带贝塞尔曲线的 Cocos 2d-x 中的圆角矩形

Question

是否可以使用DrawNode对象绘制圆角的Rect？我认为使用贝塞尔曲线是可能的，但我已经尝试过但我认为我无法处理它。

查看 API 我只发现了这两个函数：

drawQuadBezier (const Vec2 &origin, const Vec2 &control, const Vec2 &destination, unsigned int segments, const Color4F &color)

drawCubicBezier (const Vec2 &origin, const Vec2 &control1, const Vec2 &control2, const Vec2 &destination, unsigned int segments, const Color4F &color)

[回答后修改]

我在Cocos2dx中应用了答案，也许有人觉得这个有用：

（如果不需要高精度，只是做了一些转换为 int）

auto MagicConst = 0.552;
auto position = 150;
auto R = 50;

Vec2 TopLeft = Vec2(position, position + R * 2);
Vec2 TopRight = Vec2(position + R * 2, position + R * 2);
Vec2 BottomRight = Vec2(position + R * 2, position);
Vec2 BottomLeft = Vec2(position, position);

Vec2 originTL = Vec2(TopLeft.x, TopLeft.y - R);
Vec2 originTR = Vec2(TopRight.x - R, TopRight.y);
Vec2 originBR = Vec2(BottomRight.x - R, BottomRight.y);
Vec2 originBL = Vec2(BottomLeft.x, BottomLeft.y + R);

Vec2 control1TL = Vec2(TopLeft.x, (int) (TopLeft.y - R * (1 - MagicConst)));
Vec2 control1TR = Vec2((int) (TopRight.x - R * (1 - MagicConst)), TopRight.y);
Vec2 control1BR = Vec2((int) (BottomRight.x - R * (1 - MagicConst)), BottomRight.y);
Vec2 control1BL = Vec2(BottomLeft.x, (int) (BottomLeft.y + R * (1 - MagicConst)));

Vec2 control2TL = Vec2((int) (TopLeft.x + R * (1 - MagicConst)), TopLeft.y);
Vec2 control2TR = Vec2(TopRight.x, (int) (TopRight.y - R * (1 - MagicConst)));
Vec2 control2BR = Vec2(BottomRight.x, (int) (BottomRight.y + R * (1 - MagicConst)));
Vec2 control2BL = Vec2((int) (BottomLeft.x + R * (1 - MagicConst)), BottomLeft.y);

Vec2 destinationTL = Vec2(TopLeft.x + R, TopLeft.y);
Vec2 destinationTR = Vec2(TopRight.x, TopRight.y - R);
Vec2 destinationBR = Vec2(BottomRight.x, BottomRight.y + R);
Vec2 destinationBL = Vec2(BottomLeft.x + R, BottomLeft.y);

auto roundCorner = DrawNode::create();
roundCorner->drawCubicBezier(originTL, control1TL, control2TL, destinationTL, 10, Color4F::RED);
roundCorner->drawCubicBezier(originTR, control1TR, control2TR, destinationTR, 10, Color4F::GREEN);
roundCorner->drawCubicBezier(originBR, control1BR, control2BR, destinationBR, 10, Color4F::YELLOW);
roundCorner->drawCubicBezier(originBL, control1BL, control2BL, destinationBL, 10, Color4F::WHITE);
addChild(roundCorner);

这将产生：http://i.stack.imgur.com/mdEOM.png

现在您可以根据需要更改 MagicConst 圆角。

例如MagicConst = 0.9：http://i.stack.imgur.com/9V5cr.png

这就是我想要的结果！ ;) （谢谢@Mbo）

（我还不能post嵌入图片）:P

Answer 1

可以计算出近似四分之一圆的三次贝塞尔曲线来制作圆角。
轴对齐矩形左上角（点 TopLeft）和圆弧半径 R 的示例：

编辑： 更改了一些 -/+ 符号

MagicConst = 0.552
Bezier.origin.X = ToplLeft.X
Bezier.origin.Y = ToplLeft.Y + R
Bezier.control1.X = ToplLeft.X
Bezier.control1.Y = ToplLeft.Y + R * (1-MagicConst)
Bezier.control2.X = ToplLeft.X + R * (1-MagicConst)
Bezier.control2.Y = ToplLeft.Y
Bezier.destination.X = ToplLeft.X + R
Bezier.destination.Y = ToplLeft.Y

about MagicConstant

您可以轻松找到其他角的相似对称坐标。我没有考虑极短矩形边的情况 (<2*R)

带贝塞尔曲线的 Cocos 2d-x 中的圆角矩形

Rounded Corners Rect in Cocos 2d-x with Bezier

c++

geometry

cocos2d-x

bezier