  public void display() {
    System.out.println(" in boo "+a);
  }
}

public class Tester {
 public static void main(String[]args) {
  Foo f = new Bar();
  f.display();
  System.out.println(f.a);
 }
}

Here when i create a child class object with a base class reference , while invoking the method f.display() it gives me the output as in boo 8 . This is because of dynamic polymorphism which checks the object type at run time for invoking the method.

Now while printing fa it prints 3 because the variables cannot be overridden in java this is called hiding.That is why it is displaying the base variable value not the child variable value.

Now my question is f is a reference of base class which is pointing to the child class object. Then how fa points to the base variable.? what happens behind the scene ? How the reference is pointing to the base class ?

(I know the rules , but i want to know how/why ? )

Answer 1

I don't know if this is outside what you've learned, but here it is. When you compile your code, the compiler generates byte code which the JVM executes.

The reference to the field Foo.a in fa in the line

System.out.println(f.a);

gets compiled to

getfield      #6                  // Field Foo.a:I

where getfield is a bytecode instruction which

get a field value of an object objectref, where the field is identified by field reference in the constant pool index (index1 << 8 + index2)

and the constant pool

Constant pool:
   // [...]
   #6 = Fieldref           #20.#24        //  Foo.a:I
   // [...]

So the byte code is referencing the field in class Foo , the declared type of the variable, rather than in the run time class type of the instance.

---

You can look at the generated byte code with the following command

javap -c -v YourClass

Answer 2

This is because of what you have already stated in your question:

variables cannot be overridden in Java

Therefore, reference fa is resolved statically at compile time, and it brings the compiler to Foo.a , which is 3 .

Answer 3

Because variables cannot be overridden and you give the reference from base class that why its

print f.a = 3 

if you declare like this

Bar f = new Bar();

and now you print the value of fa it gives you 8