c++ - 2 차 베 지어를 3 차 베지 어로 변환

http://fontforge.sourceforge.net/bezier.html에서 :

와이즈 비즈

참고로

Any quadratic spline can be expressed as a cubic (where the cubic term is zero). The end points of the cubic will be the same as the quadratic's.

CP₀= QP₀
CP₃= QP₂

The two control points for the cubic are:

CP₁= QP₀+ 2/3 *(QP₁-QP₀)
CP₂= QP₂+ 2/3 *(QP₁-QP₂)

...There is a slight error introduced due to rounding, but it is unlikely to be noticeable.

를 구현했습니다.  위의 Owen의 답변을 기반으로 한 NSBezierPath (macOS Swift 4)의 경우

addQuadCurve

extension NSBezierPath { public func addQuadCurve(to qp2: CGPoint, controlPoint qp1: CGPoint) { let qp0 = self.currentPoint self.curve(to: qp2, controlPoint1: qp0 + (2.0/3.0)*(qp1 - qp0), controlPoint2: qp2 + (2.0/3.0)*(qp1 - qp2)) } } extension CGPoint { // Vector math public static func +(left: CGPoint, right: CGPoint) -> CGPoint { return CGPoint(x: left.x + right.x, y: left.y + right.y) } public static func -(left: CGPoint, right: CGPoint) -> CGPoint { return CGPoint(x: left.x - right.x, y: left.y - right.y) } public static func *(left: CGFloat, right: CGPoint) -> CGPoint { return CGPoint(x: left * right.x, y: left * right.y) } }