Cuando lo hago let! read = from.AsyncRead bufen F #, se bloquea y no regresa hasta que el socket TCP está muerto. ¿Por qué? ¿Y cómo lo soluciono?
Su código:
module StreamUtil
open System.IO
/// copy from 'from' stream to 'toStream'
let (|>>) (from : Stream) (toStream : Stream) =
  let buf = Array.zeroCreate<byte> 1024
  let rec doBlock () =
    async {
      let! read = from.AsyncRead buf
      if read <= 0 then
        toStream.Flush()
        return ()
      else
        do! toStream.AsyncWrite(buf, 0, read)
        return! doBlock () }
  doBlock ()
Se llama desde este código:
use fs = new FileStream(targPath, FileMode.CreateNew, FileAccess.ReadWrite)
do! req.InputStream |>> fs
y solicitado a través de HTTP con este código del emulador de Windows Phone 7.1:
public void Send()
{
    var b = new UriBuilder(_imageService.BaseUrl) {Path = "/images"};
    var req = WebRequest.CreateHttp(b.Uri);
    req.ContentType = "image/jpeg";
    req.Method = "POST";
    var imgLen = SelectedImage.ImageStream.Length;
    req.Headers[HttpRequestHeader.ContentLength] = imgLen.ToString(CultureInfo.InvariantCulture);
    req.Accept = "application/json";
    req.BeginGetRequestStream(RequestReady, new ReqState(req, imgLen));
}
void RequestReady(IAsyncResult ar)
{
    var state = (ReqState)ar.AsyncState;
    var req = state.Request;
    var reqStream = req.EndGetRequestStream(ar);
    SmartDispatcher.BeginInvoke(() =>
        {
            using (var sw = new StreamWriter(reqStream))
            using (var br = new BinaryReader(SelectedVoucher.ImageStream))
            {
                var readBytes = br.ReadBytes(state.ImgLen);
                // tried both 2
                sw.Write(readBytes);
                //sw.Write(Convert.ToBase64String(readBytes));
                sw.Flush();
                sw.Close();
            }
            req.BeginGetResponse(ResponseReady, req);
        });
}
// WHY IS IT YOU ARE NOT CALLED???
void ResponseReady(IAsyncResult ar)
{
    try
    {
        var request = (HttpWebRequest)ar.AsyncState;
        var response = request.EndGetResponse(ar);
        SmartDispatcher.BeginInvoke(() =>
            {
                var rdr = new StreamReader(response.GetResponseStream());
                var msg = rdr.ReadToEnd();
                var imageLocation = response.Headers["Location"];
                Debug.WriteLine(msg);
                Debug.WriteLine(imageLocation);
            });
    }
    catch (WebException ex)
    {
        Debug.WriteLine(ex.ToString());
    }
    catch (Exception ex)
    {
        Debug.WriteLine(ex.ToString());
    }
}
Sin éxito. La ResponseReadydevolución de llamada nunca se alcanza.
Mientras tanto, este código funciona excelente:
open System
open System.Net.Http // WebAPI nuget
let sync aw = Async.RunSynchronously aw
let postC<'a> (c : HttpClient) (r : Uri) (cont : HttpContent) =
  let response = sync <| Async.AwaitTask( c.PostAsync(r, cont) )
  let struc:'a = sync <| deserialize<'a> response
  response, struc
let withContent<'a> (fVerb : (HttpClient -> Uri -> HttpContent -> _ * 'a))=
  let c = new HttpClient()
  fVerb c
[<Test>]
let ``POST /images 201 + Location header`` () =
  let post = withContent<MyImage> postC
  let bytes = IO.File.ReadAllBytes("sample.jpg")
  let hash = SHA1.Create().ComputeHash(bytes) |> Convert.ToBase64String
  let pic = new ByteArrayContent(bytes)
  pic.Headers.Add("Content-Type", "image/jpeg")
  pic.Headers.Add("X-SHA1-Hash", hash)
  let resp, ri = (resource "/images", pic) ||> post
  resp.StatusCode =? Code.Created
  ri.sha1 =? hash
  mustHaveHeaders resp
No pude hacer que Fiddler2 funcionara con WP7.
EDITAR: Bienvenido a un yak. Yo mismo me mudé a pastos más verdes;)
                    
                        rest
                                asynchronous
                                f#
                                windows-phone
                                
                    
                    
                        Henrik
fuente
                
                fuente

from.AsyncReadbloquea, eso significa que el servidor remoto no envía ningún byte.Respuestas:
Debe poner los bytes en el antes de enviar y usar la salida de entrada de BufferStream
fuente