Method Overloading - Doubt

Hi,
what is the output of the following code snippet and plese explain the reason for the output.
class Demo
{
void eat(Object o)
{
System.out.println("Object");
}
void eat(String a)
{
System.out.println("String");
}
public static void main(String ap[])
{
new Demo().eat(null);
}
}

Quote:

---

Originally Posted by vidya lakshman

Hi,
what is the output of the following code snippet and plese explain the reason for the output.
class Demo
{
void eat(Object o)
{
System.out.println("Object");
}
void eat(String a)
{
System.out.println("String");
}
public static void main(String ap[])
{
new Demo().eat(null);
}
}

---

Did you give it a try? Were you surprised by the output? Or is this a quiz? If so, what can we win?

kind regards,

Jos

You spent all the time typing that in, and you couldn't click two buttons to compile and run it yourself?

Overloading is a simple concept, I don't want to give too much away, if you take some time and think it's easy to understand. If you want to have one method that takes multiple different types of arguments and still works, you can overload it to take a different arg list.

for example:

Code:

---

public int square(int x){
  return x * x;
}
public int square(String x){
  int y = Integer.parseInt(x);
  return y * y;
}

---

This allows you to supply the same method with different arguments and have it still work. If you did not overload it and try passing a String to the first method it will not work correctly since it requires an int.

I think it's because of this hierarchy.

java.lang.Object
|
-->java.lang.String

If you comment out String method, it'll go to Object method.
If I'm wrong, somebody else correct me.