Handler如何影响onReceiveResult(ResultReceiver)的调用方式？

Question

看,我有以下代码:

我的行动:

final Intent intent = new Intent(getApplicationContext(), MyService.class)
.putExtra(UploadService.EXTRA_RESULT_RECEIVER, new ResultReceiver(null) {
            @Override
            protected void onReceiveResult(int resultCode, Bundle resultData) {
                super.onReceiveResult(resultCode, resultData);
                String result = resultData.getString(MyService.EXTRA_RESULT_SUCCESS);
                ...
                imageView.setBackgroundDrawable(bitmap);// here my code fails
            }
        })

为MyService:

    Bundle b = new Bundle();
    b.putString(EXTRA_RESULT_SUCCESS, response.toString());
    resultReceiver.send(0, b);

我的应用程序在"imageView.setBackgroundDrawable(bitmap)"行上失败,但有以下异常:

11-13 16:25:38.986: ERROR/AndroidRuntime(3586): FATAL EXCEPTION: IntentService[MyService]
    android.view.ViewRoot$CalledFromWrongThreadException: Only the original thread that created a view hierarchy can touch its views.

但是当我像这样定义接收器(带处理程序)时,这不会发生:

new ResultReceiver(new Handler()){.../*here goes the same code as in the first example. nothing has been changed*/}

所以.当我传递默认处理程序时它不会失败.我问为什么？在这两种方式中都调用了我的代码,但是当没有Handler指定它时它会失败.Handler有什么影响？

Answer 1

Handler绑定到Android框架并确保在Handler的回调中运行的任何代码都在父Activity的主Looper线程上执行,该线程是进行所有Activity生命周期回调和UI调用的地方.如果你真的想了解它是如何工作的,你可以浏览Github上的源代码,但是在Handler中运行代码几乎可以保证把东西放在正确的位置.