Static Method - Redefine/ not Overriding

[Anvi Dixit]

Hello Ranchers,

There is rule that static method cannot be override but it can be redefine. I am confused with this.Please find below the code and let me know how it works and why.

  class Animal {
static void doStuff() {
System.out.print("a:");
}
}
class Dog extends Animal {
static void dostuff() { // it's a redefinition,
// not an override
System.out.print("d:");
}
public static void main(String [] args) {
Animal [] a = {new Animal(), new Dog(), new Animal()};
for(int x = 0; x < a.length; x++)
{
if(a[x] instanceof Dog)
{
((Dog)a[x]).doStuff();
a[x].doStuff(); // invoke the static method
System.out.println("if instance of dog");
}
else
{
a[x].doStuff();
System.out.println("if not");
}

}
}
}
  

The output for this is :
a:if not
a:a:if instance of dog
a:if not

please explain how the second output a:a:if instance of dog generates. Also let me know if i need the output as a :a:

Please help me

Thnaks in advance
Anvi Dixit

[ShivKumar Rajawat]

Hi,
this is the code that produces the output:

((Dog)a[x]).doStuff(); //Output : "a:"
a[x].doStuff(); //Output : "a:"
System.out.println("if instance of dog");

This is because array a has been defined as an array of animals.
for the 2nd statement

a[x].doStuff();

doStuff is a static method so at compile time this statement will be translated to:

Animal.doStuff();

so the statement 2 generates the out put "a:" , I am not sure how exactly complier behaves for first line . but i think it has something to do with the same concept.

[Keith Lynn]

The important thing to remember is that when a static method is invoked using a reference, the type of the reference used determines which static method is invoked.

[Andy Morris]

There is a typo in your source code!!

You have a lower case s in the dog's doStuff() method, so it never gets invoked!

If it was coded correctly the second line would have printed 'd:a:if instance of dog', because when you casted the animal that is a dog, to a dog, you would have invoked the dog's doStuff() method (not the animal's one!).

[ShivKumar Rajawat]

Thanks Andy

[Anvi Dixit]

Hi Andy,

yes you are right. there was a problem with the 's'. Lets analyse the code.

for the x=1 , a[1] = new Dog() and hence we can write
Animal a = new Dog();
a.doStuff();//which will invoke the static method of the Dog class.

output will be d:

why then its showing the output d:a:.
Please explain.


Anvi

[Keith Lynn]

Animal a = new Dog();
a.doStuff();//which will invoke the static method of the Dog class.

No, since doStuff is a static method, the doStuff method of Animal will be invoked.

[Pankaja Shinde]

i think Anvi you are confused about how static methods are invoked
just remember what Keith Lynn has written

"when a static method is invoked using a reference, the type of the reference used determines which static method is invoked."

your question is
for the x=1 , a[1] = new Dog() and hence we can write
Animal a = new Dog();
a.doStuff();
output will be d:

no you are wrong. output will be a:
as you see the reference is of Animal type so doStuff() method from Animal class is invoked. so a: will be printed. (for a[x].doStuff()
and when reference is converted from Animal class to Dog class using explicit cast, doStuff() method from Dog class is invoked printing d:
(for ((Dog)a[x]).doStuff()

Pankaj Shinde