super.start()

[ashok khetan]

hi! can u pl. explain the sequence of the statements as output in the code given below?
thanx.
ashok.
___________
class fort extends Thread {
public void start(){
super.start();
System.out.println("in start");
}
public void run() {
System.out.println("in run");
}
public static void main(String args[]){
fort t = new fort();
t.start();
}
}

[Roopa Bagur]

The output will be
in start
in run
According to the api doc the start method
"Causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread."
So my understanding is even if you call super.start() , this.Start() is executed followed by this.run()..
Anybody correct me if I am wrong.

Originally posted by ashok khetan:
hi! can u pl. explain the sequence of the statements as output in the code given below?
thanx.
ashok.
___________
class fort extends Thread {
public void start(){
super.start();
System.out.println("in start");
}
public void run() {
System.out.println("in run");
}
public static void main(String args[]){
fort t = new fort();
t.start();
}
}

[Ragu Sivaraman]

The moment you override the start() method, the ability of this method to call the run() method is GONE!!
So to make the start() initiate the run() we use super.start()
t.start() in this case, is just a normal instance member invocation and the super.start() inside does what its supposed to do
HTH
Ragu

[Cameron Park]

I don't understand why is "in start" printed before "in run"? Shouldn't super.start() print "in run" when it invoke run(), then "in start" was printed?
Thanks.

[Valentin Crettaz]

No the output is correct, but I think the reason why it is like that is that the start method returns immediately.
Val

[Roger Wang]

check this modified code:
class fort extends Thread {
public void start(){
super.start();
for (int i=0; i<100;i++) System.out.println("in start");
}
public void run() {
for (int i=0; i<100;i++) System.out.println("in run");
}
public static void main(String args[]){
fort t = new fort();
t.start();
}
}
so I think it's just because the println("in start"); get the cpu slightly earlier.

Originally posted by Cameron Park:
I don't understand why is "in start" printed before "in run"? Shouldn't super.start() print "in run" when it invoke run(), then "in start" was printed?
Thanks.

[Nisheeth Kaushal]

Hi Valentin,
I idea is not yet cleared, if you try to compile the code provided by Roger, it will sometimes print in start first ans sometimes print it later why it is so.
Please give solid concept behind this.
Nisheeth.

[R K Singh]

Originally posted by Nisheeth Kaushal:
Hi Valentin,
I idea is not yet cleared, if you try to compile the code provided by Roger, it will sometimes print in start first ans sometimes print it later why it is so.
Please give solid concept behind this.
Nisheeth.

when u call start() method of Thread , then it just registers the tread with thread schedular, now it is thread schedular's responsibility when it gives cpu to run the thread.
I think sequence of execution totally depend on thread schedular.
CMIW
------------------
Regards
Ravish

[Yin Ming]

Originally posted by ravish kumar:
when u call start() method of Thread , then it just registers the tread with thread schedular, now it is thread schedular's responsibility when it gives cpu to run the thread.
I think sequence of execution totally depend on thread schedular.

but when i call the super.start(), why does it causes executing this.run()? does the super.start() register the thread on thread schedular with this reference? so the thread schedular executes the the subclass's run() but not the super's ?

[R K Singh]

Originally posted by Yin Ming:
but when i call the super.start(), why does it causes executing this.run()? does the super.start() register the thread on thread schedular with this reference? so the thread schedular executes the the subclass's run() but not the super's ?
[/B]

I think in start() there should be some thing like..
Thread threadObj = ourObject
as our ourObject is subclass of Thread so when threadObj's run is scheduled then our run is executed coz of overriding
I am downloading source of java2sdk ,will get in touch soon..
CMIW

------------------
Regards
Ravish