android.app.Application cannot be cast to android.app.Activity

Question

I'm trying to change a LinearLayout from another class, but when i run this code:

public class IRC extends PircBot {

ArrayList<String> channels;
ArrayList<Integer> userCount;
ArrayList<String> topics;

LinearLayout channelLayout;
Context context;

public IRC(Context ctx) {
    this.setName("xxxx");
    channels = new ArrayList<String>();
    userCount = new ArrayList<Integer>();
    topics = new ArrayList<String>();

    context = ctx;

    channelLayout = (LinearLayout) ((Activity) context).findViewById(R.id.channels);
}

i get a ClassCastException

context is the Main activity that extends Activity passed with a getApplicationContext();

LOGCAT

05-08 17:53:55.102    3736-3799/g.d.allinonechat E/AndroidRuntime﹕ FATAL EXCEPTION: Thread-5357
java.lang.ClassCastException: android.app.Application cannot be cast to android.app.Activity
        at g.d.xxx.IRC.<init>(IRC.java:34)
        at g.d.xxx.MainActivity$1.run(MainActivity.java:49)
        at java.lang.Thread.run(Thread.java:856)

You shouldn't need `(Activity)` since you are using the `Activity Context`. Along with the logcat, you might want to post how you are passing the `Context`. — codeMagic, May 08 '14 at 15:48

codeMagic · Accepted Answer · 2014-05-08T16:25:31.673

132

You are passing the Application Context not the Activity Context with

getApplicationContext();

Wherever you are passing it pass this or ActivityName.this instead.

Since you are trying to cast the Context you pass (Application not Activity as you thought) to an Activity with

(Activity)

you get this exception because you can't cast the Application to Activity since Application is not a sub-class of Activity.

edited May 08 '14 at 16:25

answered May 08 '14 at 15:58

codeMagic

44,549
13
77
93

2

This is an answer? is the same that log but with a lot of lines "ClassCastException: android.app.Application cannot be cast to android.app.Activity" A solution for me is to use "this" from an Activity – vLopez Oct 19 '17 at 13:54
@vLopez I don't understand your question. But yes `this` will work in an `Activity` which is why I suggested passing that since the OP was not in an `Activity` – codeMagic Oct 19 '17 at 14:28
1

I was using context = getApplicationContext() changed it to context = this. solved the problem. – DeathRs May 07 '18 at 08:41
1

Very good answer. This was a no nonsense answer. I was getting this error for a few day and when I read this answer, I corrected the association of context to the solution provided. – Val May 11 '18 at 18:32
2

Saved all of us. Thank you! – bbruno5 Sep 04 '19 at 01:51

score 19 · Answer 2 · answered Feb 10 '21 at 08:56

19

in case your project use dagger, and then this error show up you can add this at android manifest

   <application
        ...
        android: name = ".BaseApplication"
        ...> ...

answered Feb 10 '21 at 08:56

Tri yulianto

323
2
7

3

I'm not using dagger but it works for me. Thank you – Shoaib Kakal Aug 27 '21 at 13:41

score 4 · Answer 3 · answered Mar 22 '19 at 16:48

4

In my case, when I'm in an activity that extends from AppCompatActivity, it did not work(Activity) getApplicationContext (), I just putthis in its place.

answered Mar 22 '19 at 16:48

Gerrard

819
9
7

score 2 · Answer 4 · answered Mar 12 '20 at 05:06

You are getting this error because the parameter required is Activity and you are passing it the Application. So, either you cast application to the Activity like: (Activity)getApplicationContext(); Or you can just type the Activity like: MyActivity.this

score 2 · Answer 5 · answered Apr 21 '21 at 18:03

2

In my case I just put android:name=".CustomBaseClass" inside the activity tag instead of the application tag inside the manifest, hope it helps someone.

answered Apr 21 '21 at 18:03

Basel Abuhadrous

1,444
1
14
30

score 0 · Answer 6 · answered Sep 04 '20 at 13:36

You can also try this one.

override fun registerWith( registry: PluginRegistry) {
        GeneratedPluginRegistrant.registerWith(registry as FlutterEngine)       
    //registry.registrarFor("io.flutter.plugins.firebasemessaging.FirebaseMessagingPlugin")
    }

I think this one is far better solution than creating a new class.

score 0 · Answer 7 · answered Jun 20 '22 at 11:12

I had a similar problem when checking for internet connection in my viewmodel. The error in my logcat:

java.lang.ClassCastException: android.app.Application cannot be cast to com.example.newsapplication.util.AppApplication in viewmodel

I added this line in the manifest and my application worked just fine.

  android:name="com.example.newsapplication.util.AppApplication"

score 0 · Answer 8 · answered Sep 19 '22 at 12:27

0

In my case, I just put add name property inside the application tag in the Manifest file

<application
    android:name=".TodoApplication"
    ...

answered Sep 19 '22 at 12:27

Marawan Mamdouh

584
1
6
15

score 0 · Answer 9 · edited Oct 13 '22 at 07:43

0

In my case I removed the line android:name ="<some_name>" and it worked. Looks like it was looking for a class with the name that was provided earlier whereas my Project does not have the class.

edited Oct 13 '22 at 07:43

Procrastinator

2,526
30
27
36

answered Oct 08 '22 at 10:58

Srinivas Kotamarthi

1

score 0 · Answer 10 · answered Nov 10 '22 at 12:35

0

In my case, I deleted the android:name=".TodoApplication" from application and added it inside the activity tag.

<activity android:name=".BaseApplication"/>

Ensure you delete from the Application tag

answered Nov 10 '22 at 12:35

Benjamin Sinzore

59
6

android.app.Application cannot be cast to android.app.Activity

10 Answers10

Linked

Related