【备忘】想从 Bitmap 得到 BitmapSource 还真麻烦

2011-01-13 16:18 Nana's Lich 阅读(2537) 评论(0) 编辑收藏举报

http://www.codeproject.com/KB/WPF/BitmapToBitmapSource.aspx

public static class Imaging
{
    public static BitmapSource CreateBitmapSourceFromBitmap(Bitmap bitmap)
    {
        if (bitmap == null)
            throw new ArgumentNullException("bitmap");
                 
        if (Application.Current.Dispatcher == null)
            return null; // Is it possible?
                 
        try
        {
            using (MemoryStream memoryStream = new MemoryStream())
            {
                // You need to specify the image format to fill the stream. 
                // I'm assuming it is PNG
                bitmap.Save(memoryStream, ImageFormat.Png);
                memoryStream.Seek(0, SeekOrigin.Begin);
                 
                // Make sure to create the bitmap in the UI thread
                if (InvokeRequired)
                    return (BitmapSource)Application.Current.Dispatcher.Invoke(
                        new Func<Stream, BitmapSource>(CreateBitmapSourceFromBitmap),
                        DispatcherPriority.Normal,
                        memoryStream);
                 
                return CreateBitmapSourceFromBitmap(memoryStream);
            }
        }
        catch (Exception)
        {
            return null;
        }
    }
                 
    private static bool InvokeRequired
    {
        get { return Dispatcher.CurrentDispatcher != Application.Current.Dispatcher; }
    }
                 
    private static BitmapSource CreateBitmapSourceFromBitmap(Stream stream)
    {
        BitmapDecoder bitmapDecoder = BitmapDecoder.Create(
            stream,
            BitmapCreateOptions.PreservePixelFormat,
            BitmapCacheOption.OnLoad);
                 
        // This will disconnect the stream from the image completely...
        WriteableBitmap writable = new WriteableBitmap(bitmapDecoder.Frames.Single());
        writable.Freeze();
                 
        return writable;
    }
}