Which run() method will run?

[Liang Anmian]

I got this question from Dan Chisholm's mock exam:

class A extends Thread {
  public A(Runnable r) {super(r);}
  public void run() {System.out.print("A");}
}
class B implements Runnable {
  public void run() {System.out.print("B");}
}
class C {
  public static void main(String[] args) {
    new A(new B()).start();
}}

According to Dan, this code snippet causes the run() method in class A to run, not class B. But why? I've read his explanation, but I still don't understand. If you pass a Runnable object to a Thread constructor, then shouldn't the Thread object invoke the run() method of the Runnable object?

PS: I tested this code snippet, and true enough, Dan is correct.

[rogel garcia]

Follow these steps:
Call start() on A
start() call run() on A
-- run() has been overriden by class A
run() prints A

Now other steps
supose Thread t = new Thread(new B());
Call start() on t
start() call run() on t
run() calls b.run() method
b.run() prints B

The implementation of the run method of the Thread class is the following:
public void run() {
if (target != null) {
target.run();
}
}
where target is the runnable object you passed in the construtor
if you override you don't get target run

I hope I helped

[Richard Vagner]

 
new A(new B()).start(); 

can be rewritten as

B b = new B();
A a = new A(b);
a.start(); 

[ January 24, 2005: Message edited by: Richard Vagner ]

[Richard Levy]

Hi

Because the run() method in A has no call to super.run(), a check is never made to see if it has a Runnable object to call.

If you add a call to super.run() in class A's run() method, you'll see B printed first. This is because the default implementation of run() is called first, which checks if it has a reference to a Runnable, and calls it if it does.

Cheers
Richard

p.s, I would have said that it prints B too!
[ January 25, 2005: Message edited by: Richard Levy ]

[Liang Anmian]

Originally posted by rogel garcia:
Follow these steps:
Call start() on A
start() call run() on A
-- run() has been overriden by class A
run() prints A

Now other steps
supose Thread t = new Thread(new B());
Call start() on t
start() call run() on t
run() calls b.run() method
b.run() prints B

The implementation of the run method of the Thread class is the following:
public void run() {
if (target != null) {
target.run();
}
}
where target is the runnable object you passed in the construtor
if you override you don't get target run

I hope I helped

Cool. Thank you very much. Now I understand. Due to the fact that the run() method in the Thread class has been overriden by Class A with an empty method, so the Runnable run() method has no chance to run. I hope I'm right.