我的T ic tac toe脚趾游戏理论上存在很大问题，但对我来说似乎都不错

Question

所以我进入了学习c ++的下一步，那是一个矩阵。我试图做一个简单的井字游戏，但我的游戏无法正确检查游戏是否结束。如果您在第一轮中将height = 2和width = 2表示您会赢...我不知道我可以把它搞砸了，在我看来，这一切都还不错...

   int map[3][3];
for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
        map[i][j] = 0;
    }
}

bool finished = false;
int player = 1;
while (!finished) {
    //attack
    cout << "player " << player << " it is your turn"<< endl;

    cout << "The map looks like this:" << endl;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            cout << map[i][j] << " ";
        }
        cout << endl;
    }
    bool correctMove;
    int height, width;
    do
    {
        correctMove = true;
        cout << "Where do you want to attack?" << endl;
        cout << "height = "; cin >> height;
        cout << "width = "; cin >> width;
        if (map[height][width] != 0 || width > 2 || height > 2) {
            correctMove = false;
        }
    } while (!correctMove);
    map[height][width] = player;
    //check finish game
    bool foundSequenceLine = true;
    for (int i = 0; i < 3; i++) {
        if (map[height][i] != player) {
            foundSequenceLine = false;
        }
    }
    bool foundSequenceColumn = true;
    for (int i = 0; i < 3; i++) {
        if (map[i][width] != player) {
            foundSequenceColumn = false;
        }
    }
    bool foundSequenceDiag1 = true;
    if (height == width) {
        for (int i = 0; i < 3; i++) {
            if (map[i][i] != player) {
                foundSequenceDiag1 = false;
            }
        }
    }
    bool foundSequenceDiag2 = true;
    if (height + width == 2) {
        for (int i = 0; i < 3; i++) {
            if (map[2-i][i] != player) {
                foundSequenceDiag2 = false;
            }
        }
    }
    if (foundSequenceColumn || foundSequenceLine || foundSequenceDiag1 || foundSequenceDiag2) {
        finished = true;
        cout << "Congrats player " << player << " you won!!!";
    }

    //change turn
    if (player == 1) {
        player++;
    }
    else {
        player--;
    }
}

}

Answer 1

代码进行假设，然后避免进行检查。

您的代码假设玩家已经赢了，除非您能完全证明他们没有赢。问题是您然后将证明该举动不是获胜举动的两个测试短路。

看看这段代码在做什么：

   bool foundSequenceDiag1 = true;
    if (height == width) {
        for (int i = 0; i < 3; i++) {
            if (map[i][i] != player) {
                foundSequenceDiag1 = false;
            }
        }
    }

首先，您说“玩家赢了” foundSequenceDiag1=true;。然后，您说：“是在对角线上移动吗？”，然后您才运行可以将foundSequenceDiag1设置为false的代码。

如果玩家进行的移动不在对角线上，则支票将不会运行。

修复：

    bool foundSequenceDiag1 = (height==width);  // true if the player played on diagonal
    if (foundSequenceDiag1) {  // loop code now only executes if player played on diagonal
        for (int i = 0; i < 3; i++) {
            if (map[i][i] != player) {
                foundSequenceDiag1 = false;
            }
        }
    }

[找到东西后，停止寻找。

如果我在写支票，一旦找到答案，我会使用break关键字来停止查找。

    for (int i = 0; i < 3; i++) {
        if (map[i][i] != player) {
            foundSequenceDiag1 = false;
            break; // can't be true now, so stop checking.
        }
    }