c# return zip stream from api

  
        public Stream GetDataFile(string locationId)
        {
            int cid;
            DateTime tran = DateTime.Now.Date;

            Check(string.IsNullOrWhiteSpace(locationId), "locationId field is missing.");
            Check(!int.TryParse(locationId, out cid), "locationId must be numeric.");


            DataFileRepository repository = new DataFileRepository();
            /*returning stream*/
            var resp = WebOperationContext.Current.OutgoingResponse;
            resp.ContentType = "application/octet-stream";
            resp.Headers.Add("Content-Disposition", "attachment; filename=\"myfile.zip\"");

            return ZipStream(repository.GetCSVStream(cid));
        }

        /// <summary>
        ///  Check condition and generate exception if condition met.
        /// </summary>
        /// <param name="condition"></param>
        /// <param name="exp"></param>
        public void Check(bool condition, string exp)
        {
            if (condition)
            {
                AssertionException up = new AssertionException(exp);
                throw up;
            }
        }

        /// <summary>
        ///  Method for zipping streams 
        /// </summary>
        /// <param name="stream"></param>
        /// <returns></returns>
        public Stream ZipStream(Stream stream)
        {
            stream.Position = 0;
            var ar = new ZipArchive();
            ar.AddItem("myfile.csv", stream, true, FileAttributes.Normal);

            var zip = new TempFileStream();
            ar.Save(zip, false);
            zip.Position = 0;
            return zip;
        }

interface declaration in API webservice is very simple:

      
  [OperationContract]
        [WebGet(UriTemplate = "menu/datafile/{locationId}")]
        Stream GetDataFile(string locationId);