A question for beginners

[Rahul Mahindrakar]

Hi ,
Without compiling can you tell the output of the following program

class demo extends Thread{
public void start(){
System.out.println("in start");
}
public void run(){
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}
}

and that of this program.

class demo extends Thread{
public void run(){
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}
}

and this program

class demo 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[]){
demo x=new demo();
x.start();
}
}

What are the results?? why?
Regds.
Rahul.

[This message has been edited by Rahul Mahindrakar (edited August 13, 2000).]

[Anonymous]

code1 : start() is overriden, the run() will not be called; so only "in start" printed
code2: normal way to call thread method, "in run " will be printed
code 3 : super.start() will call Thread.start(), which invoke run(), so both will be the output.
I didn't run it, so pls. tell me if I'm wrong

[Rahul Mahindrakar]

Hi Wonda,
Lets wait for some more people to respond. But you are right on track.

[vidya]

In case 1, since the start method is overridden,the output will be 'in start' and run() will not be executed.
In code 2, the execution takes its normal course,start() evokes run() and the output is ' in run '.
in code 3,the start() is overriden, and inside the method, it calls the superclass's start(),then the run() is invoked... is it true ? what happens here ?
please correct me if I am wrong and tell me what happens in case 3..
regards,
Vidya.

[V Srinivasan]

Result of the Code 1 :
" in run"
since start method of Thread class will look for only run method and in main method there is no start() method to invoke public void start() method
(Note: I assume that start is not a key or reserved word of Java, if so there will be a compile time error)
Result of the Code 2 :
" in run"
perfectly alright (is it!)
Result of the Code 3 :
" in run"
Again the same... start() method was not invoked in main method and hence super.start() will also not at all get chance to run the method.
I think Thread class start method is a key to run the run method. Since we are not directly calling the run method (like run())in main method.
Really it is interesting tough code...
Please clarify wherever I am wrong.
Thanks and regards,
[This message has been edited by V Srinivasan (edited August 17, 2000).]
[This message has been edited by V Srinivasan (edited August 17, 2000).]

[Rahul Mahindrakar]

Most of you are right on this front .
The answers are
1) Prints only "in start " as start has been overriden
2) Prints "in run" as the start method has not been overriden and the default start method of Thread calls run() by default of the current object rather than that of thread due to dynamic polymorphism, creating two theads.
3) Prints both "in start" and "in run" as the start method has been overriden but since it calls the start method of the super class which by default invokes run of the current object rather than that of the thread due to dynamic polymorphism.
for example

class Thread1{
public void start(){
run();
}
public void run(){
System.out.println("in run of thread1 class");
}
}
public class sub extends Thread1{
public static void main(String args[]){
sub xxx=new sub();
xxx.start();
}
public void start(){
super.start();
}
public void run(){
System.out.println("in subclass run");
}
}

The program above prints "out in subclass run" and not "in run of thread1 class" even though run is being called in the start method of the thread1 class.
I posted these questions to show the practical usage of overriding and how this can impact the program output. I also posted these three code questions to show the importance of the start method of the thread class. What does the start method of Thread class do. I enclose what the api says about start()
"Causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread. The result is that two threads are running concurrently: the current thread (which returns from the call to the start method) and the other thread (which executes its run method)."
The important thing to remember is that the call to start() creates two theads. The same call to run() method would not give the same results and run() would be invoked in the current thread.
Thread t=new Thread(Runnable Object);
t.start(); // creates two new threads
Thread t=new Thread(Runnable Object);
t.run();// does not create two threads
If one were to override the start method without a call to super.start() then neither would there be a call to run() be made nor a new thread created. If there is thus a need to ever override the start method then one should remember to invoke a call to the super.start().
Note : One can invoke the run method of a dead method but one cannot invoke the start method of the dead thread.
Regds.

Rahul P. Mahindrakar

[This message has been edited by Rahul Mahindrakar (edited August 20, 2000).]

[Anonymous]

Hi Rahul,
Did you say for code 2) Prints both "in start" and "in run" .
But there is no println statement for "in start". OR am I missing
something.

code:
--------------------------------------------------------------------------------
class demo extends Thread{
public void run(){
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}
}
----------------------------------------------------------------------
Correct me if I am wrong.
Regards
VR

[Rahul Mahindrakar]

hi VR
Thanks for pointing out the mistake. I have corrected it now. The answer is "in run" only for 2nd code.
Regds.
Rahul P Mahindrakar

[srinivas acharyulu]

Hi rahul!
Good support keep it up.

[Rahul Mahindrakar]

Thanks Acharya,
Good to see you here on JavaRanch.
Regds.
Rahul P. Mahindrakar

[vidya]

Thanx Rahul....
looking forward to more of this kind..
regards,
Vidya.

[V Srinivasan]

Hi Rahul,
Thank you very much. You mentoring is really helps a lot to beginners like me.
Thanks & Regards,

[Anonymous]

Originally posted by Rahul Mahindrakar:
Hi ,
[b]Without compiling can you tell the output of the following program

class demo extends Thread{
public void start(){
System.out.println("in start");
}
public void run(){
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}
}

and that of this program.

class demo extends Thread{
public void run(){
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}
}

and this program

class demo 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[]){
demo x=new demo();
x.start();
}
}

What are the results?? why?
Regds.
Rahul.
[This message has been edited by Rahul Mahindrakar (edited August 13, 2000).][/B]

The result from the first snipped code will be:
"in start";
You must expect for start() to call run() and output to be
: in start
in run.
It wont be, because since you override start it won't call run.
So the result is : "in start";
And the last two are going to work "ok".
Second output is : in run
Third: in start
in run

[Rahul Mahindrakar]

Right on Track AL

[Metal Zhang]

You are right.I compiled and ran the programs and the results are the same as you computed.

------------------
Java lover from hell!

[Girish Pednekar]

In the 3rd code snippet is the order of the 2 output statements guaranteed?
thanks,
GP

[Marcela Blei]

"wonda", "vidya", "VR", "al":
Your names are not valid as per the Javaranch name conventions. Please refer to the document located here: http://www.javaranch.com/name.jsp

[marilyn murphy]

Originally posted by Rahul Mahindrakar:
The important thing to remember is that the call to start() creates two theads.

I do not understand how start() creates two threads. Two threads are running, the original thread that contains main() and calls start(), and the new thread that is started by start().

Does start() start a thread in the superclass Thread as well as in the subclass (the one that contains run()) ??

[Anonymous]

>> In the 3rd code snippet is the order of the 2 output
>> statements guaranteed?
>> thanks,
>> GP
No. In all probability, on a single processor machine, "in start" will print before "in run." However, the two statements are being printed by two threads running concurrently. It's possible that the second thread (the one created by super.start()) will print "in run" before the original thread prints "in start"

[Rahul Mahindrakar]

Long time seeing this post..
any how marilyn murphy,
You are right actually its one additional thread gets created.

------------------

Mahindrakar
IBM Application Server & Developer Forum Moderator
Consultant - Zensar Technologies ,Pune India.
SCJP2, SCJD2 & SCJEA (Part I)

[Rahul Mahindrakar]

This output should help give you total number of threads

class demo extends Thread
{
public void start(){
super.start();
System.out.println("in start");
}
public void run(){
System.out.println(Thread.activeCount());
System.out.println(" in run");
}
public static void main(String args[]){
demo x=new demo();
x.start();
}}

------------------

Mahindrakar
IBM Application Server & Developer Forum Moderator
Consultant - Zensar Technologies ,Pune India.
SCJP2, SCJD2 & SCJEA (Part I)

[marilyn murphy]

Thank you for the clarification, Rahul

[Muhammad sibtain]

code1 : start() is overriden, the run() will not be called; so only "in start" printed
code2: normal way to call thread method, "in run " will be printed
code 3 : super.start() will call Thread.start(), which invoke run(), so after starting thread control will come back to the next statemtn and will print "in start" and then when that thread occupy its time and will print in run.
I didn't run it, so pls. tell me if I'm wrong