如何从BitmapImage创建WriteableBitmap？

Question

我可以从资源中的图片创建WriteableBitmap。

Uri imageUri1 = new Uri("ms-appx:///Assets/sample1.jpg");
WriteableBitmap writeableBmp = await new WriteableBitmap(1, 1).FromContent(imageUri1);

但是，我不能从图片目录创建WriteableBitmap，(我使用的是WinRT XAML Toolkit)

//open image
StorageFolder picturesFolder = KnownFolders.PicturesLibrary;
StorageFile file = await picturesFolder.GetFileAsync("sample2.jpg");
var stream = await file.OpenReadAsync();

//create bitmap
BitmapImage bitmap2 = new BitmapImage();
bitmap2.SetSource();
bitmap2.SetSource(stream);

//create WriteableBitmap, but cannot
WriteableBitmap writeableBmp3 = 
    await WriteableBitmapFromBitmapImageExtension.FromBitmapImage(bitmap2);

这是正确的吗？

Answer 1

这是一个完全的设计，但它似乎确实有效...

// load a jpeg, be sure to have the Pictures Library capability in your manifest
var folder = KnownFolders.PicturesLibrary;
var file = await folder.GetFileAsync("test.jpg");
var data = await FileIO.ReadBufferAsync(file);

// create a stream from the file
var ms = new InMemoryRandomAccessStream();
var dw = new Windows.Storage.Streams.DataWriter(ms);
dw.WriteBuffer(data);
await dw.StoreAsync();
ms.Seek(0);

// find out how big the image is, don't need this if you already know
var bm = new BitmapImage();
await bm.SetSourceAsync(ms);

// create a writable bitmap of the right size
var wb = new WriteableBitmap(bm.PixelWidth, bm.PixelHeight);
ms.Seek(0);

// load the writable bitpamp from the stream
await wb.SetSourceAsync(ms);

Answer 2

正如菲利普所指出的，将图像读取到WriteableBitmap的工作方式如下：

StorageFile imageFile = ...

WriteableBitmap writeableBitmap = null;
using (IRandomAccessStream imageStream = await imageFile.OpenReadAsync())
{
   BitmapDecoder bitmapDecoder = await BitmapDecoder.CreateAsync(
      imageStream);

   BitmapTransform dummyTransform = new BitmapTransform();
   PixelDataProvider pixelDataProvider =
      await bitmapDecoder.GetPixelDataAsync(BitmapPixelFormat.Bgra8, 
      BitmapAlphaMode.Premultiplied, dummyTransform, 
      ExifOrientationMode.RespectExifOrientation,
      ColorManagementMode.ColorManageToSRgb);
   byte[] pixelData = pixelDataProvider.DetachPixelData();

   writeableBitmap = new WriteableBitmap(
      (int)bitmapDecoder.OrientedPixelWidth,
      (int)bitmapDecoder.OrientedPixelHeight);
   using (Stream pixelStream = writeableBitmap.PixelBuffer.AsStream())
   {
      await pixelStream.WriteAsync(pixelData, 0, pixelData.Length);
   }
}

请注意，我使用的是像素格式和alpha模式可写位图，并且我通过了。

Answer 3

WriteableBitmapFromBitmapImageExtension.FromBitmapImage()通过使用用于加载BitmapImage和IIRC的原始Uri工作，它只适用于来自appx的BitmapImage。在你的例子中甚至没有Uri，因为从Pictures文件夹加载只能通过从stream加载来完成，所以你的选择从最快到最慢(我认为)是：

1. 从一开始就将图像作为WriteableBitmap打开，这样您就不需要重新打开它或复制比特。
2. 如果您需要两个副本，请将其作为WriteableBitmap打开，然后创建一个相同大小的新WriteableBitmap并复制像素缓冲区。如果您需要两个副本，请使用
3. 跟踪用于打开第一个位图的路径，然后通过从与原始文件相同的文件加载它来创建新的D12。<代码>H213<代码>G214

我认为选项2可能比选项3更快，因为您避免了对压缩图像进行两次解码。