Doubt Reg: initializing Variables with method call

[Ravi Pinnaboyina]

How the output is printed as Zero.......
How is the value of i being resolved ?


private int i=getData();
private int j=10;

int getData()
{
return j;
}

public static void main(String arg[])
{
System.out.println(""+new ASample().i);
}

[Keith Lynn]

The value of j is read in the method before it has been initialized.

[Ravi Pinnaboyina]

So it means that i and j will only be declared first only then assignment starts...

i.ie.,

int a=10;
int b=10;

first a nd b will be declared only then they get initialized???
Am I right??

[Keith Lynn]

The instance variables are in scope, but initialization occurs from top to bottom in the class definition. Since the call to the method accesses j before it is initialized, it has its default value.

[Anton Uwe]

public class ASample {
private int i=getData();
private int j=10;
 
int getData()
{
return j;
}
 
public static void main(String arg[]) {
System.out.println(""+new ASample().i);
}
}

is the same as

public class ASample {
private int i=0;
private int j=0;
 
public ASample() {
super();
i=getData();
j=10;
}
int getData()
{
return j;
}
 
public static void main(String arg[]) {
System.out.println(""+new ASample().i);
}
}