【问题标题】:Restart Program So My C++ game is replayable重启程序,这样我的 C++ 游戏就可以重玩了
【发布时间】:2014-12-08 05:49:59
【问题描述】:

我正在为井字游戏编写 C++ 程序。 游戏必须有一个游戏计数器来计算获胜次数 每个玩家的平局数一样长。问题 我遇到的是我似乎无法让董事会清除, 所以当一场比赛结束时,棋盘清零 X 和 O。我能得到任何帮助,我将不胜感激!


#include <iostream>

using namespace std;

void display_board();
void player_turn();
bool gameover();
int xwins, owins, ties;

char turn;
bool draw = false;
char board[3][3] = {{'1', '2', '3'}, {'4', '5', '6'}, {'7', '8', '9'}};

int main()
{
    cout << "Tic Tac Toe Game\n";
    cout << "Player 1 [X] --- Player 2 [O]\n";
    turn = 'X';
    char board[3][3] = {{'1', '2', '3'}, {'4', '5', '6'}, {'7', '8', '9'}};

    while (!gameover())
    {
        display_board();
        player_turn();
        gameover();
    }

    if (turn == 'O' && !draw)
    {
        display_board();
        cout << endl << endl << "Player 1 [X] Wins! Game Over!\n";
        xwins++;
    }
    else if (turn == 'X' && !draw)
    {
        display_board();
        cout << endl << endl << "Player 2 [O] Wins! Game Over!\n";
        owins++;
    }
    else
    {
        display_board();
        cout << endl << endl << "It's a draw! Game Over!\n";
        ties++;
    }


}

void display_board()
{
    cout << "---------------------" << endl << endl;
    cout << "     |     |     " << endl;
    cout << "  " << board[0][0] << "  |  " << board[0][1] << "  |  " << board[0][2] <<"      Player 1 wins "<<xwins<< endl;
    cout << "_____|_____|_____" <<endl;
    cout << "     |     |     " << endl;
    cout << "  " << board[1][0] << "  |  " << board[1][1] << "  |  " << board[1][2] <<"      Player 2 wins "<<owins<< endl;
    cout << "_____|_____|_____" << endl;
    cout << "     |     |     " << endl;
    cout << "  " << board[2][0] << "  |  " << board[2][1] << "  |  " << board[2][2] <<"      Ties "<<ties<< endl;
    cout << "     |     |     " << endl;
}

void player_turn()
{
    int choice;
    int row = 0, column = 0;

    if (turn == 'X')
    {
        cout << "Player 1 turn [X]: ";
    }
    else if (turn == 'O')
    {
        cout << "Player 2 turn [O]: ";
    }
    cin >> choice;

    switch (choice)
    {
    case 1: row = 0; column = 0; break;
    case 2: row = 0; column = 1; break;
    case 3: row = 0; column = 2; break;
    case 4: row = 1; column = 0; break;
    case 5: row = 1; column = 1; break;
    case 6: row = 1; column = 2; break;
    case 7: row = 2; column = 0; break;
    case 8: row = 2; column = 1; break;
    case 9: row = 2; column = 2; break;
    default:
        cout << "You didn't enter a correct number! Try again\n";
        player_turn();
    }

    if (turn == 'X' && board[row][column] != 'X' && board[row][column] != 'O')
    {
        board[row][column] = 'X';
        turn = 'O';
    }
    else if (turn == 'O' && board[row][column] != 'X' && board[row][column] != 'O')
    {
        board[row][column] = 'O';
        turn = 'X';
    }
    else
    {
        cout << "The cell you chose is used! Try again\n";
        player_turn();
    }

}

bool gameover()
{
    for (int i = 0; i < 3; i++)//Check for a win
    {
        if ((board[i][0] == board[i][1] && board[i][1] == board[i][2]) || (board[0][i] == board[1][i] && board[1][i] == board[2][i]) || (board[0][0] == board[1][1] && board[1][1] == board[2][2]) || (board[0][2] == board[1][1] && board[1][1] == board[2][0]))
        {
            return true;
        }
    }

    for (int i = 0; i < 3; i++)//Check for draw
    {
        for (int j = 0; j < 3; j++)
        {
            if (board[i][j] != 'X' && board[i][j] != 'O')
            {
                return false;
            }
        }
    }
    draw = true;
    return true;
}

【问题讨论】:

  • 为什么不避免全局变量,拥有一个可以多次调用的play_game函数...
  • 您的示例代码中没有任何内容可用于重置板。也就是说,尝试获取所有初始化代码并将其移动到 init() 函数中。如果你移动所有东西,你应该能够在你想重新启动时调用它。然后向我们展示您的代码调用了重新初始化,以便我们诊断您在做什么。

标签: c++ arrays


【解决方案1】:

我可以看到您的代码存在一些问题:

while (!gameover())
{
    display_board();
    player_turn();
    gameover();
}

循环中不需要 gameover()。您已经在条件中检查它。

此外,您不会在每个游戏实例之后重置全局变量。添加另一个 while / do-while 循环,该循环一直运行到特定条件。在此循环中,添加您的主要功能。并为循环的每个实例重置全局变量(draw、board 和 turn)。

【讨论】:

    【解决方案2】:

    如果我正确理解了您的问题,那么有两个问题:

    1) 你需要找到一个可以放置胜负平局的地方。

    bool gameover()
    {
        for (int i = 0; i < 3; i++)//Check for a win
        {
            bool row_match    = (board[i][0] == board[i][1] && board[i][1] == board[i][2]);
            bool column_match = (board[0][i] == board[1][i] && board[1][i] == board[2][i]);
            bool diagonal_match = (board[0][2] == board[1][1] && board[1][1] == board[2][0]) || (board[0][0] == board[1][1] && board[1][1] == board[2][2]);
    
            if (row_match || column_match || diagonal_match)
            {
                bool x_wins = (row_match && board[i][0] == "X") || (column_match && board[0][i] == "X") || (diagonal_match && board[1][1] == "X")
    
                if (x_wins) xwins++;
                else owins++;
                return true;
            }
        }
    
        for (int i = 0; i < 3; i++)//Check for draw
        {
            for (int j = 0; j < 3; j++)
            {
                if (board[i][j] != 'X' && board[i][j] != 'O')
                {
                    return false;
                }
            }
        }
    
        ties++;
        draw = true;
        return true;
    }
    

    2) 你需要清除棋盘。

    do
    {
        display_board();
        player_turn();
    } while (!gameover);
    clearboard();
    
    void clearboard()
    {
        for (int row = 0; row < 3; row++)
        {
            for (int col = 0; col < 3; col++)
            {
                int val = 3 ** row + col + 1;
                board[row][col] = itoa (val);
            }
        }
    }
    

    如果不能解决您的问题或者我误解了您的问题,请告诉我。

    【讨论】:

    • 我不完全了解您的透明板功能是如何工作的。你能给我解释一下吗?
    • 嗨克里斯托弗,这里,清除板功能,只是将“X”和“0”替换为1-9的整数值,这是板的重置状态。我想,您想知道变量“val”如何为棋盘条目提供正确的整数值。我们可以将棋盘视为 3x3 矩阵。使用第 0-2 行和第 0-2 列。如果您手动计算板中条目的值,给定其行和列值,您会发现变量 val 被正确计算。如果不清楚,请告诉我。
    • itoa是什么意思?
    • itoa 是将整数转换为字符串。由于 board 变量存储字符并且计算的值是 int。所以 itoa 可以弥补差距。
    • 你知道它在哪个库吗?它在 iostream 中吗?
    【解决方案3】:

    试试这个

    do{
        cin >> choice;
    
        switch (choice)
        {
            case 1: row = 0; column = 0; break;
            case 2: row = 0; column = 1; break;
            case 3: row = 0; column = 2; break;
            case 4: row = 1; column = 0; break;
            case 5: row = 1; column = 1; break;
            case 6: row = 1; column = 2; break;
            case 7: row = 2; column = 0; break;
            case 8: row = 2; column = 1; break;
            case 9: row = 2; column = 2; break;
            default:
            cout << "You didn't enter a correct number! Try again\n";
        }
    }
    while(choice<1 || choice>9);
    

    因为如果用户输入任何其他值,它会再次调用player_turn();,以便它将工作recursively(在函数内部调用相同的函数)。

    当您检查while(!gameover()) 时,无需在循环内再次调用gameover();

    比赛结束后,您需要将数组的值设置为 1,2,3... 并调用函数display_board();

    【讨论】:

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