static variable remains unchanged

[S Ali]

  public class Test extends Thread{

    static String sName = "good";
    public static void main(String argv[]){
    	Test t = new Test();
    t.nameTest(sName);
    System.out.println(sName);
    
    }
    public void nameTest(String sName){
            sName = sName + " idea ";
             start();
    }
    public void run(){
    
    for(int i=0;i  <  4; i++){
            sName = sName + " " + i;
            
            
    }
    }

}  

Hi everyone , this question is from www.javacertifications.com . My question is shouldn't sName change in the run method to become good0123 ??

[Henry Wong]

hint: did you print the result before or after the other thread changed the static variable?

[Ankur kothari]

     public class Test15 extends Thread{

    static String sName = "good";
    public static void main(String argv[]){
    	Test15 t = new Test15();
    t.nameTest(sName);
    System.out.println(sName);
    
    }
    public void nameTest(String sName){
            sName = sName + " idea ";
             start();
			 System.out.println("Ankur");
    }
    public void run(){
    
    for(int i=0;i  <  4; i++){
            sName = sName + " " + i;
            
            
    }
    }

}  

in this case the output is
Ankur
good 0 1 2 3
why?

[Henry Wong]

same hint

[Ankur kothari]

still no idea.....the name is printed only when the nameTest method completes.....in the first case a new thread runs simultaneously with the main thread.....same happens in the second case.....how does the output change by just adding one more line below start();

[S Ali]

Ankur kothari wrote:still no idea.....the name is printed only when the nameTest method completes.....in the first case a new thread runs simultaneously with the main thread.....same happens in the second case.....how does the output change by just adding one more line below start();

I'll take a wild guess . Cause it has to wait for start to begin running in order to print Ankur (before method returns). But I don't get it I mean no guarantees, right?

[Henry Wong]

S Ali wrote:I'll take a wild guess . Cause it has to wait for start to begin running in order to print Ankur (before method returns). But I don't get it I mean no guarantees, right?

Basically, you have a race condition. If the main thread prints the static variable before the newly created thread can get to it, you will have the original value. If the newly created thread gets to it first, you will have the new value.

In the original example, the main thread gets to it first -- although this too, is not guarranteed to always be so.

In the second example, the main thread now has to print out "Ankur". This has two affects. It has to print out "Ankur". And the main thread has now been slowed down enough, by printing out "Ankur", that the newly created thread can get to the static variable first.... but as before, this is not guarranteed to always be so.

Henry

[Ankur kothari]

so if in the exam there is a question to determine the output...what should the answer be in both the cases? cannot be predicted? .......

[Waclaw Borowiec]

I refresh this thread as it lacks final answer. Original options are:

A)good
B)good idea
C)good idea good idea
D)good 0 good 0 1

but the correct one is "The output is unpredictable". Right?

[Neha Daga]

in this question there is no option saying "result is not predictable", and "good" is the only answer that may be a possible output so in this case I think 1 is correct answer.

[Leonardo Carreira]

Hi..

Actually, the changes aren't reflected to the variable, the rule is same with local variable..
in this code below, i declare and define value of local variable named a.., after execute nameTest method, any changes happened to a, won't be reflected to a itself..

    
    public static void main(String argv[]){
    	String a="AAAAAA";
        Test t = new Test();
        t.nameTest(a);
        System.out.println(a);
    
    }
    public void nameTest(String sName){
            sName = sName + " idea ";
    }

Result : AAAAAA

[Leonardo Carreira]

But, if we want to change the static variable value of sName..
we can do it as bellow..

    
    static String sName = "good";
    public static void main(String argv[]){
        TestT t = new TestT();
        t.nameTest(sName);
        System.out.println(sName);
    
    }
    public void nameTest(String sName){
            this.sName = this.sName + " idea ";
    }

Result : good idea

[Leonardo Carreira]

Oopss... Sorry i've a doubt here..

    
public class Test15 extends Thread{

    static String sName = "good";
    public static void main(String argv[]){
    	Test15 t = new Test15();
    t.nameTest(sName);
    System.out.println(sName);
    
    }
    public void nameTest(String sName){
            sName = sName + " idea ";
            start();                                //I sign these two lines as "Line Breakpoint", while
	    System.out.println("Ankur");  //I use Netbeans to run this code in Debug mode 
    }
    public void run(){
          for(int i=0;i  <  4; i++){
            sName = sName + " " + i;
          }
    }
}  

Why the value is different if i use run mode and debug mode..

if i run that code in Netbeans, the result is :

Ankur
good

but, if i debug that code in Netbeans with the line breakpoints as mentioned above, the result is :

Ankur
good 0 1 2 3

of course if i run that code through command prompt, the result is :

Ankur
good

Could you describe why?..
Thanks..

[Waclaw Borowiec]

Neha Daga wrote:in this question there is no option saying "result is not predictable", and "good" is the only answer that may be a possible output so in this case I think 1 is correct answer.

What I meant is the question has no correct answer.

[Leonardo Carreira]

    
public class Test15 extends Thread{

    static String sName = "good";
    public static void main(String argv[]){
    	Test15 t = new Test15();
    t.nameTest(sName);
    System.out.println(sName);
    
    }
    public void nameTest(String sName){
            sName = sName + " idea ";
            start();                                //I sign these two lines as "Line Breakpoint", while
	    System.out.println("Ankur");  //I use Netbeans to run this code in Debug mode 
    }
    public void run(){
          for(int i=0;i  <  4; i++){
            sName = sName + " " + i;
          }
    }
}  

Why the value is different if i use run mode and debug mode..

if i run that code in Netbeans, the result is :

Ankur
good

but, if i debug that code in Netbeans with the line breakpoints as mentioned above, the result is :

Ankur
good 0 1 2 3

of course if i run that code through command prompt, the result is :

Ankur
good

Could you describe why the result different while i use debug mode and while i use Run mode in Netbeans?..

Please tell me why it occurs..
Thanks..