How to check if Receiver is registered in Android?

Question

There is no API function to check if a receiver is registered. The workaround is to put your code in a try catch block as done below.

try {

 //Register or UnRegister your broadcast receiver here

} catch(IllegalArgumentException e) {

    e.printStackTrace();
}

I am not sure the API provides directly an API, if you consider this thread:

I was wondering the same thing.
In my case I have a BroadcastReceiver implementation that calls Context#unregisterReceiver(BroadcastReceiver) passing itself as the argument after handling the Intent that it receives.
There is a small chance that the receiver's onReceive(Context, Intent) method is called more than once, since it is registered with multiple IntentFilters, creating the potential for an IllegalArgumentException being thrown from Context#unregisterReceiver(BroadcastReceiver).

In my case, I can store a private synchronized member to check before calling Context#unregisterReceiver(BroadcastReceiver), but it would be much cleaner if the API provided a check method.

simplest solution

in receiver:

public class MyReceiver extends BroadcastReceiver {   
    public boolean isRegistered;

    /**
    * register receiver
    * @param context - Context
    * @param filter - Intent Filter
    * @return see Context.registerReceiver(BroadcastReceiver,IntentFilter)
    */
    public Intent register(Context context, IntentFilter filter) {
        try {
              // ceph3us note:
              // here I propose to create 
              // a isRegistered(Contex) method 
              // as you can register receiver on different context  
              // so you need to match against the same one :) 
              // example  by storing a list of weak references  
              // see LoadedApk.class - receiver dispatcher 
              // its and ArrayMap there for example 
              return !isRegistered 
                     ? context.registerReceiver(this, filter) 
                     : null;
            } finally {
               isRegistered = true;
            }
    }

    /**
     * unregister received
     * @param context - context
     * @return true if was registered else false
     */
     public boolean unregister(Context context) {
         // additional work match on context before unregister
         // eg store weak ref in register then compare in unregister 
         // if match same instance
         return isRegistered 
                    && unregisterInternal(context);
     }

     private boolean unregisterInternal(Context context) {
         context.unregisterReceiver(this); 
         isRegistered = false;
         return true;
     }

    // rest implementation  here 
    // or make this an abstract class as template :)
    ...
}

in code:

MyReceiver myReceiver = new MyReceiver();
myReceiver.register(Context, IntentFilter); // register 
myReceiver.unregister(Context); // unregister 

ad 1

-- in reply to:

This really isn't that elegant because you have to remember to set the isRegistered flag after you register. – Stealth Rabbi

-- "more ellegant way" added method in receiver to register and set flag

this won't work If you restart the device or if your app got killed by OS. – amin 6 hours ago

@amin - see lifetime of in code (not system registered by manifest entry) registered receiver :)

I am using this solution

public class ReceiverManager {
    private WeakReference<Context> cReference;
    private static List<BroadcastReceiver> receivers = new ArrayList<BroadcastReceiver>();
    private static ReceiverManager ref;

    private ReceiverManager(Context context) {
        cReference = new WeakReference<>(context);
    }

    public static synchronized ReceiverManager init(Context context) {
        if (ref == null) ref = new ReceiverManager(context);
        return ref;
    }

    public Intent registerReceiver(BroadcastReceiver receiver, IntentFilter intentFilter) {
        receivers.add(receiver);
        Intent intent = cReference.get().registerReceiver(receiver, intentFilter);
        Log.i(getClass().getSimpleName(), "registered receiver: " + receiver + "  with filter: " + intentFilter);
        Log.i(getClass().getSimpleName(), "receiver Intent: " + intent);
        return intent;
    }

    public boolean isReceiverRegistered(BroadcastReceiver receiver) {
        boolean registered = receivers.contains(receiver);
        Log.i(getClass().getSimpleName(), "is receiver " + receiver + " registered? " + registered);
        return registered;
    }

    public void unregisterReceiver(BroadcastReceiver receiver) {
        if (isReceiverRegistered(receiver)) {
            receivers.remove(receiver);
            cReference.get().unregisterReceiver(receiver);
            Log.i(getClass().getSimpleName(), "unregistered receiver: " + receiver);
        }
    }
}

You have several options

1. You can put a flag into your class or activity. Put a boolean variable into your class and look at this flag to know if you have the Receiver registered.

2. Create a class that extends the Receiver and there you can use:

   1. Singleton pattern for only have one instance of this class in your project.

   2. Implement the methods for know if the Receiver is register.

You have to use try/catch:

try {
    if (receiver!=null) {
        Activity.this.unregisterReceiver(receiver);
    }
} catch (IllegalArgumentException e) {
    e.printStackTrace();
}