爲什麼循環不會停止迭代？

Question

我是一名初學java的學生，正在爲我的班級編寫一個gui tic-tac-toe程序。 （沒有玩家，只是計算機生成）。

我的程序中的一切都按預期工作，除了一件事;似乎我的方法調用checkWinner的位置放置不正確，因爲X和O的分配總是完成。爲什麼這個循環不會在獲勝者結束後立即結束？

它會根據方法調用返回正確的獲勝者，但for循環將繼續迭代並填充其餘部分（所以有時它看起來像x和o勝或兩勝兩次）。我一直在瘋狂，認爲這可能是我的checkWinner方法調用和if語句的位置。當我設置winner = true;不應該取消循環？我試圖把它之間，內部和外部每個for-loop沒有運氣:(

我已標記的區域，我認爲是問題//這裏有什麼問題？//關閉在右邊分開的代碼。謝謝你的任何輸入!! :)

public void actionPerformed(ActionEvent e) 
    { 
    int total = 0, i = 0; 
    boolean winner = false; 


    //stop current game if a winner is found 
    do{ 

     // Generate random # 0-1 for the labels and assign 
     // X for a 0 value and O for a 1 value 

     for (int row = 0; row < gameboard.length; row++) //rows 
     { 
     for (int col = 0; col < gameboard[row].length; col++) //columns 
     { 

      //Generate random number 
      gameboard[row][col] = (int)(Math.random() * 2); 

      //Assign proper values 
      if(gameboard[row][col] == 0) 
      { 
      labels[i].setText("X"); 
      gameboard[row][col] = 10; //this will help check for the winner 
      } 

      else if(gameboard[row][col] == 1) 
      { 
      labels[i].setText("O"); 
      gameboard[row][col] = 100; //this will help check for winner 
      }    


      /**Send the array a the method to find a winner 
      The x's are counted as 10s 
      The 0s are counted as 100s 
      if any row, column or diag = 30, X wins 
      if any row, column or diag = 300, Y wins 
      else it will be a tie 
      */ 

      total = checkWinner(gameboard);  **//Is this okay here??//** 
      if(total == 30 || total == 300)  // 
      winner = true;    //Shouldn't this cancel the do-while? 


      i++; //next label 

     } 
     }//end for 
    }while(!winner);//end while 



    //DISPLAY WINNER 
    if(total == 30) 
     JOptionPane.showMessageDialog(null, "X is the Winner!"); 
    else if(total == 300) 
     JOptionPane.showMessageDialog(null, "0 is the Winner!"); 
    else 
     JOptionPane.showMessageDialog(null, "It was a tie!"); 
    } 

Answer 1

首先，你的代碼進行迭代，併產生X的隨機標記：要允許領帶，你不需要，而在所有的外部，並且可以馬上離開這兩個維權，當獲勝者發現和O.這導致了一些非常奇怪的棋盤狀態，總是逐行填充，並且可能具有不平衡的X和O標記。

恕我直言，你應該以相反的方式組織你的代碼，以填補董事會相似的真正的遊戲。我的意思是一系列9分的'XOXOXOXOX'在整個董事會中蔓延。

令Labels labels爲9個字符的數組，初始化爲9個空格。

public int doGame(Labels labels) 
{ 
    labels = "   "; 
    int itisXmove = true;    // player X or O turn 
    for(int movesLeft = 9; movesLeft > 0; movesLeft --) 
    { 
     int position =   // 0 .. movesLeft-1 
       (int) Math.floor(Math.random() * movesLeft); 

     for(int pos = 0; pos < 9; pos ++)  // find position 
      if(labels[ pos] == " ")    // unused pos? 
       if(position-- == 0)    // countdown 
       { 
        if(itisXmove)    // use the pos 
         labels[ pos] = "X";  // for current player 
        else 
         labels[ pos] = "O"; 
        break; 
       } 

     int result = checkWinner(labels);  // who wins (non-zero)? 
     if(result != 0) 
      return result; 

     itisXmove = ! itisXmove;     // next turn 
    } 
    return 0;          // a tie 
} 

然後

public void actionPerformed(ActionEvent e) 
{ 
    Labels labels; 

    int result = doGame(labels); 

    if(result == valueForX) 
     JOptionPane.showMessageDialog(null, "X is the Winner!"); 
    else if(result == valueForO) 
     JOptionPane.showMessageDialog(null, "O is the Winner!"); 
    else 
     JOptionPane.showMessageDialog(null, "It's a tie!"); 

    for(int rowpos = 0; rowpos < 9; rowpos += 3) 
    { 
     for(int colpos = 0; colpos < 3; colpos ++) 
      /* output (char)label[ rowpos + colpos] */; 

     /* output (char)newline */; 
    } 
} 

Answer 2

你不檢查，直到兩個for循環完成的winner值。添加break設置winner = true之後，並添加

if (winner) 
{ 
    break; 
} 

你的外循環for的開頭或結尾。

Answer 3

你的問題是，你的do/while語句是圍繞for陳述。所以for報表最終在達到while聲明之前運行整個週期。溶液來解決這個問題正在檢查在for聲明贏家和斷裂：

//stop current game if a winner is found 
do { 

    for (int row = 0; row < gameboard.length; row++) //rows 
    { 
     for (int col = 0; col < gameboard[row].length; col++) //columns 
     { 

      // ... your other code ... 

      total = checkWinner(gameboard); 
      if(total == 30 || total == 300) { 
       winner = true; 
       break; // end current for-loop 
      } 

      i++; //next label 
     } 

     if (winner) break; // we have a winner so we want to kill the for-loop 
    } //end for 

} while(!winner); //end while 

所以，你應該能夠僅僅通過兩個用於語句循環並在一個勝利者打破。你的代碼似乎也不能處理一個並列的情況，但我猜你已經知道了。

Answer 4

我認爲你應該改變你的循環條件並添加一個更多的布爾。

你有一個「領帶」的條件，但目前你只檢查獲勝者。沒有checkWinner代碼的唯一解釋是你每次都遇到一條領帶。

所以......

boolean tie; 
boolean winner; 

do { 
//your stuff 
} 
while(!(tie || winner)) 

編輯：我不知道你把while循環外的for循環，你將需要爲了打破你的for循環爲while條件是檢查。

//stop current game if a winner is found 
    do{ 

     for (int row = 0; row < gameboard.length; row++) //rows 
     { 
     for (int col = 0; col < gameboard[row].length; col++) //columns 
     { 
      if(winner || tie) 
       break; 
     }//end for 

     if(winner || tie) 
      break; 
     }//end for 
    }while(!(winner || tie));//end while 
//the rest of your stuff here 

Answer 5

的最簡單辦法是打破所有循環一次。 （即使有些人不喜歡這樣）

outerwhile: while(true){ 

    // Generate random # 0-1 for the labels and assign 
    // X for a 0 value and O for a 1 value 

    for (int row = 0; row < gameboard.length; row++) //rows 
    { 
    for (int col = 0; col < gameboard[row].length; col++) //columns 
    { 

     total = checkWinner(gameboard);  
     if(total == 30 || total == 300)   
     break outerwhile; //leave outer while, implicit canceling all inner fors. 


     i++; //next label 
    } 
    }//end for 
}//end while 

這可是不允許的「領帶」的選項，因爲雖然基本上都會重新開始遊戲，如果沒有贏家已經找到。通過板

Boolean winner = false; 
    outerfor: for (int row = 0; row < gameboard.length; row++) //rows 
    { 
    for (int col = 0; col < gameboard[row].length; col++) //columns 
    { 

     total = checkWinner(gameboard);  
     if(total == 30 || total == 300){   
     winner = true;  
     break outerfor; //leave outer for, implicit canceling inner for. 

     } 

     i++; //next label 
    } 
    }//end for 

    if (winner){ 
    //winner 
    }else{ 
    //tie. 
    }