将 Swift 结构转换为 UnsafeMutablePointer<Void>

Question

有没有办法将 Swift 结构的地址转换为 void UnsafeMutablePointer？
我试过了但没有成功：

struct TheStruct {
    var a:Int = 0
}

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

谢谢！

编辑：上下文
我实际上正在尝试移植到 Swift Learning CoreAudio.
中的第一个示例这是我到目前为止所做的：

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优势，我将最终链接到 C 函数。

编辑：底线
如果您因为尝试在 Swift 中使用 CoreAudio 的 AudioQueue 而遇到这个问题...请不要。（阅读评论了解详情）

Answer 1

据我所知，最短的路是：

var myStruct = TheStruct()
var address = withUnsafeMutablePointer(&myStruct) {UnsafeMutablePointer<Void>([=10=])}

但是，你为什么需要这个？如果你想将它作为参数传递，你可以（并且应该）：

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

var myStruct = TheStruct()
foo(&myStruct)

Answer 2

随着 Swift 多年来的发展，大多数方法原型都发生了变化。这是 Swift 5:

的语法

var struct = TheStruct()

var unsafeMutablePtrToStruct = withUnsafeMutablePointer(to: &struct) {
    [=10=].withMemoryRebound(to: TheStruct.self, capacity: 1) {
        (unsafePointer: UnsafeMutablePointer<TheStruct>) in

        unsafePointer
    }
}

将 Swift 结构转换为 UnsafeMutablePointer<Void>

Cast a Swift struct to UnsafeMutablePointer<Void>

casting

unsafe-pointers

void-pointers

swift