如何从两个的交点制作矩形? [英] How to make rect from the intersection of two?

问题描述

我正在研究一个突破克隆，我一直在试图弄清楚如何获得两个碰撞矩形的交集矩形，以便我可以测量球在 x 轴和 y 轴上进入块的深度并决定哪个我将反转的速度的分量.

I'm working on a breakout clone and I've been trying to figure out how to get the intersection rect of two colliding rects so I can measure how deep the ball entered the block in both x and y axis and decide which component of the velocity I'll reverse.

我想我可以像这样计算每种情况的深度:

I figured I could calculate the depth for each case like this:

但是如果我有交叉点，那么我就不必担心球是从左/右或顶部/底部击中挡块(因为我只会分别反转 x 和 y 轴)，从而节省了我很多打字时间.

But if I had the intersection rect than I woudn't have to worry if the ball hits the block from the left/right or top/bottom (since I would be only reversing the x and y axis respectively), thus saving me a lot of typing.

我查看了 Pygame 的文档 但似乎没有有一个功能.我将如何解决这个问题?

I've looked on Pygame's docs but seems it doesn't have a function for that. How would I go about solving this problem?

推荐答案

假设您有矩形 r1 和 r2，带有 .left, .right, .顶部和底部边缘，然后

Assuming you have rectangles r1 and r2, with .left, .right, .top, and .bottom edges, then

left = max(r1.left, r2.left);
right = min(r1.right, r2.right);
top = max(r1.top, r2.top);
bottom = min(r1.bottom, r2.bottom);

(通常的惯例是坐标从上到下从左到右增加).最后，检查 left 和 top，并计算面积:

(with the usual convention that coordinates increase top to bottom and left to right). Finally, check that left<right and top<bottom, and compute the area:

Area = (right - left) * (top - bottom);

或者，您可以使用 clip() 函数.从您在问题中链接的文档:

Alternatively, you can use the clip() function. From the docs you linked in your question:

clip(Rect) -> Rect 返回一个裁剪成的新矩形完全在参数 Rect 内.如果两个矩形不重叠开始，返回大小为 0 的 Rect.

clip(Rect) -> Rect Returns a new rectangle that is cropped to be completely inside the argument Rect. If the two rectangles do not overlap to begin with, a Rect with 0 size is returned.

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