来自CGAffineTransform的UnsafePointer <CGAffineTransform>

我正在尝试在Swift中创建一个CGPath.我正在使用CGPathCreateWithRect(rect, transformPointer).

我如何UnsafePointer从一个CGAffineTransform？我试过这个:

    let transform : CGAffineTransform = CGAffineTransformIdentity
    let transformPointer : UnsafePointer = UnsafePointer(transform)

我也试过这个:

    let transform : CGAffineTransform = CGAffineTransformIdentity
    let transformPointer : UnsafePointer = &transform 

但斯威夫特抱怨说'&' with non-inout argument of type....我也试过&transform直接进入CGPathCreateWithRect但停止时出现同样的错误.

当我transform直接传入时,Swift"无法将类型'CGAffineTransform'的值转换为预期的参数类型'UnsafePointer'".

发生了什么,以及如何使用Swift 2.1？

我也试过直接传递和转换到CGPathCreateWithRect ...

你快到了.transform需要是一个变量 才能将其作为inout参数传递给&:

var transform = CGAffineTransformIdentity
let path = CGPathCreateWithRect(CGRect(...), &transform)

有关更多信息,请参阅" 使用Swift with Cocoa和Objective-C"文档中的" 与C API交互 ".

在Swift 3中,这将是

var transform = CGAffineTransform.identity
let path = CGPath(rect: rect, transform: &transform)

或者,对于身份变换,只是

let path = CGPath(rect: rect, transform: nil)