Question

那么如何从特定距离获得点的x，y坐标？

所以

public static Location2D DistanceToXY(Location2D current, Directions dir, int steps) {
            ushort x = current.X;
            ushort y = current.Y;

            for (int i = 0; i < steps; i++) {
                switch (dir) {
                    case Directions.North:
                        y--;
                        break;
                    case Directions.South:
                        y++;
                        break;
                    case Directions.East:
                        x++;
                        break;
                    case Directions.West:
                        x--;
                        break;
                    case Directions.NorthWest:
                        x--;
                        y--;
                        break;
                    case Directions.SouthWest:
                        x--;
                        y++;
                        break;
                    case Directions.NorthEast:
                        x++;
                        y--;
                        break;
                    case Directions.SouthEast:
                        x++;
                        x++;
                        break;
                }
            }
            return new Location2D(x, y);
        }

这里做的是对的吗？

Answer 1

我假设您正在寻找一般解决方案。正如其他人指出的那样，你需要一个方向作为缺失的输入。您将基本上使用方向和幅度（在这种情况下为10）并将这些极坐标转换为笛卡尔坐标。然后你将坐标加在一起X + Xoffset = Xnew，Y + Yoffset = Ynew。

转换的详细信息如下： http://www.mathsisfun.com/polar-cartesian-coordinates.html

编辑：发布代码后，答案是否定的。 NorthWest，SouthWest，NorthEast，SouthEast的案例不正确。在这些情况下，您移动1.41（aprox）像素。你不应该尝试逐步解决难题。使用极坐标数学并对总偏移求和，然后舍入到最接近的整数。

下面是解决方案的简化伪代码mod：

public static Location2D DistanceToXY(Location2D current, Directions dir, int steps) {
        ushort x = current.X;
        ushort y = current.Y;

        switch (dir) {
            case Directions.North:
                y=y+steps;
                break;
            case Directions.South:
                y=y-steps;
                break;
            case Directions.East:
                x=x+steps;
                break;
            case Directions.West:
                x=x-steps;
                break;
            case Directions.NorthWest:
                float sqrt2 = 2 ^ 0.5
                x=x+int((sqrt2 * steps) + 0.5);
                y=y-int((sqrt2 * steps) + 0.5);
                break;
            case Directions.SouthWest:
                x=x-int((sqrt2 * steps) + 0.5);
                y=y-int((sqrt2 * steps) + 0.5);
                break;
            case Directions.NorthEast:
                x=x+int((sqrt2 * steps) + 0.5);
                y=y+int((sqrt2 * steps) + 0.5);
                break;
            case Directions.SouthEast:
                x=x-int((sqrt2 * steps) + 0.5);
                y=y+int((sqrt2 * steps) + 0.5);
                break;
        }
        return new Location2D(x, y);
    }

如何从特定距离获取x，y坐标

1 个答案: