• cocos2dx Geometry Size和Rect


    Size

    代码都是基础代码不注释了,写一些特别的
    1、赋值时可以接收Size和Vec2类型的值,保证的类型的兼容性
    2、对运算符进行了重载,可以按照正常的数学逻辑运算
    3.、可以使用equals对比大小

    Rect

    1、Rect(x, y, width, height) 四个参数分别表示起点的 xy坐标 和 宽高
    2、可以直接接受Vec2和Size作为参数,如 Rect(Vec2,Size)

    //是否包含在矩形内
    bool Rect::containsPoint(const Vec2& point) const
    {
        return (point.x >= getMinX() &&
                point.x <= getMaxX() &&
                point.y >= getMinY() &&
                point.y <= getMaxY());
    }
    
    //判断是否相交
    bool Rect::intersectsRect(const Rect& rect) const
    {
        return !(     getMaxX() < rect.getMinX() ||
                 rect.getMaxX() <      getMinX() ||
                      getMaxY() < rect.getMinY() ||
                 rect.getMaxY() <      getMinY());
    }
    
    //与圆是否相交
    bool Rect::intersectsCircle(const Vec2& center, float radius) const
    {
        //矩形中心
        Vec2 rectangleCenter((origin.x + size.width / 2),
                             (origin.y + size.height / 2));
        
        //矩形宽高的一半
        float w = size.width / 2;
        float h = size.height / 2;
        
        //圆心和矩形中心的距离
        float dx = std::abs(center.x - rectangleCenter.x);
        float dy = std::abs(center.y - rectangleCenter.y);
        
        //保证圆和矩形相交
        //短板效应
        //如果两个中点距离大于圆半径+宽或高的一半就返回false 因为如果这样圆和矩形不会相交 ?
        if (dx > (radius + w) || dy > (radius + h))
        {
            return false;
        }
        
        //获取两心的最小距离
        Vec2 circleDistance(std::abs(center.x - origin.x - w),
                            std::abs(center.y - origin.y - h));
        
        //圆在矩形内
        if (circleDistance.x <= (w))
        {
            return true;
        }
        
        if (circleDistance.y <= (h))
        {
            return true;
        }
        
        //矩形在圆内
        float cornerDistanceSq = powf(circleDistance.x - w, 2) + powf(circleDistance.y - h, 2);
        
        return (cornerDistanceSq <= (powf(radius, 2)));
    }
    
    //找到可以容纳两种图形的最小矩形,并将当前rect设置为计算得到的矩形
    void Rect::merge(const Rect& rect)
    {
        float minX = std::min(getMinX(), rect.getMinX());
        float minY = std::min(getMinY(), rect.getMinY());
        float maxX = std::max(getMaxX(), rect.getMaxX());
        float maxY = std::max(getMaxY(), rect.getMaxY());
        setRect(minX, minY, maxX - minX, maxY - minY);
    }
    
    //计算可以容纳两个矩形的最小矩形并返回
    Rect Rect::unionWithRect(const Rect & rect) const
    {
        //获取四个顶点坐标
        float thisLeftX = origin.x;
        float thisRightX = origin.x + size.width;
        float thisTopY = origin.y + size.height;
        float thisBottomY = origin.y;
        
        //交换大小
        if (thisRightX < thisLeftX)
        {
            std::swap(thisRightX, thisLeftX);   // This rect has negative width
        }
        
        if (thisTopY < thisBottomY)
        {
            std::swap(thisTopY, thisBottomY);   // This rect has negative height
        }
        
        //获取目标四个顶点坐标
        float otherLeftX = rect.origin.x;
        float otherRightX = rect.origin.x + rect.size.width;
        float otherTopY = rect.origin.y + rect.size.height;
        float otherBottomY = rect.origin.y;
        
        //交换大小
        if (otherRightX < otherLeftX)
        {
            std::swap(otherRightX, otherLeftX);   // Other rect has negative width
        }
        
        if (otherTopY < otherBottomY)
        {
            std::swap(otherTopY, otherBottomY);   // Other rect has negative height
        }
        
        //找到可以容纳两个矩形的最小矩形,并返回这个最小矩形
        float combinedLeftX = std::min(thisLeftX, otherLeftX);
        float combinedRightX = std::max(thisRightX, otherRightX);
        float combinedTopY = std::max(thisTopY, otherTopY);
        float combinedBottomY = std::min(thisBottomY, otherBottomY);
        
        return Rect(combinedLeftX, combinedBottomY, combinedRightX - combinedLeftX, combinedTopY - combinedBottomY);
    }
    
    
    const Rect Rect::ZERO = Rect(0, 0, 0, 0);
    
    NS_CC_END
    
    

    主要用来判断是否相交、包含,可能会用在碰撞检测中。

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  • 原文地址:https://www.cnblogs.com/sakuraneo/p/11992051.html
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