Rounded Corners Rect in Cocos 2d-x with Bezier

Question

Is it possible to draw a Rect with rounded corners using a DrawNode object? I think that something is possible using Bezier curves, but I have made some tries and I think I can't handle it.

Looking at API I've found only these 2 functions:

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)

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[Modified after answer]

I have applied the answer in Cocos2dx, maybe somebody find this useful:

(just done some casting to int if you don't need high precision)

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);

This will produce: http://i.stack.imgur.com/mdEOM.png

Now you can change MagicConst to round the corners as you want.

For example with MagicConst = 0.9: http://i.stack.imgur.com/9V5cr.png

That is the result I wanted! ;) (thank you @Mbo)

(I can't post embedded image yet) :P

Answer 1

It is possible to calculate cubic Bezier curve that approximates a quarter of circle to make round corner.
Example for top left corner of axis-aligned rectangle (point TopLeft) and arc radius R:

Edit: Changed some -/+ signs

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

You can easily find similar symmetric coordinates for other corners. I did not consider case of extreme short rectangle edges (<2*R)