将2个vector2点转换为xna/monogame中的矩形 [英] Convert 2 vector2 points to a rectangle in xna/monogame

问题描述

我有一些代码可以检测单击并拖动动作的起点和终点，并将其保存到2个vector2点.然后，我使用以下代码进行转换:

I have some code which will detect the start and end point of a click-and-drag action, and will save it to 2 vector2 points. I then use this code to convert:

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，无法使用Google搜索，到目前为止，尚无定论. 任何人都可以向我提供一些代码或公式来正确转换此代码吗?
注意:vector2具有一个x和y，一个矩形具有x，一个y，一个宽度和一个高度.

The code above does not work and googling, so far has come up inconclusive. Could anyone please provide me with some code or a formula to properly convert this?
Note: a vector2 has an x and a y, and a rectangle has an x, a y, a width, and a height.

感谢您的帮助！谢谢

推荐答案

我认为您需要在其中具有附加逻辑，以便确定将哪个向量用作左上角，并将哪个向量用作右下角.

I think you need to have additional logic in there to decide which vector to use as the top left and which to use as the bottom right.

尝试一下:

    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);
    }

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