java - Tic Tac Toe 程序未确定获胜者

Question

这个问题已经有答案了:

Is Java "pass-by-reference" or "pass-by-value"? (91 个回答)

已关闭 7 年前。

我的井字棋程序有一个小问题。我有一个嵌套计数器，用于检查 X block 的垂直和水平胜利(我还没有为 O 完成它)。由于某种原因，它永远不会结束，让我可以放置越来越多的碎片。

对于某些上下文，这里有一些代码片段。

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 是检查玩家是否已经下棋的方法。我认为问题出在 WinCheck 方法中。如果有人能帮助我，那就太好了!第一次发帖，如有错误请多多包涵。

Answer 1

您正在修改参数win在方法winCheck中并且您期望这会更改局部变量 win在方法game中。但事实并非如此 - Java 按值传递方法参数，因此 win在winCheck是一个副本，任何更改都不会反射(reflect)回来。您可以将方法的返回类型设置为 winCheck一个boolean .

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

    return win;
}

在你的 game 中方法，您更改方法的任何调用 winCheck至:

win = winCheck(...);