我是线程世界的新手，并试图让我的应用程序与线程一起工作。这是我得到的:

public static void ThreadProc()
{
    Thread.Sleep(500);
    MemoryMappedFile mmf = MemoryMappedFile.OpenExisting("SuperMMFofDoom", MemoryMappedFileRights.ReadWrite);
    MemoryMappedViewAccessor accessor = mmf.CreateViewAccessor(0, sizeof(double)*3 + sizeof(int) *2);
    Image imgS = new Image();
    ImageTrigger myMessage;
    Mutex imgMutex = new Mutex(false, "imgMutex");

    while (threadRunning)
    {
        imgMutex.WaitOne();

        accessor.Read(0, out myMessage);

        // [...]

        Dispatcher.Invoke(DispatcherPriority.Normal,
            new Action(delegate()
                {
                    // [...]
                }),
            new object[] { imgS, myMessage.performance }
           );

        imgMutex.ReleaseMutex();
    }
}

using System;
using System.Collections.Generic;
using System.Linq;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Media;
using System.Windows.Media.Imaging;
using System.Windows.Threading;
using Microsoft.Win32;
using System.Runtime.InteropServices;
using System.ComponentModel;
using System.IO;
using System.Reflection;
using System.Diagnostics;
using System.Threading;
using System.IO.MemoryMappedFiles;
using Common;

最佳答案

你不应该创建一个新的 Action只是将您的委托(delegate)作为一个:

异步:

Dispatcher.BeginInvoke(DispatcherPriority.Background, (Action)delegate()
{

});

Dispatcher.Invoke(DispatcherPriority.Background, (Action)delegate()
{

});

Application.Current.Dispatcher.Invoke(DispatcherPriority.Background, (Action)delegate()
{

});

https://stackoverflow.com/questions/13716708/