Is there a way to convert a System.IO.Stream to a Windows.Storage.Streams.IRandomAccessStream?

Question

In Windows 8; I would like to pass the contents of a MemoryStream to a class that accepts a parameter of type Windows.Storage.Streams.IRandomAccessStream. Is there any way to convert this MemoryStream to an IRandomAccessStream?

Answer 1

To use the extensions: you must add "using System.IO"

In Windows8, .NET and WinRT types are generally converted to/from compatible types under the hood so you don't have to care about it.

For streams, however, there are helper methods to convert between WinRT and .NET streams: For converting from WinRT streams -> .NET streams:

InMemoryRandomAccessStream win8Stream = GetData(); // Get a data stream from somewhere.
System.IO.Stream inputStream = win8Stream.AsStream()

For converting from .NET streams -> WinRT streams:

Windows.Storage.Streams.IInputStream inStream = stream.AsInputStream();
Windows.Storage.Streams.IOutputStream outStream = stream.AsOutputStream();

UPDATE: 2013-09-01

Let it not be said that Microsoft doesn't listen to it's developer community ;)

In the announcement for .NET FX 4.5.1, Microsoft states that:

Many of you have been wanting a way to convert a .NET Stream to a Windows Runtime IRandomAccessStream. Let’s just call it an AsRandomAccessStream extension method. We weren't able to get this feature into Windows 8, but it was one of our first additions to Windows 8.1 Preview.

You can now write the following code, to download an image with HttpClient, load it in a BitmapImage and then set as the source for a Xaml Image control.

    //access image via networking i/o
    var imageUrl = "http://www.microsoft.com/global/en-us/news/publishingimages/logos/MSFT_logo_Web.jpg";
    var client = new HttpClient();
    Stream stream = await client.GetStreamAsync(imageUrl);
    var memStream = new MemoryStream();
    await stream.CopyToAsync(memStream);
    memStream.Position = 0;
    var bitmap = new BitmapImage();
    bitmap.SetSource(memStream.AsRandomAccessStream());
    image.Source = bitmap;

HTH.