Question

我只是想知道是否有更简单的方法：

for (int i = 0; i < 1; i++)
{
    for (int j = 0; i < 8; j+2)
    {
        board[ i, j ] = 2;
        board[( i + 1 ), j ] = 2;
        board[( i + 2 ), j ] = 2;
    }
}

我要做的是将棋子放在实际的棋盘上。所以这是将黑色部分放在顶部。

P.S。如果你也可以给我一些关于底部件（白色）的帮助。

Answer 1

除了修复循环外，您还可以明确地放置碎片，使其更具可读性

int[,] board = new[,]{{1,0,1,0,1,0,1,0},
                      {0,1,0,1,0,1,0,1},
                      {1,0,1,0,1,0,1,0},
                      {0,0,0,0,0,0,0,0},
                      {0,0,0,0,0,0,0,0},
                      {0,1,0,1,0,1,0,1},
                      {1,0,1,0,1,0,1,0},
                      {0,1,0,1,0,1,0,1}};

Answer 2

您错过了告诉我们您想要实现的目标。我在你的代码中看到了几个大的错误，所以我假设你没有完全掌握循环的工作原理。我希望我在这里解释它并不是太冒昧

For循环

For循环用于多次执行相同部分的代码。执行次数取决于您设置的条件。通常，您会以这种格式看到它：

for (int i = 0; i < n; i++)
{
    // Some code
}

这个for循环执行代码在括号内{和}）n次。这不仅是定义循环的方法。循环的更全面定义如下：

for (<initialization>; <condition>; <afterthought>)

初始化 - 您可以使用循环所需的一些变量。在第一次执行循环中的代码之前执行一次。这是可选的，您可以将其保留为空并使用之前在 condition 中声明的变量。
条件 - 这是在循环中每次执行代码之前执行的。如果条件表达式求值为true，则执行循环。执行循环并执行 afterterthought 后，将一次又一次地评估条件，直到它评估为false。条件也是可选的。如果你把它放在外面，在C＃循环中将再次执行，直到你以不同的方式打破循环。
事后补充 - 每次循环中的代码执行完毕时执行此操作。这通常用于增加循环所依赖的变量。这也是可选的。

修复代码

我假设您想在棋盘格中标记8x8矩阵中的字段，尽管您的问题中没有说明这一点。你可以这样做：

// For each row in a board (start with 0, go until 7, increase by 1)
for (int i = 0; i < 8; i++)
{
    // start coloring the row. Determine which field within the row needs
    // to be black. In first row, first field is black, in second second
    // field is black, in third row first field is black again and so on.
    // So, even rows have black field in first blace, odd rows have black
    // on second place.
    // We calculate this by determining division remained when dividing by 2.
    int firstBlack = i % 2;
    // Starting with calculated field, and until last field color fields black.
    // Skip every second field. (start with firstBlack, go until 7, increase by 2)
    for (int j = firstBlack; j < 8; j += 2)
    {
        // Color the field black (set to 2)
        board[i][j] = 2;
    }
}

您可以在线查看我的评论。

代码中的大错误

// This loop would be executed only once. It goes from 0 to less than 1 and is increased
// after first execution. You might as well done it without the loop.
for (int i = 0; i < 1; i++)
{
    // This doesn't make sense, because you use i in condition, and initialize
    // and update j.
    // Also, you are trying to update j, but you are not doing so. You are not
    // assigning anything to you. You should do j+=2 to increase by two. Or you
    // can do j = j + 2
    for (int j = 0; i < 8; j+2)
    {
        // I didn't really understand what you were trying to achieve here
        board[ i, j ] = 2;
        board[( i + 1 ), j ] = 2;
        board[( i + 2 ), j ] = 2;
    }
}

Answer 3

Modulo可以做到这一点，但我认为TonyS的anwser是我的第一反应，我宁愿把它放在下面显示的那个。

  char[,] board = new char[8,8];
    private void InitializeBoard()
    {
        string BoardRepresentation = "";
        //for every board cell
        for (int i = 0; i < 8; i++)
        {
            for (int j = 0; j < 8; j++)
            {
                //initialize board cell
                board[i, j] = '0';

                if (j <= 2)//black top
                {
                    //Modulo is the trick
                    if ((j - i) == 0 || ((j - i) % 2) == 0)
                    {
                        board[i, j] = 'B';
                    }
                }
                else if (j >= 5) //white bot
                {
                    if ((j - i) == 0 || ((j - i) % 2) == 0)
                    {
                        board[i, j] = 'W';
                    }
                }
            }
        }

        for (int j = 0; j < 8; j++)
        {
            for (int i = 0; i < 8; i++)
            {
                BoardRepresentation += board[i, j] + " ";
            }
            BoardRepresentation += Environment.NewLine;
        }
    }

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3 个答案: