Why this difference of handling method ambiguity in Java & c#? [duplicate]

Question

I have two classes A & B, where B is derived from A.

Both the classes have a method with same signature. They are called in the following manner in Java & c#-->

---

In case of JAVA :

class A
{
    public void print()
    {
        System.out.println("Inside Parent");
    }
}

class B extends A
{
    public void print()
    {
        System.out.println("Inside Child");
    }
}

class test4
{
    public static void main(String args[])
    {
        B b1=new B();
        b1.print();
        A a1=new B();
        a1.print();
    }
}

This program generates the following output:-

Inside Child
Inside Child

---

In case of C# :

class A
{
    public void print()
    {
        System.Console.WriteLine("Inside Parent");
    }
}

class B : A
{
    public void print()
    {
        System.Console.WriteLine("Inside Child");
    }
}

class Program
{
    public static void Main(string[] args)
    {
        B b1=new B();
        b1.print();
        A a1=new B();
        a1.print();
        System.Console.Read();
    }
}

This program generates the following output:-

Inside Child
Inside Parent

Why does the output differ ?

Answer 1

In Java, methods are virtual by default.
In C#, methods are not virtual by default.

So, in order for the C# code to behave the same as the Java code, make the method virtual in the base class and override in the derived class.

Or, in order for the Java code to behave the same as the C# code, make the method final in the base class.

Answer 2

In case of c# you need to make the parent method as virtual and child method as an override

class A
{
public virtual void print()
{
System.Console.WriteLine("Inside Parent");
}
}
class B : A
{
public override void print()
{
System.Console.WriteLine("Inside Child");
}
}
class Program
{
public static void Main(string[] args)
{
B b1=new B();
b1.print();
A a1=new B();
a1.print();
System.Console.Read();
}
}