最新测试版中的Swift C void*和void**指针(4)

Question

在以前的betas中，我可以将指向任何对象的指针传递给C，执行以下操作

var aString : NSString = "This is a string" // create an NSString
var anUnmanaged = Unmanaged.passUnretained(aString) // take an unmanaged pointer
var opaque : COpaquePointer = anUnmanaged.toOpaque() // convert it to a COpaquePointer
var mut : CMutablePointer = &opaque // this is a CMutablePointer
doSomething(mut) // pass the pointer to an helper function written in C

使用像这样的C函数：

void doSomething(void ** ptr)
{
   // do something with the pointer, change it and make it point to another nsstring for example
}

现在，(当然在Beta4中，甚至在Beta3中)从C到Swift的语法已经发生了变化

以前，doSomething函数期望一个CMutablePointer (现在已从Swift中删除)，现在它期望一个：

UnsafePointer<UnsafePointer<()>>

()是一个要作废的类型别名

因此，从句法上讲更清晰。然而，现在我不能理解如何像在以前的betas中那样获取任何对象的空指针。

有什么建议吗？

Answer 1

这对我很有效：

var str = "Hello, playground"
var p: UnsafePointer<Void> = unsafeAddressOf(str)

Answer 2

不确定这是否是正确的答案，我希望有人有更好的方法来做这件事。主要的问题是在Swift对象和空UnsafePointer ( UnsafePointer<()> )之间没有直接的转换，但是我设法通过一个COpaquePointer实现了它

    // since I can't go directly from NSMutableString to UnsafePointer<()> (which is UnsafePointer<Void>) i pass through a COpaquePointer
    var aString : NSMutableString = NSMutableString(string: "test") // create an NSString
    var anUnmanaged = Unmanaged.passUnretained(aString) // take an unmanaged pointer
    var opaque : COpaquePointer = anUnmanaged.toOpaque() // convert it to a COpaquePointer
    var anUnsafe : UnsafePointer<()> = UnsafePointer(opaque) // convert it to an UnsafePointer to Void ( () is a Typealias to Void) using the init(_ other: COpaquePointer) initializer

    doSomething(&anotherUnsafe)

Answer 3

你能试试这个吗：

let strPtr = unsafeBitCast(aString, UnsafePointer<Void>.self)
let strPtrPtr = unsafeBitCast(strPtr, UnsafePointer<Void>.self) 

然后：

doSomething(strPtrPtr)