Question

有没有办法知道libgdx中两个Rectangle之间的交叉矩形区域，比如c＃http://msdn.microsoft.com/en-us/library/microsoft.xna.framework.rectangle.intersect.aspx中的Rectangle？

我需要获得两个矩形之间的交叉矩形区域，但libgdx中的overlap方法只返回两个矩形相交的布尔值。我已阅读过Intersector类，但它没有提供任何功能。

Answer 1

实际上，LibGDX没有内置此功能，所以我会做这样的事情：

/** Determines whether the supplied rectangles intersect and, if they do,
 *  sets the supplied {@code intersection} rectangle to the area of overlap.
 * 
 * @return whether the rectangles intersect
 */
static public boolean intersect(Rectangle rectangle1, Rectangle rectangle2, Rectangle intersection) {
    if (rectangle1.overlaps(rectangle2)) {
        intersection.x = Math.max(rectangle1.x, rectangle2.x);
        intersection.width = Math.min(rectangle1.x + rectangle1.width, rectangle2.x + rectangle2.width) - intersection.x;
        intersection.y = Math.max(rectangle1.y, rectangle2.y);
        intersection.height = Math.min(rectangle1.y + rectangle1.height, rectangle2.y + rectangle2.height) - intersection.y;
        return true;
    }
    return false;
}

Answer 2

您可以使用Intersector类。

import com.badlogic.gdx.math.Intersector;

Intersector.intersectRectangles(rectangle1, rectangle2, intersection);

Answer 3

我想在nEx软件的答案中添加一些细微的变化。即使您想将结果值存储在其中一个源矩形中，这个也会起作用：

public static boolean intersect(Rectangle r1, Rectangle r2, Rectangle intersection) {
    if (!r1.overlaps(r2)) {
        return false;
    }

    float x = Math.max(r1.x, r2.x);
    float y = Math.max(r1.y, r2.y);
    float width = Math.min(r1.x + r1.width, r2.x + r2.width) - x;
    float height = Math.min(r1.y + r1.height, r2.y + r2.height) - y;
    intersection.set(x, y, width, height);

    return true;
}

以下是usege的示例：

Rectangle r1 = new Rectangle();
Rectangle r2 = new Rectangle();

// ...

intersect(r1, r2, r1);

Libgdx - 从Rectangle.overlap（Rectangle）获取交叉矩形

3 个答案: