這個KnockKnockProtocol類的構造函數代碼中的Java錯誤在哪裏？

我正在研究以下客戶端/服務器代碼，名爲KnockKnockServer and KnockKnockClient.它還有一個名爲KnockKnockProtcol的助手類，它負責KnockKnock笑話的順序。這個KnockKnockProtocol類的構造函數代碼中的Java錯誤在哪裏？

我想修改它，以便您可以在特定的笑話中啓動程序。就像現在，你必須從第一個笑話開始。

這是試圖爲KnockKnockProtocol：

public class KnockKnockProtocol { 

    int ourstep; 

    public KnockKnockProtocol (int step) { 
     this.ourstep = step; 
    } 

    private static final int WAITING = 0; 
    private static final int SENTKNOCKKNOCK = 1; 
    private static final int SENTCLUE = 2; 
    private static final int ANOTHER = 3; 

    private static final int NUMJOKES = 5; 

    private int state = WAITING; 
    int currentJoke = ourstep; //we initialize the step here 

    private String[] clues = { "Turnip", "Little Old Lady", "Atch", "Who", "Who" }; 
    private String[] answers = { "Turnip the heat, it's cold in here!", 
           "I didn't know you could yodel!", 
           "Bless you!", 
           "Is there an owl in here?", 
           "Is there an echo in here?" }; 

    public String processInput(String theInput) { 
     String theOutput = null; 

     if (state == WAITING) { 
      theOutput = "Knock! Knock!"; 
      state = SENTKNOCKKNOCK; 
     } else if (state == SENTKNOCKKNOCK) { 
      if (theInput.equalsIgnoreCase("Who's there?")) { 
       theOutput = clues[currentJoke]; 
       state = SENTCLUE; 
      } else { 
       theOutput = "You're supposed to say \"Who's there?\"! " + 
       "Try again. Knock! Knock!"; 
      } 
     } else if (state == SENTCLUE) { 
      if (theInput.equalsIgnoreCase(clues[currentJoke] + " who?")) { 
       theOutput = answers[currentJoke] + " Want another? (y/n)"; 
       state = ANOTHER; 
      } else { 
       theOutput = "You're supposed to say \"" + 
       clues[currentJoke] + 
       " who?\"" + 
       "! Try again. Knock! Knock!"; 
       state = SENTKNOCKKNOCK; 
      } 
     } else if (state == ANOTHER) { 
      if (theInput.equalsIgnoreCase("y")) { 
       theOutput = "Knock! Knock!"; 
       if (currentJoke == (NUMJOKES - 1)) 
        currentJoke = 0; 
       else 
        currentJoke++; 
       state = SENTKNOCKKNOCK; 
      } else { 
       theOutput = "Bye."; 
       state = WAITING; 
      } 
     } 
     return theOutput; 
    } 
}

而在服務器級別的，我稱之爲KnockKnockProtocol這樣的：

/* omitting boilerplate code */ 
    String inputLine, outputLine; 

    // Initiate conversation with client 
    KnockKnockProtocol kkp = new KnockKnockProtocol(2); 
    outputLine = kkp.processInput(null); 
    out.println(outputLine); 

    while ((inputLine = in.readLine()) != null) { 
     outputLine = kkp.processInput(inputLine); 
     out.println(outputLine); 
     if (outputLine.equals("Bye.")) 
      break;

的問題是，當我運行我的代碼我總是從「蘿蔔」的笑話開始。我該如何做到這一點，以便我可以從笑話列表中的任意一個笑話開始？我可以看到這個笑話由clues數組控制，但之後我沒有看到它。感謝

來源

2015-07-21 Coffee

如果你看一下輸入KnockKnockProtocol構造函數，你會發現它直接影響通過this.ourstep的currentJoke計數器，然後將其用於引用clues陣列。

因此，從不同的位置開始，您可以在程序開始時在構造函數中傳入不同的數字。

希望有幫助！

來源

2015-07-21 02:36:24

您確定這是否行？你參考的代碼是我添加的..哪些沒有工作 – Coffee

這可能是因爲currentJoke設置的位置。刪除步驟類變量並將其替換爲currentJoke。在構造函數中設置currentJoke，並在你的其他函數中使用它（確保你使用this.currentJoke）。只有你的數組等靜態元素應該在構造函數之外初始化。 –

這個KnockKnockProtocol類的構造函數代碼中的Java錯誤在哪裏？

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