forward referencing for static variable??

[Richa Jeetah]

Hi,
Can you please explain why the answer is 4 and not 1?
as i understant at line #7 we try to print value of p which should be intialised only after the return atatement at line #8.
tough i know that static variable are initialized automatically(if not given an initializing value, like if it were static int p; at line#3)
...BUT are they intialized automatically even if we specify..as on line#3 ?
1 class Q30
2 {
3 static int p=abc();
4 static public int abc()
5 {
6 int i=123;
7 System.out.println(p);
8 return i;
9 }
10 public static void main(String arg[])11
12 {
13 }
14 }
1. Compile time error for forward referencing the variable p at line no. 7.
2. Run time error for forward referencing the variable p at line no. 7.
3. Program compiles correctly and prints 123 when executed.
4. Program compiles correctly and prints 0 when executed.

[Thomas Paul]

Because all instance and class variables are provided with a default Value. If you run the program, the output is 0.

[Alfred Kemety]

Well I think the trick is:
static int p = abc();
is declaration + initialization to the variable, you should look at at as:
static int p;
p = abc();
so my guess is that once the compiler see the declaration part, it gives the default value to the p as it's a member variable, then it does the evaluation of abc() and then assign it to p.
When abc() is executed, the value of x is already set to the default (0). if you add a line in the main method, p = abc(); it will print 0 then 123.
if you want to see it clear, then make abc()
static int abc(int i){
System.out.println(p);
return i;
}
pass 2 in the first call and 3 in the second and it will print 0 then 2
hope this help.

[Richa Jeetah]

thanx, now its seems more clear