polymorphism - Behavior of method overloading in java -

this question has answer here:

• two methods same name in java 3 answers

i tried following code

public class helloworld {      public void printdata(test t) {         system.out.println("reached 1");     }      public void printdata(newtest t) {         system.out.println("reached 2");     }      public void printdata(newtest1 t) {         system.out.println("reached 3");     }      public static void main(string args[]) {         test t1 = new test();         helloworld h = new helloworld();         h.printdata(t1);          newtest t2 = new newtest();         h.printdata(t2);          newtest1 t3 = new newtest1();         h.printdata(t3);          test t4 = new newtest();         h.printdata(t4);          test t5 = new newtest1();         h.printdata(t5);     } } 

and have simple classes

class test { }  class newtest extends test { }  class newtest1 extends test { } 

and output got is

reached 1 reached 2 reached 3 reached 1 reached 1 

from output looks when jvm decides function execute takes consideration only type of reference , not actual type of object.

why happen? why can't jvm take consideration type of actual object rather type of reference pointing it?

function overloading compile time polymorphism , here compiler decide version of method called.for compiler it's difficult know actual object run time check reference type irrespective of object it's going point.