Thread - Please explain!

[Sridhar Srinivasan]

I get the output for the following code as "vandeleur".Can anybody explain please?Thanks.

 public class Tux extends Thread{        static String sName = "vandeleur";        public static void main(String argv[]){        Tux t = new Tux();        t.piggy(sName);        System.out.println(sName);                }        public void piggy(String sName){                sName = sName + " wiggy";        start();        }        public void run(){                for(int i=0;i  <  4; i++){                sName = sName + " " + i;                        }        }}
 

[fethi makhlouf]

Hi Sridhar
that's ur code in a more "readable" way ....
public class Tux extends Thread{
static String sName = "vandeleur";
public static void main(String argv[]){
Tux t = new Tux();
t.piggy(sName);
System.out.println(sName);
}
public void piggy(String sName){
sName = sName + " wiggy";
start();
}
public void run(){
for(int i=0;i < 4; i++){
sName = sName + " " + i;
}
}
}

i compiled it and it run the result is vandeleur 0 1 2 3
wher is problem?

[Sridhar Srinivasan]

Hi!
Why shouldn't it be vandeluerwiggy 0 1 ..like that>can u please explain.Thanks

[Giray �nver]

Hi Sridhar,
Because the "piggy(String sName)" method uses just a copy of the string object refrenced by instance variable "static String sName". So any modifications you make on it (in piggy method), does not affect the actual object.
You may consider giving a return type to the piggy method and changing the code in your main method accordingly.
Rgds.

[Kadiyala Naga Sai]

Hi Sridhar,
Try this code , u may understand...If still can't , post reply.I will explain.This is very long to explain.But I will patiently do that.
public class Tux extends Thread{
static String sName = "vandeleur";
public static void main(String argv[]){
Tux t = new Tux();
t.piggy(sName);
System.out.println("Main method print>>"+sName);
}
public String piggy(String sName){
Tux.sName = Tux.sName + " wiggy";
System.out.println("Piggy method print>>"+Tux.sName);
start();
return Tux.sName;
}
public void run(){
System.out.println("Print1 from run method>>"+sName);
for(int i=0;i < 4; i++){
sName=sName + " " + i;
}
Tux.sName=sName;
System.out.println("Print2 from run method>>"+sName);
}
}

[Bijesh Krishnadas]

I just got confused. I thought changes made to an object passed to a method are reflected outside the method, i.e. the method actually receives a copy of the reference (not object).
Here sName is an object, then why doesn't the change get reflected.
Consider code below

public class Test 
{
  String s = "Test";
  
  public static void main(String argv[]){
        
        Test t = new Test();
        System.out.println(t.s);
        method1(t);
        System.out.println(t.s);
        
        String s1 = "Test";
        System.out.println(s1);
        method2(s1);
        System.out.println(s1);
        
        
   }
    public static void method1(Test t) 
    {
      t.s = "Welcome";
    }
    
    public static void method2(String s) 
    {
      s = "This is weird";
    }  
}
Output:
-------
Test 
Welcome
Test
Test

When the change to Test object in method1 is seen outside the method, why doesn't the same thing happen with the String object. I know I am missing something very simple here but cannot figure it out.
TIA,
Bijesh

[Shahfazal Mohammed]

I believe what's happening there is you are not updating String s1 with the return string of ur method2(). that way the string "This is wierd" is lost. now if you'd said

 s1 = method2(s1) 

s1 would have been changed to "this is wierd". i'm not able to look at ur code while im posting this answer right now, guess i got my variables wrong, but hope you got the idea

[Bijesh Krishnadas]

Okie I got it figured out...
From Java Tutorial on Sun:

In Java methods, arguments are passed by value. When invoked, the method receives the value of the variable passed in. When the argument is of primitive type, pass-by-value means that the method cannot change its value. When the argument is of reference type, pass-by-value means that the method cannot change the object reference, but can invoke the object's methods and modify the accessible variables within the object.

So method2() was trying to change the reference of the String object and this change will not be available outside of the method... Pretty simple, huh?

[fethi makhlouf]

Hello
i think u can get it more simplly....
public class Tux extends Thread{
static String sName = "vandeleur"; // class member Called sName
public static void main(String argv[]){
Tux t = new Tux();
t.piggy(sName);
System.out.println(sName);
}
public void piggy(String sName){ // argument is called also sName, it will put static member in shadow!
sName = sName + " wiggy"; // All changement done to sName here will affect that variable but not the class one!
start();
}
public void run(){
for(int i=0;i < 4; i++){
sName = sName + " " + i;
}
}
}

Hope it's clear for you!

[Bob Chandler]

public class Tux extends Thread{
static String sName = "vandeleur"; // class member Called sName
public static void main(String argv[]){
Tux t = new Tux();
t.piggy(sName); // if this line was changed to.. t.sName = t.piggy(sName);
// and piggy had the statement.... return sName;
// would the value of the class variable sName be altered?

System.out.println(sName);
}
public void piggy(String sName){ // argument is called also sName, it will put static member in shadow!
sName = sName + " wiggy"; // All changement done to sName here will affect that variable but not the class one!
start();
}
public void run(){
for(int i=0;i < 4; i++){
sName = sName + " " + i;
}
}
}