How to create a Looper thread, then send it a message immediately?

Solution 1:

Eventual solution (minus error checking), thanks to CommonsWare:

class Worker extends HandlerThread {

    // ...

    public synchronized void waitUntilReady() {
        d_handler = new Handler(getLooper(), d_messageHandler);
    }

}

And from the main thread:

Worker worker = new Worker();
worker.start();
worker.waitUntilReady(); // <- ADDED
worker.handler.sendMessage(...);

This works thanks to the semantics of HandlerThread.getLooper() which blocks until the looper has been initialized.

Incidentally, this is similar to my solution #1 above, since the HandlerThread is implemented roughly as follows (gotta love open source):

public void run() {
    Looper.prepare();
    synchronized (this) {
        mLooper = Looper.myLooper();
        notifyAll();
    }
    Looper.loop();
}

public Looper getLooper() {
    synchronized (this) {
        while (mLooper == null) {
            try {
                wait();
            } catch (InterruptedException e) {
            }
        }
    }
    return mLooper;
}

The key difference is that it doesn't check whether the worker thread is running, but that it has actually created a looper; and the way to do so is to store the looper in a private field. Nice!

Solution 2:

take a look at the source code of HandlerThread

@Override
     public void run() {
         mTid = Process.myTid();
         Looper.prepare();
         synchronized (this) {
             mLooper = Looper.myLooper();
             notifyAll();
         }
         Process.setThreadPriority(mPriority);
         onLooperPrepared();
         Looper.loop();
         mTid = -1;
     }

Basically, if you are extending Thread in worker and implementing your own Looper, then your main thread class should extend worker and set your handler there.