java - Confusing method bindings -

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the output of simple program this base.

public class mainapp{     private void func(){       system.out.println("this base");      }      public static void main(string[] args){         mainapp newobj = new derived();         newobj.func();     } }  class derived extends mainapp{     public void func(){        system.out.println("this derived");      } }

  • my question when using line mainapp newobj = new derived(); not creating object of derived class using reference of base class mainapp. so, when using object call it's method why don't method derived class? why method base class.

  • using line, mainapp newobj = new derived();, working reference of mainapp or working object of derived class. 1 correct?

the reason you're getting base class method base class version of func declared private, , java not allow private methods overridden subclasses. means if extend base class , pure coincidence decide name private member function same name private member function in base class, don't accidentally change behavior of base class's member functions. indeed working object of type derived, because reference statically typed mainapp, calling func interpreted calling private method func in mainapp rather public method func in derived. changing code read

derived d = new derived(); d.func();

fixes this, because static type of d derived , call func has different meaning.

at jvm bytecode level, instruction used call private member functions invokespecial whereas instruction used call normal, override-able member functions invokevirtual. 2 have totally different semantics; invokespecial begins looking in current class (or base class things constructors), whereas invokevirtual looks in class corresponding object's type @ runtime.


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