android - 为什么可以使用在后台线程中创建的处理程序来更新UI？

Question

我有一个 worker 类，其中创建了一个处理程序。当我尝试使用在后台线程中创建的处理程序更新我的UI时，它会更新UI，并且这里没有崩溃。理想情况下，它应在后台线程中运行时崩溃。请在下面找到代码:

class WorkerThread(val mainThreadHandler: Handler, handlerThreadName: String) :
  HandlerThread(handlerThreadName) {

  lateinit var workerThreadHandler: Handler

override fun onLooperPrepared() {
    super.onLooperPrepared()
    Log.d("LLoyd", "onLooperPrepared")


}


override fun run() {
    super.run()
    Log.d("LLoyd", "run method called")

}

fun startTask(){
    workerThreadHandler = object : Handler(this.looper) {
        override fun handleMessage(msg: Message) {
            super.handleMessage(msg)
            Log.d(
                "Lloyd",
                "Message from Main Thread received " + msg.what + "  Thread name  " + 
currentThread().name
            )
            val message = Message()
            message.what = msg.what
            mainThreadHandler.sendMessage(message)
        }
    }

 }
}

只需单击以下按钮，即可从 Activity 中调用此startTask():

val workerThread = WorkerThread(mainThreadHandler, "Lloyd")
    workerThread.start()
    btn_task1.setOnClickListener {
        workerThread.startTask()
        val message = Message()
        message.what = FROM_MAINTHREAD
        workerThread.workerThreadHandler.sendMessage(message)
        val message1 = Message()
        message.what = 55
        Log.d("Lloyd ", "Outside Thread name " + Thread.currentThread().name)

        workerThread.workerThreadHandler.post {
            Log.d("Lloyd ", "Runnable Thread name " + currentThread().name)
            btn_task1.text = "Task 2"
            tv.text= "YAAAAAAAAAAAAAAAAAAAAAAy"
        }
    }

任何形式的帮助，我们感激不尽!

Answer 1

When you connect a Handler to your UI thread, the code that handles messages runs on the UI thread.

取自 the docs

我相信，因为您是通过按钮单击(即UI线程)调用 workerThread.startTask()的，所以处理程序也只是将结果返回给UI线程，这就是UI也将更新的原因