Reputation: 7792
So let's say I'm creating and starting a bunch of threads in a for loop, that is being executed in the run method of a launcher thread. Let's also say that I want to be able to interrupt the launcher thread and all threads that the thread has created, and I do this through a button.
So something like this -
try{
for(int i = 0; i < n;i++){
Worker currThread = new Worker(someArgs);
workerThreads.add(currThread);
currThread.start();
}
} catch (InterruptedException e){
e.printStackTrace();
}
BUTTON-
public void actionPerformed(ActionEvent arg0) {
List<Worker> threads = launchThread.getWorkerThreads();
for(int i = 0; i < threads.size();i++){
threads.get(i).interrupt();
}
launchThread.interrupt();
}
Now, let's say that I want to make it so that the interrupts cannot occur at the same time as thread creation. I think a way to do this would be to construct a dummy object and put both pieces of code inside a lock
synchronized(dummyObject){
//thread creation or interruption code here (shown above)
}
Will this way work? I ask because I'm not sure how to test to see if it will.
Upvotes: 0
Views: 43
Reputation: 5252
The concept of synchronization remains the same however complicated are the underlying operations to be executed.
As you specified, there are two types of mutually exclusive tasks (thread creation and interruption). So locking is pretty much the canonical tool for the job.
Upvotes: 0
Reputation: 360046
Start the threads separately from creating them.
for(int i = 0; i < n; i++) {
Worker currThread = new Worker(someArgs);
workerThreads.add(currThread);
}
// later
for (Worker w : workerThreads) {
w.start();
}
If that's still not enough, your dummyObject
synchronization should work just fine.
// You probably need to make this a (private final) field
Object lock = new Object();
// later
synchronized (lock) {
for(int i = 0; i < n; i++) {
Worker currThread = new Worker(someArgs);
workerThreads.add(currThread);
w.start();
}
}
// later still
public void actionPerformed(ActionEvent arg0) {
synchronized (lock) {
// interruption code here
}
}
Upvotes: 1