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如何在井字游戏中检查获胜者的二维阵列

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

I am creating a game of tic tac toe as a project with a 2dimensional array and I am having some trouble checking the spots to see if there is a winner. 我正在创建一个井字游戏,作为一个具有二维数组的项目,但在检查位置以查看是否有赢家时遇到了一些麻烦。 Can someone help me debug this? 有人可以帮我调试吗? Here is the checkwinner function and the main function. 这是检查赢家功能和主要功能。

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; 

You can optimize your conditional code with the following: 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;
}

This is little bit cleaner. 这有点干净。 And it does 6 loops instead of 9. 它执行6个循环而不是9个循环。

You immediately assign the CheckWinner result to Win , then play the game out, then call CheckWinner again without assigning its result to Win . 您立即将CheckWinner结果分配给Win ,然后玩游戏,然后再次调用CheckWinner而不将结果分配给Win Thus when you check Win in the following line, you have the original result from when the board wasn't even initialized. 因此,当您在下一行中选中Win时,您将获得甚至没有初始化板时的原始结果。

Probably has to do with assigning a literal O or X into a return value that is supposed to be a char. 可能与将文字O或X分配给应该是char的返回值有关。 Thus instead of O or X it should be 'O' or 'X'. 因此,代替O或X的应该是'O'或'X'。 Unless O or X is defined as a char variable outside the posted code. 除非将O或X定义为发布代码之外的char变量。

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