java - 如何使用返回 2 个以上值的 boolean 方法

Question

此方法返回 true、false 和另一个值。

public boolean Intersection (Circle circle, Rectangle rectangle) {

... // test something
... return true;
... // test another thing
... return false;
... 
... return xCornerDistSq + yCornerDistSq <= maxCornerDistSq; //Third return value
}

这是一款 2D 游戏，球应该从矩形(包括矩形的边缘)弹起。我在上面链接的第三个返回值应该检测球和矩形边缘之间的碰撞。问题是，一旦调用该方法，我就不知道如何在代码中使用它。

我目前拥有的是这个

这是该方法的完整代码:

        public boolean Intersection (Circle circle, Rectangle rectangle) {
            double cx = Math.abs(circle.getLayoutX() - rectangle.getLayoutX() - rectangle.getWidth() / 2);
            double xDist = rectangle.getWidth() / 2 + circle.getRadius();

            if (cx > xDist) { return false; }

            double cy = Math.abs(circle.getLayoutY() - rectangle.getLayoutY() - rectangle.getHeight() / 2) ;
            double yDist = rectangle.getHeight() / 2 + circle.getRadius();

            if (cy > yDist) { return false; }

            if (cx <= rectangle.getWidth() / 2 || cy <= rectangle.getHeight() / 2) { return true; }

            double xCornerDist = cx - rectangle.getWidth() / 2;
            double yCornerDist = cy - rectangle.getHeight() / 2;
            double xCornerDistSq = xCornerDist * xCornerDist;
            double yCornerDistSq = yCornerDist * yCornerDist;
            double maxCornerDistSq = circle.getRadius() * circle.getRadius();
            return xCornerDistSq + yCornerDistSq <= maxCornerDistSq;
        }

那么，当我调用该函数时，如何实现它呢？我希望我的球也从边缘反弹，但我不知道如何使用此方法调用它。

我目前拥有的是:

                boolean intersection = Intersection(circle1, rect1);
                if (intersection == true) {
                    double x = (rect1.getLayoutX() + rect1.getWidth()  / 2) - (circle1.getLayoutX() + circle1.getRadius());
                    double y = (rect1.getLayoutY() + rect1.getHeight() / 2) - (circle1.getLayoutY() + circle1.getRadius());
                    if (Math.abs(x) > Math.abs(y)) {
                        c1SpeedX = -c1SpeedX;
                    } else {
                        c1SpeedY = -c1SpeedY;
                    }
                }
            }
        }

Answer 1

“假设的”枚举可能看起来像

public enum Ternary {
  TRUE, FALSE, MAYBE;
}