Question

当运行此代码时，游戏永远不会宣布胜利者或领带，它会不断要求下一个玩家移动......

示例输出：

Choose a position to play
1 | 2 | 3
---------
4 | 5 | 6
---------
7 | 8 | 9
Player X (Enter a position or -1 to resign): 1

Choose a position to play.
1 | 2 | 3
---------
4 | 5 | 6
---------
7 | 8 | 9
Player O (Enter a position or -1 to resign): 1

它不会改变角色的空间数量，甚至不会记住它甚至宣布获胜者。

这是我的代码中的runStandardGame（）部分，我认为我的问题就在那里？

//***RUN STANDARD GAME******************************************************

static void runStandardGame() {

    int position = QUIT;
    char playerChar = 'X';
    boolean isGameOver = false;

    fillPosition();

    System.out.println("Choose a position to play.\n");


    displayGrid();

    do {
        System.out.print("Player " + playerChar +
                         " (Enter a position or " + QUIT + " to resign): ");
        position = keyboard.nextInt();

        System.out.println("Choose a position to play.\n");

        displayGrid();


        if(isWin()) {
            System.out.print("\nPlayer " + playerChar + " WINS!!!\n");
            isGameOver = true;
        }
        else if (isTie()) {
            System.out.print("\nTIE.\n");
            isGameOver = true;
        }
        else {
            //switch players because one of the players has just played
            //and not won;, so, it is the other player's turn
            if (playerChar == 'X') {
                playerChar = 'O';
            }
            else{
                playerChar = 'X';
            }
        }
    }while(!isGameOver && position != QUIT);

}//end of runStandardGame

甚至可能只是游戏部分，因为这是分配给数组的地方......

//***PLAY*******************************************************************    

static void play(int cell, char playerChar) {
    //the player has entered a number 1 through 9 for the position they want
    //to play, so, figure out which row and which column of the array this is
    //For example, if the player wants to play 'X' in position 7, this means
    //the 'X' must go into row 2, column 0 of the array
    int row = 0;
    int column = 0;

    if (cell > 0 && cell <= (STANDARD_GRID_ROWS * STANDARD_GRID_COLUMNS)){
        row = (cell - 1) / STANDARD_GRID_ROWS;
        column = (cell - 1) % STANDARD_GRID_COLUMNS;
        grid[row][column] = playerChar;
    }
}

感谢任何帮助。谢谢。

Answer 1

在我看来，

position = keyboard.nextInt();

在显示和继续之前需要对位置（将其写入电路板）做一些事情吗？

Answer 2

我看到你正在通过

获得用户输入

position = keyboard.nextInt();

但是我没有看到你实际上通过调用

来使用它

play(position, playerChar);

TicTacToe Java游戏不会结束

2 个答案: