rotation - 如何围绕其自身边界的中心旋转 UIBezierPath？

Question

假设我们有一个 UIBezierPath...它的边界是完全正方形的...像这样:

func getExponentPath(rotate180: Bool) -> UIBezierPath {

    // establish unit of measure (grid) based on this containing view's bounds... (not to be confused with this bezierpath's bounds)

    let G = bounds.width / 5

    let exponentPath = UIBezierPath()

    let sstartPoint = CGPoint(x:(3.8)*G,y:(1.2)*G)
    exponentPath.move(to: sstartPoint)
    exponentPath.addLine(to: CGPoint(x:(5)*G,y:(1.2)*G))
    exponentPath.addLine(to: CGPoint(x:(4.4)*G,y:(0.2)*G))
    exponentPath.addLine(to: CGPoint(x:(5)*G,y:(0.2)*G))
    exponentPath.addLine(to: CGPoint(x:(5)*G,y:(0)*G))
    exponentPath.addLine(to: CGPoint(x:(3.8)*G,y:(0)*G))
    exponentPath.addLine(to: CGPoint(x:(3.8)*G,y:(0.2)*G))
    exponentPath.addLine(to: CGPoint(x:(4.4)*G,y:(0.2)*G))
    exponentPath.addLine(to: sstartPoint)

    exponentPath.close()

    // this does not work:
    // if rotate180 { exponentPath.apply(CGAffineTransform(rotationAngle: CGFloat.pi)) }

    return exponentPath

}

如果旋转，此 bezierpath 仍需要在其包含 View 中占据完全相同的区域。

我只能假设这是行不通的，因为旋转中心存在一些问题，不是我想要的……尽管即使说“旋转 0”，我也会得到相同(错误)的结果。

那么路径如何围绕它自己的中心点旋转呢？

似乎应该有一个简单的线性代数矩阵乘法类型可以应用于点集。 =T

Answer 1

extension UIBezierPath
{
    func rotateAroundCenter(angle: CGFloat)
    {
        let center = self.bounds.getCenter()
        var transform = CGAffineTransform.identity
        transform = transform.translatedBy(x: center.x, y: center.y)
        transform = transform.rotated(by: angle)
        transform = transform.translatedBy(x: -center.x, y: -center.y)
        self.apply(transform)
    }
}