请问如何在Swift中将self转换为UnsafeMutablePointer <Void>类型

Question

如何在Swift中将self转换为UnsafeMutablePointer <Void>类型

在调用以下代码时，尝试将“self”传递给swift中的C函数：

var callbackStruct : AURenderCallbackStruct = 
    AURenderCallbackStruct.init(
      inputProc: recordingCallback,
      inputProcRefCon: UnsafeMutablePointer<Void>
    )

在这里将“self”转换为UnsafeMutablePointer类型的理想方法是什么？

Answer 1

一个对象的指针（即，实例引用类型）可以被转换为一个UnsafePointer<Void>（的夫特映射const void *，UnsafeRawPointer在夫特3）和背部。在Objective-C中你会写void&nbsp;*voidPtr&nbsp;=&nbsp;(__bridge&nbsp;void*)self;//&nbsp;MyType&nbsp;*mySelf&nbsp;=&nbsp;(__bridge&nbsp;MyType&nbsp;*)voidPtr;（有关这些演员阵容的确切含义，请参阅Clang ARC文档中的3.2.4桥接演员表。）Swift有一种Unmanaged用于此目的的类型。使用起来有点麻烦，因为它可以COpaquePointer&nbsp;代替使用UnsafePointer<Void>。这里有两个辅助方法（以Objective-C&nbsp;__bridge强制转换命名）：func&nbsp;bridge<T&nbsp;:&nbsp;AnyObject>(obj&nbsp;:&nbsp;T)&nbsp;->&nbsp;UnsafePointer<Void>&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;UnsafePointer(Unmanaged.passUnretained(obj).toOpaque()) &nbsp;&nbsp;&nbsp;&nbsp;//&nbsp;return&nbsp;unsafeAddressOf(obj)&nbsp;//&nbsp;***}func&nbsp;bridge<T&nbsp;:&nbsp;AnyObject>(ptr&nbsp;:&nbsp;UnsafePointer<Void>)&nbsp;->&nbsp;T&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;Unmanaged<T>.fromOpaque(COpaquePointer(ptr)).takeUnretainedValue() &nbsp;&nbsp;&nbsp;&nbsp;//&nbsp;return&nbsp;unsafeBitCast(ptr,&nbsp;T.self)&nbsp;//&nbsp;***}“复杂”表达式只是满足Swifts严格类型系统所必需的。在编译的代码中，这只是指针之间的转换。（***如果您愿意使用“不安全”方法，可以在注释中指示更短，但编译后的代码是相同的。）使用此辅助方法，您可以将其传递self给C函数&nbsp;let&nbsp;voidPtr&nbsp;=&nbsp;bridge(self)（或者UnsafeMutablePointer<Void>(bridge(self))如果C函数需要一个可变指针），并将其转换回对象指针 - 例如在回调函数中 -&nbsp;let&nbsp;mySelf&nbsp;:&nbsp;MyType&nbsp;=&nbsp;bridge(voidPtr)不会发生所有权转移，因此self&nbsp;只要使用了void指针，就必须确保存在。并且为了完整起见，Swift等同于Objective-C&nbsp;__bridge_retained和__bridge_transferfunc&nbsp;bridgeRetained<T&nbsp;:&nbsp;AnyObject>(obj&nbsp;:&nbsp;T)&nbsp;->&nbsp;UnsafePointer<Void>&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;UnsafePointer(Unmanaged.passRetained(obj).toOpaque())}func&nbsp;bridgeTransfer<T&nbsp;:&nbsp;AnyObject>(ptr&nbsp;:&nbsp;UnsafePointer<Void>)&nbsp;->&nbsp;T&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;Unmanaged<T>.fromOpaque(COpaquePointer(ptr)).takeRetainedValue()}bridgeRetained()将对象指针强制转换为void指针并保留该对象。bridgeTransfer()将void指针转换回对象指针并使用retain。优点是在调用之间不能释放对象，因为保持了强引用。缺点是调用必须适当平衡，并且很容易导致保留周期。更新Swift 3（Xcode 8）：func&nbsp;bridge<T&nbsp;:&nbsp;AnyObject>(obj&nbsp;:&nbsp;T)&nbsp;->&nbsp;UnsafeRawPointer&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;UnsafeRawPointer(Unmanaged.passUnretained(obj).toOpaque())}func&nbsp;bridge<T&nbsp;:&nbsp;AnyObject>(ptr&nbsp;:&nbsp;UnsafeRawPointer)&nbsp;->&nbsp;T&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;Unmanaged<T>.fromOpaque(ptr).takeUnretainedValue()}func&nbsp;bridgeRetained<T&nbsp;:&nbsp;AnyObject>(obj&nbsp;:&nbsp;T)&nbsp;->&nbsp;UnsafeRawPointer&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;UnsafeRawPointer(Unmanaged.passRetained(obj).toOpaque())}func&nbsp;bridgeTransfer<T&nbsp;:&nbsp;AnyObject>(ptr&nbsp;:&nbsp;UnsafeRawPointer)&nbsp;->&nbsp;T&nbsp;{ &nbsp;&nbsp;&nbsp;&nbsp;return&nbsp;Unmanaged<T>.fromOpaque(ptr).takeRetainedValue()}对“不安全指针”的相关更改在中描述SE-0017更改Unmanaged以使用UnsafePointerSE-0107 UnsafeRawPointer API

Answer 2

在我看来，这是什么withUnsafeMutablePointer- 将任意Swift指针转换为C指针。所以大概你可以做到这一点（我没试过，但我测试的代码安全地工作）：var&nbsp;mself&nbsp;=&nbsp;self&nbsp;withUnsafeMutablePointer(&mself)&nbsp;{&nbsp;v&nbsp;in &nbsp;&nbsp;&nbsp;&nbsp;let&nbsp;v2&nbsp;=&nbsp;UnsafeMutablePointer<Void>(v) &nbsp;&nbsp;&nbsp;&nbsp;myStruct.inputProcRefCon&nbsp;=&nbsp;v2}

Answer 3

func bridge<T : AnyObject>(obj : T) -> UnsafePointer<Void> {return UnsafePointer(Unmanaged.passUnretained(obj).toOpaque())}func bridge<T : AnyObject>(ptr : UnsafePointer<Void>) -> T {return Unmanaged<T>.fromOpaque(ptr).takeUnretainedValue()}func bridgeRetained<T : AnyObject>(obj : T) -> UnsafePointer<Void> {return UnsafePointer( Unmanaged.passRetained(obj).toOpaque())}func bridgeTransfer<T : AnyObject>(ptr : UnsafePointer<Void>) -> T {return Unmanaged<T>.fromOpaque(ptr).takeRetainedValue()}