将UnsafeMutablePointer<Void>指向自身

Question

在C中，将指针指向分配给指针的内存位置是有效的：

static void * SomeContext = &SomeContext;

我已经看到了在带有KVO的Objective代码库中使用这个指针作为context在observeValueForKeyPath:ofObject:change:context:中的用法。

是否可以将指针指向Swift中为该指针分配的内存空间，或者手动为指针分配内存？在初始化过程中试图将指针指向自身的内存值会导致错误(如预期的那样)：

var somePointer: UnsafeMutablePointer<Void> = &somePointer // Error: Variable used within its own initial value

CMutableVoidPointer不再存在于该语言中，因此不推荐初始化指针(UnsafeMutablePointer<Void>())，只需设置指向nil的指针。

也许是这样：

var ptr = malloc(sizeof(UnsafeMutablePointer<Void>))
ptr = &ptr // Error: error: cannot assign value of type 'inout UnsafeMutablePointer<Void>' (aka 'inout UnsafeMutablePointer<()>') to type 'UnsafeMutablePointer<Void>' (aka 'UnsafeMutablePointer<()>')

Answer 1

关于Swift 3：

var somePointer = UnsafeMutableRawPointer(bitPattern: 1)!
somePointer = UnsafeMutableRawPointer(mutating: &somePointer)

print(somePointer)
print(somePointer.load(as: UnsafeMutableRawPointer.self))
// Output of both should be the same.

Swift 2：

withUnsafePointer以获取指向变量的指针。然后把它分配给自己。

var somePointer: UnsafeMutablePointer<Void> = nil
withUnsafePointer(&somePointer) { somePointer = UnsafeMutablePointer($0) }