Why does AsyncTask run in the main Thread of an application?

20,115

Solution 1

An AsyncTask has several parts that you can override: a doInBackground method that does, in fact, run on a separate thread, and three methods—onPreExecute, onProgressUpdate, and onPostExecute—that run on the UI thread. (The default implementation of these methods do nothing and onProgressUpdate only runs if you call publishProgress, usually from within doInBackground.) The purpose of onPostExecute is to publish results (such as updating the view hierarchy, or setting text in a text view) that must be done on the UI thread. It also can post progress updates. In order for this to all work properly, the AsyncTask must be created, and the execute method called, on the UI thread.

You must not call UI actions from within doInBackground -- doing so will crash your application.

Solution 2

In fact, I have checked the doc of android, onPreExecute(), onProgressUpdate(Progress...) and onPostExecute(Result) are all "invoked on the UI thread".

Only the doInBackground(Params...) method is "invoked on the background thread".

Share:
20,115
Mathieu
Author by

Mathieu

Mobile app developer, both iOS and Android :-)

Updated on October 22, 2020

Comments

  • Mathieu
    Mathieu over 3 years

    in my application, I have a class for UI stuff, the name of which is "SettingActivity".

    Then for doing some jobs in background, I bind this UI class(SettingActivity) to a Service. There are two predefined methods in that Service(defined in .aidl file), one is startTask(), the other one is stopTask().

    Within startTask(), I made a call to an AsyncTask. But when I checked the name of the Looper of this AsyncTask. It is "main". In my opinion, an AsyncTask should start an another Thread other than the main Thread.

    So does somebody know why this happens?

    The codes are as follows:

    @Override
        protected void onPreExecute() {
            super.onPreExecute();
            Log.d(TAG, "onPreExecute "+Looper.myLooper().getThread().getName());
        }
    

    Then I will get main as the output.