If 2 broadcasts, A and B, and sent in that order, does Android guarantee that all interested receivers will receive them in the same order?
You could have the behavior check for broadcast A and if isn't there, don't use B until A happens? So they might be received backwards, but still acted on properly?
You might want to use sendOrderedBroadcast (if you have control of this) to guarantee order.
Some more info about this question may be found here:
Broadcast receiver execution.
https://groups.google.com/forum/#!topic/android-developers/ClIGNuGJUts
In the above link Dianne Hackborn writes ...
"A particular receiver can only process one broadcast at a time. As each broadcast happens, it is processed to determine the targets it should go to, and dispatched into the message queue for each target. When a later broadcast is sent, it will generally not be moved ahead of the queue.
For the screen on/off broadcasts, you can rely on the most recent broadcast you get being the current state of the screen. once everything has settled down. You can't rely on ordering with any other broadcasts, or other broadcast receivers that may also be receiving the broadcasts."
This article suggests possibly using sendBroadcastSync() for synchronous delivery, but it is not guaranteed.
Does LocalBroadcastManager deliver events in the order in which the events wer sent?
I believe you are asking if there is any way to guarantee that each interested receiver will receive its respective Broadcast A before getting Broadcast B. The best answer I can give is that it is highly likely, because all broadcasts are passed to the ActivityManager
which should handle them in turn. But I don't believe there is anything in the framework that "guarantees" this behavior (I can't find a queue on ActivityManager where they are all posted or anything like that). Also, BroadcastReceiver
will only handle one Intent at a time, which helps.
If instead you meant "can I control the order of receivers in which each broadcast is sent", then your answer lies with sendOrderedBroadcast()
as other have eluded to.