C# socket发送图片

Question

  //打开文件
                OpenFileDialog dlg = new OpenFileDialog();
                dlg.Filter = "图片文件(*.jpg;*.bmp;*.png;*.gif)|*.jpg;*.bmp;*.png;*.gif";
                if (dlg.ShowDialog() == DialogResult.OK)
                {
                    FileInfo fileInfo = new FileInfo(dlg.FileName);
                    byte[] buffer = new byte[fileInfo.Length];
                    using (FileStream stream = fileInfo.OpenRead())
                    {
                        stream.Read(buffer, 0, buffer.Length);                        if (sendSocket.Connected)
                        {
                            sendSocket.Send(buffer);
                        }
                    }
                       }发送的时候出现值不在预期的范围内。
还有窗体程序接收时，文字能接收成功，不知道图片怎么接收，并显示在picturebox中 
请高人指点

Answer 1

try
            {
                Socket rSocket = (Socket)ar.AsyncState;
                int rEnd = rSocket.EndReceive(ar);
                for (int i = 0; i < clientNum; i++)
                {                    if (clientSocket[i].Connected)
                    {
                        clientSocket[i].Send(msgBuffer, 0, rEnd, 0);
                                              
                        int size =0;
                        byte[] buf = new byte[1024];
                        do
                        {
                            size = rSocket.Receive(buf, 1024, SocketFlags.None);
                            if (size > 0)
                            {
                                
                                //String tmp = Encoding.ASCII.GetString(buf, 0, size);
                                String tmp = Encoding.GetEncoding("gb2312").GetString(buf, 0, size);
                                textBox1.Text = DateTime.Now+"收到："+clientSocket[clientNum].RemoteEndPoint.ToString()+"："+tmp + "\r\n";                                 }
                        } while (size > 0);                             }
                    rSocket.BeginReceive(msgBuffer, 0, msgBuffer.Length, 0, new AsyncCallback(RecieveCallBack), rSocket);
                }
             
            }
这是接收文字的代码，不知道怎么把接收图片的代码写入！！！

Answer 2

试看看
FileInfo fileInfo = new FileInfo(dlg.FileName);
  byte[] buffer = null;
  using (FileStream stream = fileInfo.OpenRead())
  {
buffer = new byte[stream.Length];