How to call the m1() method of Base class in below snippet ?
class Base
{
public void m1()
{
System.out.println("Base m1 ");
}
public void m2()
{
System.out.println("Base m1 ");
}
}
======================
class Derived extends Base
{
public void m1()
{
System.out.println("Derived m1");
}
public void m3()
{
System.out.println("Derived m3");
}
public static void main(String[] args)
{
Base ob=new Derived();
ob.m1();
//System.out.println("Hello
World!"+ob.m1());
}
}
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Answer / mitu
class Derived extends Base
{
public void m1()
{
System.out.println("Derived m1");
}
public void m3()
{
System.out.println("Derived m3");
}
public void m26()
{
super.m1();
}
public static void main(String[] args)
{
Derived ob=new Derived();
ob.m1();
ob.m26();
}
}
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / kishore
Nand asking in the about snippet is saying like: The base class is referenced to derived class object (i.e Polymorphism). As a polymorphism point of view only overidden methods of Derived class will be called. But how is it possible to access the m1() of Base class using the reference of Base Class ?
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / surendra
in main() create an object to derived class and call m1() by using this obj since all super class members are available to
subclass object.
Derived d=new Derived();
d.m1();
| Is This Answer Correct ? | 1 Yes | 2 No |
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How to call the m1() method of Base class in below snippet ? class Base { public void m1() { System.out.println("Base m1 "); } public void m2() { System.out.println("Base m1 "); } } ====================== class Derived extends Base { public void m1() { System.out.println("Derived m1"); } public void m3() { System.out.println("Derived m3"); } public static void main(String[] args) { Base ob=new Derived(); ob.m1(); //System.out.println("Hello World!"+ob.m1()); } }
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