c# - WriteableBitmap 的异步操作

标签 c# multithreading asynchronous writeablebitmap

我正在用 WPF (C#) 编写一个应用程序,它对位图集合执行长时间操作。为了让我的应用程序保持响应,我决定使用另一个线程来执行位图操作并在主 UI 线程的进度条上报告进度。我原以为 BackgroundWorker 会为我做任何事情,但看起来并没有那么容易。

我有以下代码:

public class ImageProcessor
{
    public Collection<WriteableBitmap> Pictures { get; private set; }
    private BackgroundWorker _worker = new BackgroundWorker();

    public ImageProcessor()
    {
        _worker.DoWork += DoWork;
    }

    public void DoLotsOfOperations()
    {
        _worker.RunWorkerAsync();
    }

    private void DoWork(object sender, DoWorkEventArgs e)
    {
        // operations on Pictures collection
    }
}

在运行时,我使用标准的打开文件对话框将图像加载到图片集合中,然后调用 DoLotsOfOperations() 方法。但是,一旦我尝试访问单个位图的任何属性,我就会收到 InvalidOperationException:“调用线程无法访问该对象,因为不同的线程拥有它”。

这显然是正确的 - 我在 UI 线程中加载位图并填充集合,然后尝试在另一个线程中读取集合元素。所以我尝试了不同的方法:

  • 我将整个集合作为 RunWorkerAsync 方法的参数传递,并在 DoWork 方法中从 e.Argument 取回它,但是当我尝试读取单个位图的属性时,我仍然遇到相同的异常。
  • 我尝试了同样的事情,这次将单个位图作为 backgroundworker 的参数传递,但我仍然无法获得位图的任何属性,更不用说位图的像素了。

那么如何在另一个线程中访问位图数据(最好使用 BackgroundWorker)?

我不知道,也许我的整个方法是错误的。我想实现的总体思路是:

  1. 用户加载位图,然后显示在窗口中。
  2. 用户点击一个按钮并在位图上执行长时间操作,但 UI 是响应式的(例如允许用户取消操作)并且进度条会报告进度。

在此先感谢您的帮助。

最佳答案

1)代理类(无线程限制)

    public class WriteableBitmapProxy
    {
        public IntPtr BackBuffer { get; set; }
        public int BackBufferStride { get; set; }
        public int PixelHeight { get; set; }
        public int PixelWidth { get; set; }
    }

2) 扩展方法(不安全)

    public class RGBColor
    {
        public byte R { get; set; }
        public byte G { get; set; }
        public byte B { get; set; }
        public uint Value
        {
            get
            {
                return (uint)(((uint)R << 16) + ((uint)G << 8) + (B) + ((uint)255 << 24));
            }
        }
    }

   public static RGBColor GetPixel(this WriteableBitmap bmp, uint x, uint y)
    {
        unsafe
        {
            if (y >= bmp.PixelHeight) return default(RGBColor);
            if (x >= bmp.PixelWidth) return default(RGBColor);


            // Get a pointer to the back buffer.
            uint pBackBuffer = (uint)bmp.BackBuffer;

            // Find the address of the pixel to draw.
            pBackBuffer += y * (uint)bmp.BackBufferStride;
            pBackBuffer += x * 4;

            byte* pCol = (byte*)pBackBuffer;
            return new RGBColor() { B = pCol[0], G = pCol[1], R = pCol[2] };
        }
    }

    public static void SetPixel(this WriteableBitmapProxy bmp, uint x, uint y, RGBColor col)
    {
        SetPixel(bmp, x, y, col.Value);
    }

    public static void SetPixel(this WriteableBitmapProxy bmp, uint x, uint y, uint value)
    {
        unsafe
        {
            if (y >= bmp.PixelHeight) return;
            if (x >= bmp.PixelWidth) return;

            // Get a pointer to the back buffer.
            uint pBackBuffer = (uint)bmp.BackBuffer;

            // Find the address of the pixel to draw.
            pBackBuffer += y * (uint)bmp.BackBufferStride;
            pBackBuffer += x * 4;

            // Assign the color data to the pixel.
            *((uint*)pBackBuffer) = value;
        }
    }

3) 在不同线程中触发长时间运行的操作的程序

      var image = sender as Image;
        var bitmap = image.Source as WriteableBitmap;

        var prx = new WpfImage.MyToolkit.WriteableBitmapProxy()
        {
            BackBuffer = bitmap.BackBuffer,
            BackBufferStride = bitmap.BackBufferStride,
            PixelHeight = bitmap.PixelHeight,
            PixelWidth = bitmap.PixelWidth
        };

        bitmap.Lock();

        Thread loader = new Thread(new ThreadStart(() => 
        {


            Global_Histogramm(prx);

            Dispatcher.BeginInvoke(DispatcherPriority.Background,
                  (SendOrPostCallback)delegate { bitmap.AddDirtyRect(new Int32Rect(0, 0, prx.PixelWidth - 1, prx.PixelHeight - 1)); bitmap.Unlock(); }, null);

        }

        ));
        loader.Priority = ThreadPriority.Lowest;
        loader.Start();

4) 长时间运行的操作实现

    void Global_Histogramm(WpfImage.MyToolkit.WriteableBitmapProxy src)
    {
        int SrcX = src.PixelWidth;
        int SrcY = src.PixelHeight;

        double[] HR = new double[256];
        double[] HG = new double[256];
        double[] HB = new double[256];
        double[] DR = new double[256];
        double[] DG = new double[256];
        double[] DB = new double[256];
        uint i, x, y;

        //  wyzeruj tablice
        for (i = 0; i < 256; i++) HB[i] = HG[i] = HR[i] = 0;

        //  wypelnij histogramy R G B
        for (y = 0; y < SrcY; y++)
            for (x = 0; x < SrcX; x++)
            {
                var color = src.GetPixel(x, y);
                HB[color.B]++;
                HG[color.G]++;
                HR[color.R]++;
            };

        // oblicz histogramy znormalizowane i przygotuj dystrybuanty
        int ilosc_punktow = SrcX * SrcY;
        double sumaR = 0, sumaG = 0, sumaB = 0;

        for (i = 0; i < 256; i++)
        {
            DB[i] = sumaB + HB[i] / ilosc_punktow;
            DG[i] = sumaG + HG[i] / ilosc_punktow;
            DR[i] = sumaR + HR[i] / ilosc_punktow;
            sumaB = DB[i];
            sumaG = DG[i];
            sumaR = DR[i];
        };

        Dispatcher.BeginInvoke(DispatcherPriority.Background,
              (SendOrPostCallback)delegate { progressBar1.Maximum = SrcY - 1; }, null);



        // aktualizuj bitmape
        for (y = 0; y < SrcY; y++)
        {
            for (x = 0; x < SrcX; x++)
            {

                var stmp = src.GetPixel(x, y);
                var val = new WpfImage.MyToolkit.RGBColor()
                {
                    B = (byte)(DB[stmp.B] * 255),
                    G = (byte)(DG[stmp.G] * 255),
                    R = (byte)(DR[stmp.R] * 255)
                };
                src.SetPixel(x, y, val);                    
            };

            Dispatcher.BeginInvoke(DispatcherPriority.Background,
                  (SendOrPostCallback)delegate { progressBar1.Value = y; }, null);


        }
    }

5) 希望它能证明这一点。

关于c# - WriteableBitmap 的异步操作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1855308/

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