如何检查二维数组在井字游戏的赢家

How to check 2 Dimensional array for a winner in a game of tic tac toe

本文关键字：游戏 赢家 二维数组 何检查 检查      更新时间：2023-10-16

我正在创建一个游戏井字游戏作为一个项目与一个二维数组，我有一些麻烦检查点，看看是否有一个赢家。有人能帮我调试这个吗?下面是checkwinner函数和main函数。

char CheckWinner( char board[3][3])
{
    int i = 0;
    int j = 0;
                    if ( board[i][j] =='X' && board[i][j+1] == 'X' && board[i][j+2]== 'X' )
                    { W = X;}
                    else if (board[i+1][j] && board[i+1][i+1] && board[i+1][j+2]== 'X' )
                    { W = X;}
                    else if (board[i+2][j] && board[i+2][j+1] && board[i+2][j+2]== 'X')
                    { W = X;}
                    else if (board[i][j] && board[i+1][j] && board[i+2][j]== 'X')
                    { W = X;}
                    else if (board[i][j+1] && board[i+1][j+1] && board[i+2][j+1]== 'X')
                    { W = X;}
                    else if (board[i][j+2] && board[i+1][j+2] && board[i+2][j+2]== 'X')
                    { W =X;}
                    else if (board[i][j] && board[i+1][j+1] && board[i+2][j+2]=='X')
                    { W = X;}
                    else if (board[i+2][j] && board[i+1][j+1] && board[i][j+2]== 'X')
                    { W = X;}
                    else if (board[i][j] && board[i][j+1] && board[i][j+2]== 'O' )
                    { W = O;}
                    else if (board[i+1][j] && board[i+1][j+1] && board[i+1][j+2]== 'O')
                    { W = O;}
                    else if (board[i+2][j] && board[i+2][j+1] && board[i+2][j+2]== 'O')
                    { W = O;}
                    else if (board[i][j] && board[i+1][j] && board[i+2][j]== 'O')
                    { W = O;}
                    else if (board[i][j+1] && board[i+1][j+1] && board[i+2][j+1]== 'O')
                    { W = O;}
                    else if (board[i][j+2] && board[i+1][j+2] && board[i+2][j+2]== 'O')
                    { W = O;}
                    else if (board[i][j] && board[i+1][j+1] && board[i+2][j+2]== 'O')
                    { W = O;}
                    else if (board[i+2][j] && board[i+1][j+1] && board[i][j+2]== 'O')
                    { W = O;}

    return W;
}
int main () 
{ 
    char board[3][3];
    char Win = CheckWinner(board);
    int r = 0;

    InitializeBoard(board);
     for (int r = 0; r < 4 ; r++)
     { 
        PlayX(board);
        PlayO(board); 
        PrintBoard(board);
    }
     CheckWinner(board);
     cout << Win ;
    if (Win == X)
    {
        cout << "The winner is Player 1.";
    }
    else if (Win == O)
    {
        cout << "The winner is Player 2.";
    }
    else if (Win == TIE)
    {
        cout << " IT'S A TIE";
    }
    else; 

    system("PAUSE");
return 0;

您可以使用以下方法优化条件代码:

typedef char piece;
piece iswin() const
{
    piece ret = 'T';
    // Checks for horizontal win
    for (int i = 0; i < 3; ++i)
        if (*arr[i] == arr[i][1] && arr[i][1] == arr[i][2])
            if ((ret = *arr[i]) != 'T')
                return ret;
    // Checks for vertical win
    for (int i = 0; i < 3; ++i)
        if (arr[0][i] == arr[1][i] && arr[1][i] == arr[2][i])
            if ((ret = arr[0][i]) != 'T')
                return ret;
    // Check for diagonal win (upper left to bottom right)
    if (**arr == arr[1][1] && arr[1][1] == arr[2][2])
        if ((ret = **arr) != 'T')
            return ret;
    // Check for diagonal win (upper right to bottom left)
    if (arr[0][2] == arr[1][1] && arr[1][1] == arr[2][0])
        if ((ret = arr[0][2]) != 'T')
            return ret;
    return ret;
}
// checking the result:
switch (winner) {
    case 'T': /* tie */ break;
    case 'X': /* X won */ break;
    case 'O': /* O won */ break;
}

这个更清晰一些。它循环了6次而不是9次。

您立即将CheckWinner的结果分配给Win，然后进行游戏，然后再次调用CheckWinner，而不将其结果分配给Win。因此，当您检查下一行中的Win时，您将获得板甚至未初始化时的原始结果。

可能与将文字0或X赋值到应该是char的返回值有关。因此，它应该是'O'或'X'，而不是'O'或'X'。除非O或X在发布的代码之外被定义为字符变量

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