Reputation: 27375
The code I've witten doesn't work as I expected.
static Integer sync = 1;
static void m() throws Exception {
synchronized (sync) {
System.err.println("First");
sync.notify();
sync.wait(1000L);
System.err.println("Second");
System.err.println("Third");
}
}
public static void main(String[] args) throws Exception {
Runnable r = new Runnable() {
@Override
public void run() {
try {
m();
} catch (Exception ex) {
Logger.getLogger(IO.class.getName()).log(Level.SEVERE, null, ex);
}
}
};
Runnable t = new Runnable() {
@Override
public void run() {
try {
m();
} catch (Exception ex) {
Logger.getLogger(IO.class.getName()).log(Level.SEVERE, null, ex);
}
}
};
Thread th1 = new Thread(r);
Thread th2 = new Thread(t);
th1.run();
th2.run();
}
We have two threads which execute m()
's syncjronized statement. When the first thread executes one and come across the wait()
it'll be added to the wait set. After this, the second thread is starting to execute the synchronized statement, and perform notify(). Since the output must be
First
First
....
But actually it is
First
Second
Third
First
Second
Third
Why?
Upvotes: 0
Views: 117
Reputation: 6753
Use .start()
instead of run()
to add runables to the queue instead of running them immediately
Documentation says that wait
with timeout waits for any notify
on this object or the timeout. In your case when runnables are being executed one by one it goes:
PS. calling run()
and not setting timeout will cause deadlock on t's wait, cause it already has the object but will wait never be notified about it.
Hope this helps.
Upvotes: 1
Reputation: 173
Firstly, To actually create new threads please use
th1.start() th2.start()
inplace of run() , which is just a regular method call on the thread object.
Secondly, it is possible that the second thread 'th2' did not start running by the time 1000 ms was fninshed , so the first thread finished wait(1000) and executed the remainging lines of code.
if you want the output like so :
first
first
second
third
second
third
then remove the time interval for wait() which will make the threads wait until notified. as in :
static void m() throws Exception {
synchronized (sync) {
System.err.println("First");
sync.notify();
sync.wait();
System.err.println("Second");
System.err.println("Third");
}
}
Upvotes: 1
Reputation: 27115
First of all, your program is not creating any threads. You must call th1.start() and th2.start() to create threads.
t.start() is the method that the library provides for your code to call when you want to start a thread. run() is the method that you provide for the library to call in the new thread. Your run() method defines what the thread will do. IMO, run() was a really misleading name.
Second, notify() and wait() don't do what it looks like you think they will do. In particular, sync.notify() will not do anything at all if there are no other threads currently in sync.wait().
The correct way to use notify() and wait() is, one thread does this:
synchronized(lock) {
while (! someCondition()) {
lock.wait()
}
doSomethingThatRequiresSomeConditionToBeTrue();
}
The other thread does this
synchronized(lock) {
doSomethingThatMakesSomeConditionTrue();
lock.notify();
}
When you use this pattern, no thread should ever change the result of someCondition() except from inside a synchronized(lock) block.
Upvotes: 4