C#在线程之间传递数据，很容易

Question

我目前正在从事一个项目，我将帧作为字节数组接收并在 GUI (WPF) 上显示它们。现在我对性能不满意，因为帧显示延迟。 所以我想到了做多线程，让socket和GUI彼此独立工作。但是，如果我尝试在单独的线程中运行 socketRoutine，我会收到一个错误，即一个线程无法访问另一个线程的资源。

我不确定是指哪个资源。我猜它要么是传递给 GUI 的字节数组，要么是必须访问的 GUI 组件。

这是我现在的代码。

namespace Subscriber_WPF{
/// <summary>
/// Interaction logic for MainWindow.xaml
/// </summary>
/// 

public partial class MainWindow : Window{

    /*Kinect Datentypen*/
    private WriteableBitmap colorBitmap = null;
    static readonly String publisherID = "TYPE1";

    SocketHandler socketHandler;
    Thread socketThread;

    public MainWindow(){     
        ConsoleManager.Show();
        colorBitmap = new WriteableBitmap(1920, 1080, 96.0, 96.0, PixelFormats.Bgra32, null);
        this.DataContext = this;

        //initializeKinectComponents();
       socketHandler = new SocketHandler(5555, publisherID, this);
       socketThread = new Thread(()=>socketHandler.runSocketRoutine());

        Console.WriteLine("GUI-Components initialized. Press Enter to start receiving Frames.");

        this.KeyDown += MainWindow_KeyDown;

    }

    private void MainWindow_KeyDown(object sender, KeyEventArgs e){
        if (e.Key.Equals(Key.Return)) {
            socketThread.Start();
        }
    }

    public void refreshGUI(byte[]content) {
       Action EmptyDelegate = delegate () { };
       BitmapSource source = BitmapSource.Create(1920, 1080, 72, 72, PixelFormats.Bgra32, BitmapPalettes.Gray256, content, 1920 * 4);
       videoView.Source = source;

        /*interrupt the socket-Loop to update GUI=> that is the current method without multithreading*/
       //this.Dispatcher.Invoke(DispatcherPriority.Render, EmptyDelegate);

    }


}

套接字处理程序

namespace Subscriber_WPF{

class SocketHandler{

    private int port;
    private string publisherID;
    private MainWindow window;
    private static ZContext context;
    private static ZSocket subscriber;

    public SocketHandler(int port, string publisherID, MainWindow window) {
        this.port = port;
        this.publisherID = publisherID;
        this.window = window;
        this.initializeZMQSocket(this.port, this.publisherID);
    }

    private void initializeZMQSocket(int port, String publishID){
        context = new ZContext();
        subscriber = new ZSocket(context, ZSocketType.SUB);
        /*initialize sockets*/
        subscriber.Connect("tcp://127.0.0.1:" + port);
        subscriber.Subscribe(publishID);
        Console.WriteLine("subscriber is ready!");
    }


    public void runSocketRoutine(){
        Console.WriteLine("Waiting for Messages.");

        while (true)
        {
            byte[] content = new byte[8294400];

            using (ZMessage message = subscriber.ReceiveMessage())
            {
                Console.WriteLine("Message received!");
                string pubID = message[0].ReadString();
                /**/
                if (pubID.Equals(publisherID))
                {

                    content = message[1].Read();
                    Console.WriteLine("size of content: " + message[1].Length);
                    window.refreshGUI(content);
                }

            }
        }

    }


}

如果有人知道如何停止延迟显示或如何轻松处理该线程问题，我将不胜感激！

许多问候！

Answer 1

我过去曾使用出色的RX framework 做过类似的事情。您的运行套接字例程变为 IObservable，其中 T 是您返回的类型，在您的情况下为 byte[]。然后，您可以在 UI 线程上订阅此流。使用 RX 框架，您还可以做一些很酷的事情，例如速率限制，因此如果您的例程吐出大量“记录”，您可以将其限制为每个时间跨度 1 个。