将Swift结构转换为UnsafeMutablePointer <Void>

Question

有没有办法将Swift结构的地址强制转换为void UnsafeMutablePointer？
我尝试了这个没有成功:

struct TheStruct {
    var a:Int = 0
}

var myStruct = TheStruct()
var address = UnsafeMutablePointer<Void>(&myStruct)

谢谢!

编辑:
我实际上尝试移植到Swift 的上下文是Learning CoreAudio中的第一个示例.
这是我到目前为止所做的事情:

func myAQInputCallback(inUserData:UnsafeMutablePointer<Void>,
    inQueue:AudioQueueRef,
    inBuffer:AudioQueueBufferRef,
    inStartTime:UnsafePointer<AudioTimeStamp>,
    inNumPackets:UInt32,
    inPacketDesc:UnsafePointer<AudioStreamPacketDescription>)
 { }

struct MyRecorder {
    var recordFile:     AudioFileID = AudioFileID()
    var recordPacket:   Int64       = 0
    var running:        Boolean     = 0
}

var queue:AudioQueueRef = AudioQueueRef()
AudioQueueNewInput(&asbd,
    myAQInputCallback,
    &recorder,  // <- this is where I *think* a void pointer is demanded
    nil,
    nil,
    UInt32(0),
    &queue)

我正在努力留在Swift,但如果事实证明这是一个问题而不是一个优势,我将最终链接到一个C函数.

编辑:bottome line
如果你来到这个问题,因为你试图在Swift中使用CoreAudio的AudioQueue ...不要.(阅读评论了解详情)

Answer 1

据我所知,最短的方法是:

var myStruct = TheStruct()
var address = withUnsafeMutablePointer(&myStruct) {UnsafeMutablePointer<Void>($0)}

但是,为什么你需要这个？如果你想把它作为参数传递,你可以(并且应该):

func foo(arg:UnsafeMutablePointer<Void>) {
    //...
}

var myStruct = TheStruct()
foo(&myStruct)