Synchronization question from Exam Lab

[Maren Fisher]

Hi,

This is question 37 from the diagnostic quiz in Exam Lab

public class TWorks extends Thread{
     public void run(){
          synchronized(new String("a")){
               System.out.print("-A-");
               System.out.print("-B-");
               System.out.print("-C-");
          }
     }
     public static void main(String args[]){
          TWorks tw=new TWorks();
          Thread rb1=new Thread(tw);
          Thread rb2=new Thread(tw);
          rb1.start();
          rb2.start();
     }
}

The possible answers are :
A) compilation fails
B) Prints A-A-B-B-C-C
C) Prints A-B-C-A-B-C
D) Result is unpredictable
E) Exception at runtime.


The answer is D. The explanation give is "a new string object is created every time because of the keyword new. Therefore the lock of an instance of the given String will be owned by the first thread; and another instance of the given String will be owned by the second thread.

However, in the K&B book page 433-434 it says "If a match is found, the reference to the new literal is directed to the existing String, and no new String literal object is created."

In the above example, wouldn't the string "a" be in the string pool, meaning it's the same object, and so synchronization would work and the output would be A-B-C-A-B-C

Does the new String("a") causes a new String in the string pool? So if you did the following:
new String("a");
new String("a");
new String("a);

would you end up with 3 String objects in the String pool all representing the letter a?

[Imad Aydarooos]

Dear Maren Fisher , it works like this:
1- if you assign a String literal to a String variable the JVM will check the String Pool for a similar String object, if it finds one it will be assigned to your variable, else the String object will be created using your String literal and will be added to the String Pool.
e.g: String s = "if not in the pool, create one. If in the pool assign to s";

2- When using the new keyword you are forcing the JVM to create your String object in the heap, anyway the JVM will chick the String Pool after creating your String object in the heap: if there is a similar String object nothing will happen, else the new created String object will be added to the String Pool.
Regards

[Maren Fisher]

Ah, ok. Thank you.

[Imad Aydarooos]

Imad Aydarooos wrote:Dear Maren Fisher , it works like this:
1- if you assign a String literal to a String variable the JVM will check the String Pool for a similar String object, if it finds one it will be assigned to your variable, else the String object will be created using your String literal and will be added to the String Pool.
e.g: String s = "if not in the pool, create one. If in the pool assign to s";

2- When using the new keyword you are forcing the JVM to create your String object in the heap, anyway the JVM will chick the String Pool after creating your String object in the heap: if there is a similar String object nothing will happen, else the new created String object will be added to the String Pool.
Regards

Sorry Maren Fisher and other readers, I'm wrong in statment 2, and if you used the new keyword nothing will be added to the String Pool, objects will be created in the heap.
please read this :
Strings, Literally