GOBang对战
一、學習目標
參考可以通過因特網對弈的“吃棋子”游戲程序，按以下要求進行改編，要求如下：
1）將游戲改為雙方對弈，而不是系統自動下棋；
2）修改游戲規則，如五子棋的游戲規則；
3）同步改異步；
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二、設計思路
1）放置棋子。先不考慮輪流下子的情況和輸贏的問題，重新把服務器端和客戶端的發送信息和處理收到的信息處理一次。
客戶端：修改玩家對鼠標點擊的事件處理，設置顏色color變量屬性表示玩家下的棋子的顏色，和side變量相同，下的棋子的顏色會相同，在下棋的地方會出現玩家相應顏色的棋子。
private int color; //edit this.color = side;
/// <summary>在pictureBox1中按下鼠標觸發的事件</summary>
??????? private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
??????? {
??????????? int x = e.X / 20;
??????????? int y = e.Y / 20;
??????????? if (!(x < 1 || x > 15 || y < 1 || y > 15))
??????????? {
??????????????? if (grid[x - 1, y - 1] == DotColor.None)//edit
??????????????? {
??????????????????? //int color = (int)grid[x - 1, y - 1];
??????????????????? //發送格式：setDot,桌號,座位號,行,列,顏色
?????????????????? ?service.SendToServer(string.Format(
?????????????????????? "SetDot,{0},{1},{2},{3},{4}", tableIndex, side, x - 1, y - 1, this.color));//edit
??????????????? }
??????????? }
??????? }
服務器端，去掉課本例子中根據事件間隔隨機交替的發送兩種顏色的棋子那部分功能。每當一方玩家下棋后，給同桌的玩家兩人發送新的棋子的位置。
public int turn;?????????????????? //輪流玩,0為黑，1為白
/// <summary>發送產生的棋子信息</summary>
??????? /// <param name="i">指定棋盤的第幾行</param>
??????? /// <param name="j">指定棋盤的第幾列</param>
??????? /// <param name="dotColor">棋子顏色</param>
??????? public void SetDot(int i, int j, int dotColor)//edit private to public
??????? {
??????????? if (dotColor == turn)//edit
??????????? {
??????????????? //向兩個用戶發送產生的棋子信息，并判斷是否有相鄰棋子
??????????????? //發送格式：SetDot,行,列,顏色
??????????????? grid[i, j] = dotColor;
??????????????? service.SendToBoth(this, string.Format("SetDot,{0},{1},{2}", i, j, dotColor));
??????????????? if (win(dotColor))//edit
??????????????? {
??????????????????? ShowWin(dotColor);
??????????????? }
??????????????? turn = (turn + 1)%2;
??????????? }
??????? }
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2）解決輸贏判斷的問題。五子棋游戲的規則為是先有相同顏色的五子相連的那方的玩家贏，棋子可以按照橫豎斜三個方向，但是考慮邊界的情況。
修改游戲勝利的條件。每當一個玩家下子完畢后就判斷。GameTable類中更改勝利條件
//是否勝利。
??????? private bool win(int dotColor)//edit
??????? {
??????????? int num = 15;
??????????? int checkPoint = 0;
??????????? int i = 0;
??????????? int j = 0;
??????????? //橫著檢查。
??????????? for (int x = 0; x < num * num; x++)
??????????? {
??????????????? int consecutive = 0;
??????????????? checkPoint = x;
??????????????? for (int y = 0; y < 5; y++)
??????????????? {
??????????????????? i = (checkPoint + y) % num;
??????????????????? j = (int)checkPoint / num;
??????????????????? if (checkPoint > (num * num - 1) || i > num - 1)
??????????????????????? break;
??????????????????? if (grid[i, j] == dotColor)
??????????????????? {
??????????????????????? consecutive++;
??????????????????? }
??????????????????? //checkPoint++;
??????????????? }
??????????????? if (consecutive == 5)
??????????????????? return true;
??????????? }
??????????? //豎著檢查
??????????? for (int x = 0; x < num * num; x++)
??????????? {
??????????????? int consecutive = 0;
??????????????? checkPoint = x;
??????????????? for (int y = 0; y < 5; y++)
??????????????? {
??????????????????? i = (checkPoint + y) % num;
??????? ????????????j = (int)checkPoint / num;
??????????????????? if (checkPoint > (num * num - 1) || i > num - 1)
??????????????????????? break;
??????????????????? if (grid[j, i] == dotColor)
??????????????????? {
??????????????????????? consecutive++;
??????? ????????????}
??????????????????? //checkPoint++;
??????????????? }
??????????????? if (consecutive == 5)
??????????????????? return true;
??????????? }
??????????? //正斜
??????????? for (int x = 0; x < num * num; x++)
??????????? {
??????????????? int consecutive = 0;
??????????????? checkPoint = x;
??????????????? for (int y = 0; y < 5; y++)
??????????????? {
??????????????????? i = checkPoint % num + y;
??????????? ????????j = ((int)checkPoint / num) + y;
??????????????????? if (i > num - 1 || j > num - 1)
??????????????????????? break;
??????????????????? if (grid[i, j] == dotColor)
??????????????????? {
??????????????????????? consecutive++;
??????????????????? }
??????????????? }
??????????????? if (consecutive == 5)
??????????????????? return true;
??????????? }
??????????? //反斜
??????????? for (int x = 0; x < num * num; x++)
??????????? {
??????????????? int consecutive = 0;
??????????????? checkPoint = x;
??????????????? for (int y = 0; y < 5; y++)
??????????????? {
??????????????????? i = checkPoint % num - y;
??????????? ????????j = (int)checkPoint / num + y;
??????????????????? if (i > num - 1 || i < 0 || j > num - 1 || j < 0)
??????????????????????? break;
??????????????????? if (grid[i, j] == dotColor)
??????????????????? {
??????????????????????? consecutive++;
?????? ?????????????}
??????????????? }
??????????????? if (consecutive == 5)
??????????????????? return true;
??????????? }
??????????? return false;
??????? }
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勝利后的提示信息改為
case "win":
??????????????????????? //格式：Win,相鄰棋子的顏色,黑方成績,白方成績
??????????????????????? string winner = "";
??????????????????????? if ((DotColor)int.Parse(splitString[1]) == DotColor.Black)
??????????????????????? {
??????????????????????????? winner = "黑方出現五子相連，黑方勝利！";//edit
??????????????????????? }
??????????????????????? else
??????????????????????? {
??????????????????????????? winner = "白方出現五子相連，白方勝利！";//edit
??????????????????????? }
??????????????????????? formPlaying.ShowMessage(winner);
??????????????????????? formPlaying.Restart(winner);
??????????????????????? break;
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3）雙方輪流的下棋子。在雙方都開始后游戲正式開始。
在前面代碼中已經添加了turn變量，控制下棋方玩家。初始化的時候是-1，表示如果雙方沒有開始游戲，都不能下子。FormServer類收到的message處理start時，更改條件，都開始后，把turn改為0，表示黑子先下，符合五子棋規則。關鍵代碼如下：
if (gameTable[tableIndex].gamePlayer[anotherSide].started == true)
??????????????????????? {
??????????????????????????? gameTable[tableIndex].ResetGrid();
??????????????????????????? //gameTable[tableIndex].StartTimer();
??????????????????????????? gameTable[tableIndex].Turn = 0;
??????????????????????? }
GameTable中，只有當輪到自己時，即到自己的顏色時才能下子。
public int Turn{
??????????? get{return turn;}
??????????? set { turn = value; }
??????? }
下子后，變換下子方。
turn = (turn + 1)%2;
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4）同步改異步
程序在調用Begin…后，可以在調用線程上繼續執行其下面的指令，同時異步操作在另一個線程上執行。Begin…方法開始異步操作，并返回一個實現 IAsyncResult接口的對象。IAsyncResult對象存儲有關異步操作的狀態信息。這些信息包括：IAsyncState：可選的特定的對象，包含異步操作需要的信息。syncWaitHandle：用于在異步操作完成前阻止程序執行。CompletedSynchronously：指示異步操作是否在用于調用Begin…的線程上完成，而不是在單獨的ThreadPool線程上完成。IsCompleted：一個布爾值，指示異步操作是否已完成。
服務器端：監聽客戶端和接收客戶端消息類似，使用異步方式調用同步方法，通過輪詢方式檢查異步調用是否完成，使用EndInvoke結束異步調用，EndInvoke方法用于檢索異步調用的結果，并結束異步調用。調用BeginInvoke之后，隨時可以調用該方法。如果異步調用尚未完成，則EndInvoke會一直阻止調用線程，直到異步調用完成。
private delegate void ListenClientDelegate(out TcpClient client);
??????? /// <summary>接受掛起的客戶端連接請求</summary>
??????? private void ListenClient(out TcpClient newClient)
??????? {
??????????? try
??????????? {
??????????????? newClient = myListener.AcceptTcpClient();
??????????? }
??????????? catch
??????????? {
??????????????? newClient = null;
??????????? }
??????? }
/// <summary>監聽客戶端請求</summary>
??????? private void ListenClientConnect()
??????? {
??????????? TcpClient newClient = null;
??????????? while (true)
??????????? {
??????????????? ListenClientDelegate d = new ListenClientDelegate(ListenClient);
??????????????? IAsyncResult result = d.BeginInvoke(out newClient, null, null);
??????????????? //使用輪詢方式來判斷異步操作是否完成
??????????????? while (result.IsCompleted == false)
??????????????? {
??????????????????? if (isExit)
??????????????????? {
??????????????????????? break;
??????????????????? }
??? ????????????????Thread.Sleep(250);
??????????????? }
??????????????? //獲取Begin方法的返回值和所有輸入/輸出參數
??????????????? d.EndInvoke(out newClient, result);
??????????????? if (newClient != null)
??????????????? {
??????????????????? //每接受一個客戶端連接,就創建一個對應的線程循環接收該客戶端發來的信息
??????????????????? User user = new User(newClient);
??????????????????? Thread threadReceive = new Thread(ReceiveData);
??????????????????? threadReceive.Start(user);
??????????????????? userList.Add(user);
??????????????????? service.AddItem(string.Format("[{0}]進入", newClient.Client.RemoteEndPoint));
??????????????????? service.AddItem(string.Format("當前連接用戶數：{0}", userList.Count));
??????????????? }
??????????????? else
??????????????? {
??????????????????? break;
??????????????? }
??????????? }
??? ????}
服務器端發送消息也類似上面的方法：
/// <summary>異步發送message給user</summary>
??????? public void AsyncSendToOne(User user, string message)
??????? {
??????????? SendToOneDelegate d = new SendToOneDelegate(SendToOne);
??????????? IAsyncResult result = d.BeginInvoke(user, message, null, null);
??????????? while (result.IsCompleted == false)
??????????? {
??????????????? Thread.Sleep(250);
??????????? }
??????????? d.EndInvoke(result);
??????? }
??????? public delegate void SendToOneDelegate(User user, string str);
?????? ?/// <summary>向某個客戶端發送信息</summary>
??????? /// <param name="gameTable">指定客戶</param>
??????? /// <param name="gameTable">信息</param>
??????? public void SendToOne(User user, string str)
??????? {
??????????? try
??????????? {
??????????????? user.sw.WriteLine(str);
??????????????? user.sw.Flush();
??????????????? AddItem(string.Format("向{0}發送{1}", user.userName, str));
??????????? }
??????????? catch
??????????? {
??????????????? AddItem(string.Format("向{0}發送信息失敗", user.userName));
??????????? }
??????? }
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客戶端：用BackgroundWork組件實現異步編程功能
BackgroundWorker connectWork = new BackgroundWorker();
connectWork.DoWork += new DoWorkEventHandler(connectWork_DoWork);
????? connectWork.RunWorkerCompleted +=
??????????????? new RunWorkerCompletedEventHandler(connectWork_RunWorkerCompleted);
??????? /// <summary>異步方式與服務器進行連接</summary>
??????? void connectWork_DoWork(object sender, DoWorkEventArgs e)
??????? {
??????????? client = new TcpClient();
??????????? //此處為方便演示，實際使用時要將Dns.GetHostName()改為服務器域名
??????????? IAsyncResult result = client.BeginConnect(Dns.GetHostName(), 51888, null, null);
??????????? while (result.IsCompleted == false)
??????????? {
??????????????? Thread.Sleep(100);
??????????? }
??????????? try
??????????? {
??????????????? client.EndConnect(result);
????? ??????????e.Result = "success";
??????????? }
??????????? catch (Exception ex)
??????????? {
??????????????? e.Result = ex.Message;
??????????????? return;
??????????? }
??????? }
??????? /// <summary>異步方式與服務器完成連接操作后的處理</summary>
??????? void connectWork_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e)
??????? {
??????????? if (e.Result.ToString() == "success")
??????????? {
??????????????? groupBox1.Visible = true;
??????????????? textBoxLocal.Text = client.Client.LocalEndPoint.ToString();
??????????????? textBoxServer.Text = client.Client.RemoteEndPoint.ToString();
??????????????? buttonConnect.Enabled = false;
??????????????? //獲取網絡流
??????????????? NetworkStream netStream = client.GetStream();
??????????????? //將網絡流作為二進制讀寫對象
???????????? ???sr = new StreamReader(netStream, System.Text.Encoding.UTF8);
??????????????? sw = new StreamWriter(netStream, System.Text.Encoding.UTF8);
??????????????? service = new Service(listBox1, sw);
??????????????? service.SendToServer("Login," + textBoxName.Text.Trim());
??????????????? Thread threadReceive = new Thread(new ThreadStart(ReceiveData));
??????????????? threadReceive.IsBackground = true;
??????????????? threadReceive.Start();
??????????? }
??????????? else
??????????? {
??????????????? MessageBox.Show("與服務器連接失敗 "+e.Result, "",
?????????????????? MessageBoxButtons.OK, MessageBoxIcon.Information);
??????????????? buttonConnect.Enabled = true;
??????????? }
??????? }
??????? /// <summary>【登錄】按鈕的Click事件</summary>
??????? private void buttonConnect_Click(object sender, EventArgs e)
??????? {
??????????? buttonConnect.Enabled = false;
??????????? connectWork.RunWorkerAsync();
??????? }
接受消息的方法跟服務器端相同，不重復描述。
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三、個人總結
???? 由于這個實驗是參考書上已有的例子修改的，所以沒有出現之前無從下手的情況。在實驗的過程中，遇到了一些問題。首先，是輪流下子的問題，不知道怎么通知對方。后來的解決方案是當服務器端發送下子坐標的時候同時更新同一桌玩家雙方的棋盤布局，只需要在特定棋盤上用turn變量斷定下棋的玩家。第二是邊界的判斷比較復雜，我采取了最笨的方法，每次下子的時候都是從新判斷整個棋盤的橫豎正斜反斜，這樣程序的運行效率降低了。第三是同步改異步的問題，。改成異步以后，一直在想，在Form中生成了一個Service類，那么終止條件是什么。請教了同學之后，發現其實在Form退出的時候，service就自動消失了。通過本次試驗，我感覺受益匪淺。
轉載于:https://www.cnblogs.com/maomiyy/p/3762183.html
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