swift - 将 nil 作为 UnsafePointer 传递

Question

在 Swift 2.2 中，我能够传入 nil作为需要 UnsafePointer<UInt8> 的函数的有效参数.在 Swift 3 中，我不能再这样做了:

func myFuncThatTakesAPointer(buffer: UnsafePointer<UInt8>, length: Int) { /** **/ }

myFuncThatTakesAPointer(buffer: nil, length: 0)

Playground execution failed: error: Xcode8Playground-iOS.playground:62:33: error: nil is not compatible with expected argument type 'UnsafePointer<UInt8>'
myFuncThatTakesAPointer(buffer: nil, length: 0)
                                ^

我现在是否需要在我的函数中将指针声明指定为可选的？

Answer 1

Am I required to specify the pointer declaration in my function as optional now?

一句话，是的。来自release notes :

The types UnsafePointer, UnsafeMutablePointer, AutoreleasingUnsafeMutablePointer, OpaquePointer, Selector, and NSZone now represent non-nullable pointers—that is, pointers that are never nil. A nullable pointer is now represented using Optional, for example, UnsafePointer<Int>?.