TCP同步连续传输大量数据出现数据不一致现象，求解决方法。
有两个类TCPService与TCPClient，这是我自己封装的两个Socket类，实现的原理是通过多线程实现同步数据通信
第一个类TCPService：

using System;
using System.Collections.Generic;
using System.Text;
using System.Net;
using System.Net.Sockets;
using System.Windows.Forms;
using PDSCommon;
using System.Threading;

namespace BottomCommunicate
{
  public class TCPService
  {
  Socket soc;
  const string State = "State";
  const string Next = "Next";
  const string End = "End";
  const string Again = "Again";
  IPEndPoint ips = null;
  List<Socket> listSocket = new List<Socket>();
  private TCPService() { }
  public TCPService(string ip, int port)
  {
  //IPEndPoint ips = null;
  try
  {
  ips = new IPEndPoint(IPAddress.Parse(ip), port);
  }
  catch
  {
  MessageBox.Show("IP或端口错误！");
  }
  soc = new Socket(ips.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
  }
  public TCPService(IPEndPoint ip)
  {
  ips = ip;
  soc = new Socket(ip.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
  }
  #region ITcpService 成员
  void Listen()
  {
  soc.Bind(ips);
  soc.Listen(100);
  ThreadPool.QueueUserWorkItem(ListenThread,soc);
  }
  void ListenThread(object obj)
  {
  Socket s = (Socket)obj;
  Socket target = s.Accept();//当等待客户端连接时会阻塞
  AcceptEvent(target);//与客户端建立连接，通知用户
  listSocket.Add(target);
  }
  public bool Accept()
  {
  try
  {
  Listen();
  return true;
  }
  catch (Exception ex)
  {
  MessageBox.Show("<TCPService,Accept>:" + ex.Message);
  return false;
  }
  }

  public event acceptSocket AcceptSocket;
  void AcceptEvent(Socket target)
  {
  if (AcceptSocket != null)
  AcceptSocket(target);
  }

  public bool Send(Socket target, byte[] data)
  {
  return Send(target, data, SocketFlags.None);
  }

  public bool Send(Socket target, byte[] data, SocketFlags socketflags)
  {
  lock (new object())
  {
  try
  {
  StateObject stateobject = new StateObject();
  stateobject.target = target;
  stateobject.data = data;
  stateobject.socketflags = socketflags;
  ThreadPool.QueueUserWorkItem(SendData, stateobject);
  return true;
  }
  catch (Exception ex)
  {
  MessageBox.Show(ex.Message);
  return false;
  }