我是 Java的初学者,已经创建了一个简单的Java Android代码片段,在1.5秒后我在Runnable中将TextView从Hello World更改为Hola Mundo.它完美无缺,基本上WeakReference应该防止这种内存泄漏发生吗？我怀疑是否在设备方向发生时绝对没有内存泄漏.我很想检查这个,但无法在我模拟的Android中改变方向. @H_403_2@这是代码：

package com.example.helloworld;

import android.app.Activity;
import android.os.Bundle;
import android.os.Handler;
import android.widget.TextView;
import android.util.Log;
import java.lang.ref.WeakReference;

public class HelloWorldActivity extends Activity
{
    private Handler h = new Handler();
    private static TextView txtview;
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState)
    {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        txtview = (TextView) findViewById(R.id.mainview);

        h.postDelayed(new WeakRunnable(txtview),1500);
    }

    private static final class WeakRunnable implements Runnable {
        private final WeakReference<TextView> mtextview;

        protected WeakRunnable(TextView textview){
            mtextview = new WeakReference<TextView>(textview);
        }

            @Override
            public void run() {
                TextView textview = mtextview.get();
                if (textview != null) {
                    txtview.setText("Hola Mundo");
                    textview = null; // No idea if setting to null afterwards is a good idea
                }
                Log.d("com.example.helloworld","" + textview);
            }
    }           

}

@H_403_2@编辑

@H_403_2@内存泄漏是安全的,但是一些答案也与UI线程阻塞有关.实际上,此代码在主(UI)线程中运行Handler.为了生成一个新线程,我手动生成一个线程,如下所示：

package com.example.helloworld;

import android.app.Activity;
import android.os.Bundle;
import android.os.Handler;
import android.widget.TextView;
import android.util.Log;
import java.lang.ref.WeakReference;

public class HelloWorldActivity extends Activity
{

    private static TextView txtview;
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState)
    {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        txtview = (TextView) findViewById(R.id.mainview);

        Thread t = new Thread(new WeakRunnable(txtview));
        t.start();
    }

    private static final class WeakRunnable implements Runnable {
        private final WeakReference<TextView> mtextview;

        protected WeakRunnable(TextView textview){
            mtextview = new WeakReference<TextView>(textview);
        }

            @Override
            public void run() {
                TextView textview = mtextview.get();
                if (textview != null) {
                    /*
                    try {
                        Thread.sleep(1500);
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    */
                    txtview.setText("Hola Mundo");
                    textview = null;
                }
                Log.d("com.example.helloworld","" + Thread.currentThread().getName()); // Outputs "Thread-<num>" if not running on UI thread
            }
    }           

}

@H_403_2@现在的问题是我似乎无法以任何方式延迟产生的线程,否则它的工作原理.

@H_403_2@这个：

try {
    Thread.sleep(1500);
} catch (InterruptedException e) {
    e.printStackTrace();
}

@H_403_2@使应用程序自行退出,我不明白为什么.有些东西告诉我,我是以错误的方式推迟它.

@H_403_2@EDIT2

@H_403_2@感谢链接@EugenMatynov给我：update ui from another thread in android我理解为什么应用程序退出了.这一切都归结为您无法从主线程以外的线程调用UI方法的原因.从另一个线程更新UI是不好的做法.

private static Handler h = new Handler();

txtview.postDelayed(new WeakRunnable(txtview),1500);

txtview.postDelayed(new WeakRunnable(txtview),1500);