什么时候我的BackgroundWorker实例将被垃圾收集

时间:2008-10-28 14:39:20

标签: c# .net events garbage-collection backgroundworker

考虑这个代码块

public void ManageInstalledComponentsUpdate()
        {
            IUpdateView view = new UpdaterForm();
            BackgroundWorker worker = new BackgroundWorker();
            Update update = new Update();
            worker.WorkerReportsProgress = true;
            worker.WorkerSupportsCancellation = true;
            worker.DoWork += new DoWorkEventHandler(update.DoUpdate);
            worker.ProgressChanged += new ProgressChangedEventHandler(view.ProgressCallback);
            worker.RunWorkerCompleted += new RunWorkerCompletedEventHandler(view.CompletionCallback);            
            worker.RunWorkerAsync();
            Application.Run(view as UpdaterForm);     
        }

一切都很好,但我想了解为什么对象(工人,视图和更新)不会被垃圾收集

3 个答案:

答案 0 :(得分:7)

线程计为根对象;我不确切知道BackgroundWorker是如何操作的,但似乎主线程方法将在worker实例上访问状态;因此,工作线程本身将使BackgroundWorker实例保持活动状态,直到(至少)线程退出。

当然;集合还要求所有其他(实时)对象都取消引用工作对象;另请注意,堆栈变量的集合在调试/发布中可能有所不同,并且附加/不附带调试器。

[编辑] 还有人指出; worker(在你的代码中)上的事件处理程序将使“视图”和“更新”对象保持活动状态(通过委托),但不是相反。只要工作者的生命比“视图”和“更新”更短,您就不需要对取消订阅事件感到偏执。我编辑了代码以包含一个只由工作者引用的“SomeTarget”对象:你应该看到这种效果(即目标与工人一起死亡)。

当线程死亡时,工人被收集:这是证据;在工人报告退出后,你应该看到“工人最终确定”:

using System;
using System.ComponentModel;
using System.Threading;
using System.Windows.Forms;
class Demo : Form
{
    class ChattyWorker : BackgroundWorker
    {
        ~ChattyWorker()
        {
            Console.WriteLine("Worker finalized");
        }
    }
    class SomeTarget
    {
        ~SomeTarget()
        {
            Console.WriteLine("Target finalized");
        }
        public SomeTarget()
        {
            Console.WriteLine("Target created");
        }
        public void Foo(object sender, EventArgs args)
        {
            Console.WriteLine("Foo");
        }
    }
    static void Collect(object sender, EventArgs args)
    {
        Console.WriteLine("Collecting...");
        GC.Collect(GC.MaxGeneration, GCCollectionMode.Forced);
    }
    protected override void OnLoad(EventArgs e)
    {
        base.OnLoad(e);

        System.Windows.Forms.Timer timer = new System.Windows.Forms.Timer();
        timer.Interval = 100;
        timer.Tick += Collect;
        timer.Start();

        ChattyWorker worker = new ChattyWorker();
        worker.RunWorkerCompleted += new SomeTarget().Foo;
        worker.DoWork += delegate
        {
            Console.WriteLine("Worker starting");
            for (int i = 0; i < 10; i++)
            {
                Thread.Sleep(250);
                Console.WriteLine(i);
            }
            Console.WriteLine("Worker exiting");
        };
        worker.RunWorkerAsync();
    }
    [STAThread]
    static void Main()
    { // using a form to force a sync context
        Application.Run(new Demo());
    }
}

答案 1 :(得分:0)

事件处理程序是引用,因此在将事件处理程序附加到worker之前,它不会被视为“无法访问”。

在ComplitionCallback中注意取消挂钩事件处理程序。

答案 2 :(得分:-1)

这些局部变量对象保持活动状态,直到函数退出,即表单退出时。因此在调用Run之前将它们清空,或将它们移动到不同的上下文。

public void ManageInstalledComponentsUpdate() {
    UpdaterForm form = new UpdaterForm();
    FireAndForgetWorker( form );
    Application.Run( form );  //does not return until form exits
}

void FireAndForgetWorker( IUpdateView view ) {
    BackgroundWorker worker = new BackgroundWorker();
    Update update = new Update();
    worker.WorkerReportsProgress = true;
    worker.WorkerSupportsCancellation = true;
    worker.DoWork += new DoWorkEventHandler(update.DoUpdate);
    worker.ProgressChanged += new ProgressChangedEventHandler(view.ProgressCallback);
    worker.RunWorkerCompleted += new RunWorkerCompletedEventHandler(view.CompletionCallback);
    worker.RunWorkerAsync();
}

对vsick的说明:

尝试运行以下程序,你会惊讶于x永远存在。

使用System;

class FailsOnGarbageCollection  
{ ~FailsOnGarbageCollection() { throw new NotSupportedException(); } }

class Program{
    static void WaitForever() { while (true) { var o = new object(); } }

    static void Main(string[] args)
    {
        var x = new FailsOnGarbageCollection();
        //x = null; //use this line to release x and cause the above exception
        WaitForever();
    }
}