synchronized run()

[Meena R. Krishnan]

 
public class Test extends Thread
{
 static int x, y;
 public synchronized void run(){  for(; :wink: { x++; y++; System.out.println(x+" "+y);} }
 public static void main(String[] args)
 {
   new Test().start();
   new Test().start();
 }
}

What will the above code print?

I chose:
"It will keep on printin same values for x and y incrementing by 1 on each line."


but the correct answer given is:
"You cannot say anything about the values."

When I tried the program, I see output matching my answer. Any thoughts?

source :Enthuware

[Keith Lynn]

When answering questions about what multiple threads output, you cannot always rely on what happens when you run the program on your machine a few times.

Sometimes you might get the output you saw, but it isn't guaranteed.

The main point to notice in that program is that the synchronized keyword modifies an instance method in the class definition.

That means that when that method is executed, the object that is locked is the current instance of the class.

Since there are two different instances of the class, when one instance executes run(), it does not block the other instance from executing its run() method.

[Meena R. Krishnan]

Since there are two different instances of the class, when one instance executes run(), it does not block the other instance from executing its run() method.

Thanks Keith.

[Sasha Ruehmkorf]

For me, it's always a big help, to modify such a code by suggesting to yield the running thread to the JVM. So:

public class Test extends Thread{
static int x, y;
public synchronized void run(){  
for(int i = 0;i < 100;i++){ 
x++; 
yield();
y++; 
System.out.println(x+" "+y);
} 
}

public static void main(String[] args){
new Test().start();
new Test().start();
}
}

shows, what may happen if the running thread loses his running-state at a bad moment.