在Swift中是否等效于CGPathApply？

Question

在预发布文档中，似乎没有Swift版本的CGPathApply。是否有同等的或替代的？我正在尝试获取CGPath的所有子路径，以便可以从其他起点重绘它。

Answer 1

斯威夫特3.0

在Swift 3.0中，您可以这样使用CGPath.apply：

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)")
}

斯威夫特2.2

通过添加@convention(c)，您现在可以CGPathApply直接从Swift进行调用。这是做必要魔术的包装器：

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，@convention(c)我的代码中没有提到，但是CGPathApply在Core Graphics模块的声明中使用了它。）

用法示例：

let path = UIBezierPath(roundedRect: CGRectMake(0, 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()")
    }
}