a question on thread flow

[Vad Fogel]

Hi all, can you tell me how to approach questions like this?

class CountStrings implements Runnable{
    public static void main(String[] s){
        String str1, str2;
        Thread t=new Thread(new CountStrings());
        t.start();
        str1="abc";
        str2="cde";
    }
    
    public void run() {
        String y,z;
        //K&B's code fragment goes here
        String x = new String("abc");
        y = "cde";
        z = x + y;  //line 1  
    }   
}
/**
 *  How many String objects are created by the time line 1 completes?
 *  A. 2
 *  B. 3
 *  C. 4
 *  D. Impossible to tell.
 *  E. None of the above.
 */

[Barkat Mardhani]

Hi Vad,
I would say 4 objects per notes in the code below:

  
class CountStrings implements Runnable{    
public static void main(String[] s){        
String str1, str2;        
Thread t=new Thread(new CountStrings());        
t.start();        
str1="abc"; // first object in the constant pool        
str2="cde"; // second object in the constant pool   
}        
public void run() {        
String y,z;        
//K&B's code fragment goes here        
String x = new String("abc"); // third object on the heap        
y = "cde"; //      // already in the pool
z = x + y;  //line 1  // forth object "abccde"    
}   
}

[ October 07, 2003: Message edited by: Barkat Mardhani ]

[Vad Fogel]

My doubt is what if the scheduler decides to finish off the main() thread first and only then execute the new thread? Then, 2 String constants would be put in the constant pool to be reused in the run() method (then, "abc" and "cde" objects wouldn't be created?)
Yeah, after thinking awhile on my own stuff, I see you're right
4 objects will be created anyway. The difference would appear should I have different String constants in the main(). Or would it?
[ October 07, 2003: Message edited by: Vad Fogel ]