最新测试版中的Swift C void *和void **指针(4)

在以前的测试版中,我能够将指向任何东西的指针传递给C做类似的事情

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C函数如下：

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

现在,(肯定在Beta4中,甚至可能在Beta3中)从C转换为Swift的语法已经改变

之前,doSomething函数期待CMutablePointer(现在已从Swift中删除)
现在,它期望：

UnsafePointer<UnsafePointer<()>>

()是Void的类型

所以,语法上的说话更清楚.
但是,现在我无法理解如何获取任何对象的void **指针,就像我在之前的测试版中所做的那样.

有什么建议吗？

这对我有用：

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