java - 井字游戏程序无法识别获胜者

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Is Java “pass-by-reference” or “pass-by-value”?

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4年前关闭。




我的Tic Tac Toe程序有这个小问题。我有一个嵌套的计数器，用于检查X片的垂直和水平胜利（我还没有为O做过）。由于某种原因，它永无止境，让我可以放置越来越多的作品。

对于某些上下文，这是一些代码片段。

public static void game (String winner, Board pointer)
{
    boolean win = false;
    int turnCount = 1;
    winner = "Tie";
    while (win == false || turnCount < 9)
    {
        int row = Integer.parseInt(JOptionPane.showInputDialog(pointer.getName1() + ", it is your turn. " + "\n" + "What row would you like your 'x' to be in?"));
        int col = Integer.parseInt(JOptionPane.showInputDialog("What column would you like it to be in?"));
        pointer.play[row] [col] = pointer.x;
        SpotCheck1(winner, pointer, row, col);
        System.out.println(BoardStat(pointer));
        winCheck(win, pointer, winner);
        int row2 = Integer.parseInt(JOptionPane.showInputDialog(pointer.getName2() + ", it is your turn. " + "\n" + "What row would you like your 'o' to be in?"));
        int col2 = Integer.parseInt(JOptionPane.showInputDialog("What column would you like it to be in?"));
        SpotCheck2(winner, pointer, row2, col2);
        pointer.play[row2] [col2] = pointer.o;
        System.out.println(BoardStat(pointer));
        winCheck(win, pointer, winner);
        turnCount++;
    }
    //return winner;
}

public static void winCheck (boolean win, Board pointer, String winner)
{
    for (int counter = 1; counter <= 3; counter++)
    {
        if ( pointer.play [counter] [1].compareTo(pointer.x) > 0 && pointer.play [counter] [2].compareTo(pointer.x) > 0 && pointer.play [counter] [3].compareTo(pointer.x) > 0)
        {
            win = true;
            winner = pointer.getName1();
            win(winner, pointer);
        }
        else if (pointer.play [1] [counter].compareTo(pointer.x) > 0 && pointer.play [2] [counter].compareTo(pointer.x) > 0 && pointer.play [3] [counter].compareTo(pointer.x) > 0)
        {
            win = true;
            winner = pointer.getName1();
            win(winner, pointer);
        }
    }
    //return win;

}

SpotChecks 1和2是检查玩家是否已放置棋子的方法。我认为问题出在WinCheck方法中。如果有人可以帮助我，那就太好了！第一篇文章，如果我做错了事，我深表歉意。

最佳答案

您正在修改方法win中的参数winCheck，并且期望这会更改方法win中的局部变量game。但是事实并非如此-Java按值传递方法参数，因此win中的winCheck是副本，任何更改都不会反映出来。您可以将方法winCheck的返回类型设置为boolean。

public static boolean winCheck (boolean win, Board pointer, String winner)
{
    // Other code is unchanged

    return win;
}

然后在您的game方法中，将对方法winCheck的任何调用更改为：

win = winCheck(...);

关于java - 井字游戏程序无法识别获胜者，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/37461238/