Question

有没有办法将Swift结构地址强制转换为void UnsafeMutablePointer？
我试了这个没有成功：

struct TheStruct {
    var a:Int = 0
}

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

谢谢！

编辑：上下文
我实际上是在尝试将 Learning CoreAudio 中的第一个示例移植到Swift 这就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，但如果这更像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：底线
如果你来到这个问题是因为你试图在Swift中使用CoreAudio的AudioQueue ......不要。（阅读评论以获取详细信息）

Answer 1

据我所知，最简短的方法是：

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)

Answer 2

随着多年来Swift的发展，大多数方法原型都已更改。这是 Swift 5 的语法：

    var struct = TheStruct()

    var unsafeMutablePtrToStruct = withUnsafeMutablePointer(to: &struct) {
        $0.withMemoryRebound(to: TheStruct.self, capacity: 1) {
            (unsafePointer: UnsafeMutablePointer<TheStruct>) in

            unsafePointer
        }
    }

将Swift结构转换为UnsafeMutablePointer <void> </void>

2 个答案: