static block

[anshi kohli]

The following code will give

1: public class Q8
2: {
3: int i = 20;
4: static
5: {
6: int i = 10;
7:
8: }
9: public static void main(String[] args)
10: {
11: Q8 a = new Q8();
12: System.out.println(a.i);
13: }
14: }


in my opinion it should be 10 as static block is loaded first bt it was 20 ....Please explain

[Manfred Klug]

Hi anshi,

have a look at the highlighted part:

public class Q8
{
    int i = 20;
    static
    {
        [B]int[/B] i = 10;
    }
    public static void main(String[] args)
    {
        Q8 a = new Q8();
        System.out.println(a.i);
    }
}

Edit: Hint, there are two problems which prevent your expected output.
[ July 14, 2007: Message edited by: Manfred Klug ]

[anshi kohli]

int is unknown at the time of loading???

[Manfred Klug]

Originally posted by anshi kohli:
int is unknown at the time of loading?

i is a local variable in the sttic initializer block.

[saravana.T kumar]

public class Q8
{
int i = 20;
static
{
int i = 10;
}
public static void main(String[] args)
{
Q8 a = new Q8();
System.out.println(a.i);
}
}

Here this code compiles fine

1.int i 20 (instance block)
2.when the static block is run here the int i=10; is intiliazed but lost
after the exceution of the STATIC BLOCK.
3.o/p is 20
ABout static blocks K&B pg.224-226


public class Q8
{
{
int i = 20;
}
static
{
int i = 10;

}

public static void main(String[] args)
{
Q8 a = new Q8();
System.out.println(a.i);
}
}

But compile this it wont(for your understanding)
because the i is within instance block which will not be accessible as
Instance block and static block are excution oriented.

[saravana.T kumar]

Originally posted by saravana.T kumar:
public class Q8
{
int i = 20;
static
{
int i = 10;
}
public static void main(String[] args)
{
Q8 a = new Q8();
System.out.println(a.i);
}
}

Here this code compiles fine

1.int i 20 (instance )
2.when the static block is run here the int i=10; is intiliazed but lost
after the exceution of the STATIC BLOCK.
3.o/p is 20
ABout static blocks K&B pg.224-226


public class Q8
{
{
int i = 20;
}
static
{
int i = 10;

}

public static void main(String[] args)
{
Q8 a = new Q8();
System.out.println(a.i);
}
}

But compile this it wont(for your understanding)
because the i is within instance block which will not be accessible as
Instance block and static block are excution oriented.

[Jared Folino]

Static blocks cannot access instance variables. Stop and think about it for a second. If their are 7 instances of the class, which instance would it access, umm, arrr, yeah, as such remember it like this

static{}
can only call static methods and access static variables

{} //Initialisation block
can access static methods, static variables, instance variable (this) and instance methods (this)

[madhu v pe]

Anshi,

1: public class Q8
2: {
3: int i = 20;
4: static
5: {
6: int i = 10;
7:
8: }
9: public static void main(String[] args)
10: {
11: Q8 a = new Q8(); //Line 1
12: System.out.println(a.i);
13: }
14: }
[/QB]

you are right that static block will be executed first,
but called before creating object.

here there is a need to remember a single point
1.static variable cannot be referenced with object. so a.i cannot be a static variable

Thanks
[ July 15, 2007: Message edited by: madhu v pe ]

[Manfred Klug]

Originally posted by madhu v pe:
1.static variable cannot be referenced with object. so a.i cannot be a static variable

That is not correct.

class Test {
    static int i = 77;
 
    public static void main (String[] args) {
        Test a = null;
        System.out.println(a.i);
    }
}

In this case, the important point to remember is, that the compiler only uses the type of the reference variable, and not the value.

[Ganesh Kumar]

When a variable is declared in the initializer block it is only within the scope of that block.You check with giving any type of variables.Scope is the thing which matters.Put a out.print in that block..Got it!!

[Chetan Parekh]

Originally posted by Ganesh Kumar:
When a variable is declared in the initializer block it is only within the scope of that block.You check with giving any type of variables.Scope is the thing which matters.Put a out.print in that block..Got it!!

True and that is the reason following code will not compile.

  
public class MyClass {
//int i = 20;
static {
int i = 10;
 
}
 
public static void main(String[] args) {
MyClass a = new MyClass();
System.out.println(a.i);
}
}