Swift中的CGPathApply等价或替代？

在 pre-release documentation,似乎没有Swift版本的CGPathApply.有等同还是替代？我试图获取CGPath的所有子路径,以便我可以从不同的起始点重新绘制.

Swift 3.0

在Swift 3.0中,您可以使用CGPath.apply,如下所示：

let path: CGPath = ...
// or let path: CGMutablePath

path.apply(info: nil) { (_,elementPointer) in
    let element = elementPointer.pointee
    let command: String
    let pointCount: Int
    switch element.type {
    case .movetoPoint: command = "moveto"; pointCount = 1
    case .addLinetoPoint: command = "lineto"; pointCount = 1
    case .addQuadCurvetoPoint: command = "quadCurveto"; pointCount = 2
    case .addCurvetoPoint: command = "curveto"; pointCount = 3
    case .closeSubpath: command = "close"; pointCount = 0
    }
    let points = Array(UnsafeBufferPointer(start: element.points,count: pointCount))
    Swift.print("\(command) \(points)")
}

Swift 2.2

通过添加@convention(c),您现在可以直接从Swift调用CGPathApply.这是一个需要魔法的包装：

extension CGPath {
    func forEach(@noescape body: @convention(block) (CGpathelement) -> Void) {
        typealias Body = @convention(block) (CGpathelement) -> Void
        func callback(info: UnsafeMutablePointer<Void>,element: UnsafePointer<CGpathelement>) {
            let body = unsafeBitCast(info,Body.self)
            body(element.memory)
        }
        print(sizeofValue(body))
        let unsafeBody = unsafeBitCast(body,UnsafeMutablePointer<Void>.self)
        CGPathApply(self,unsafeBody,callback)
    }
}

(请注意,我的代码中没有提及@convention(c),而是在Core Graphics模块的CGPathApply声明中使用.)

使用示例

let path = UIBezierPath(roundedRect: CGRectMake(0,200,100),cornerRadius: 15)
path.CGPath.forEach { element in
    switch (element.type) {
    case CGpathelementType.MovetoPoint:
        print("move(\(element.points[0]))")
    case .AddLinetoPoint:
        print("line(\(element.points[0]))")
    case .AddQuadCurvetoPoint:
        print("quadCurve(\(element.points[0]),\(element.points[1]))")
    case .AddCurvetoPoint:
        print("curve(\(element.points[0]),\(element.points[1]),\(element.points[2]))")
    case .CloseSubpath:
        print("close()")
    }
}