Question

我有一些代码可以检测点击并拖动动作的起点和终点，并将其保存到2个vector2点。然后我使用此代码转换：

public Rectangle toRect(Vector2 a, Vector2 b)
{
    return new Rectangle((int)a.X, (int)a.Y, (int)(b.X - a.X), (int)(b.Y - a.Y));
}

上面的代码不起作用，谷歌搜索，到目前为止尚未确定。有谁可以请我提供一些代码或公式来正确转换这个？
注意：vector2有x和y，矩形有x，y，宽度和高度。

任何帮助表示赞赏！感谢

Answer 1

我认为你需要在那里有额外的逻辑来决定哪个矢量用作左上角，哪个矢量用作右下角。

试试这个：

    public Rectangle toRect(Vector2 a, Vector2 b)
    {
        //we need to figure out the top left and bottom right coordinates
        //we need to account for the fact that a and b could be any two opposite points of a rectangle, not always coming into this method as topleft and bottomright already.
        int smallestX = (int)Math.Min(a.X, b.X); //Smallest X
        int smallestY = (int)Math.Min(a.Y, b.Y); //Smallest Y
        int largestX = (int)Math.Max(a.X, b.X);  //Largest X
        int largestY = (int)Math.Max(a.Y, b.Y);  //Largest Y

        //calc the width and height
        int width = largestX - smallestX;
        int height = largestY - smallestY;

        //assuming Y is small at the top of screen
        return new Rectangle(smallestX, smallestY, width, height);
    }

将2个vector2点转换为xna / monogame中的矩形

1 个答案: