Static variable roles!

[rex tony]

Hello to everyone.
Usually the static variable that we never to use "this" pointer concepts.
Is it true?whereas i saw one programme where static variable used this pointer.
Even as i've written the same code.
-----------------
public class Staticthis {
public static int i = 22;
public void drive() {
static int i = 10;

System.out.println(this.i);

}


public static void main(String argv[]) {
int i = 99;
Staticthis h = new Staticthis();
h.drive();
}

}
anyone can you explain this code.
regards
rex

[Lalit Bansal]

Rex,

You cannot used 'static' modifier for a method's local variable. For them only 'final' modifier is permitted.
Hence you will get a compiler error at line:
static int i = 10;

If we remove the static modifier and then run the program, the o/p is 22.
However if we remove 'this' keyword sysout, the o/p is 10.

I too am not sure on the reason for this. Might be related to shadowing of variables.
Other member(s)/moderator(s), please respond with the solution.

[ravin kacha]

'this' actually returns the current object.

but as far as static is concern it means it belongs to the class. so it doesn't matter you use 'this' with static or not.

It will always remain same for all other objects.

[dhwani mathur]

Hi ravi!!

you have said that

but as far as static is concern it means it belongs to the class. so it doesn't matter you use 'this' with static or not.

what it actualy means ,please correct if i am wrong

public class Staticthis {
public static int i = 22;//static variable
public void drive() {
int i = 10;//local variable
 
[B]System.out.println(i);//my doubt is here if i dont use "this" 
 
//will the value of i will be printed 22?[/B]
 
}
 

public static void main(String argv[]) {
int i = 99;
Staticthis h = new Staticthis();
h.drive();
}
 
}

i am not sure with above ,please explain............

Thanks in advance!!!

Preparing SCJP 1.5

[ravin kacha]

no the value will be 10, because when you declare the variable i in your method, which is local to that method only, sysout will find the local one first.
cause you have override the variable i, which is unfortunately having the same name as your static variable.

now you will be having another question that static members can not be override.....
yes, that is true static members can't be override but here it's been redefined and not override.

[Lalit Bansal]

Hi Ravin,

Understood your logic. But what exactly is the reason that on using 'this' it goes for the static variable and not local?

Had the static 'i' been an instance variable, then the logic works well. But in case of static 'this' doesn't have a meaning.

Pls explain in detail.

Thanks in advance.

[dhwani mathur]

hey Ravin,

thanks for the explanation now
i am clear with the doubt but still
want to conform one thing please let me know
if i am right or not?

Does the above concept is related to Shadowing of
variables since both instance and local variable
has got the same name?

[ravin kacha]

@Lalit:

actually using this with any member is like below example

class A{
   int i = 1;
   static int j = 2;
   public static void main(String... args){
      A a = new A();
      a.i = 10;
      a.j = 20;
      //this.j = 200; //in static you can't use this
   }
}

in the above example when you assign a value to j either with object 'a' or with 'this' it doesn't have to do anything with the instance. because it is a class variable.

it just a way to confuse you by assigning it a value with the instance.

see all the things are one and same for a static variable which i have described below:

a.j = 20;
this.j = 20;
A.j = 20; // you can reference a static member with its class name

all the statements above are one and same.

@Dhawani:

yes dhawani you are right, it is the concept of shadowing hope the above example will give you a better picture of static concepts.
[ August 30, 2007: Message edited by: ravin kacha ]

[dhwani mathur]

Thanks ravin !!!