ios - Swift 中的 SecAccessControlCreateWithFlags()

Question

我正在尝试从 swift 代码调用 Security.framework 中的函数。暂时忘记“错误输出”(最后一个)参数，如果我这样调用函数:

let accessControlRef = SecAccessControlCreateFlags(
            kCFAllocatorDefault,
            kSecAttrAccessibleWhenUnlockedThisDeviceOnly,
            SecAccessControlCreateFlags.UserPresence,
            nil
        )

我收到以下错误:

Cannot find an initializer for type 'SecAccessControlCreateFlags' that accepts an argument list of type '(CFAllocator!, CFStringRef, SecAccessControlCreateFlags, nil)'

...但是，如果我将代码重新格式化为以下内容:

let allocator:CFAllocatorRef! = kCFAllocatorDefault

let protection:AnyObject!     = kSecAttrAccessibleWhenUnlockedThisDeviceOnly

let flags:SecAccessControlCreateFlags = SecAccessControlCreateFlags.UserPresence


let accessControlRef = SecAccessControlCreateWithFlags(
    allocator,
    protection,
    flags,
    nil
)

(特定类型 - 例如，CFAllocatorRef - 取自 Xcode 自动完成的函数原型(prototype))...它编译没有问题。 这是怎么回事？

接下来，错误参数。我应该通过什么？迁移我的 Objective-C 代码时，我很想传递以下变量(当然，前缀是 &):

var accessControlError:CFErrorRef! = nil

...这给出了错误:

Cannot invoke 'SecAccessControlCreateWithFlags' with an argument list of type '(CFAllocatorRef!, AnyObject!, SecAccessControlCreateFlags, inout CFErrorRef!)'

相反，如果我传递以下变量(同样，前面加上地址运算符):

var accessControlError:UnsafeMutablePointer<Unmanaged<CFError>?>

(与原型(prototype)自动完成建议的类型相同)，我得到:

Cannot invoke 'SecAccessControlCreateWithFlags' with an argument list of type '(CFAllocatorRef!, AnyObject!, SecAccessControlCreateFlags, inout UnsafeMutablePointer?>)'

...那么，是什么给了？

编辑:忘记错误参数。我似乎两次使用地址(即指向指针的指针)。相反，我应该这样做:

var accessControlError:UnsafeMutablePointer<Unmanaged<CFError>?> = nil
// ^ Already a 'pointer'

let allocator:CFAllocatorRef!         = kCFAllocatorDefault
let protection:AnyObject!             = kSecAttrAccessibleWhenUnlockedThisDeviceOnly
let flags:SecAccessControlCreateFlags = SecAccessControlCreateFlags.UserPresence 

let accessControlRef = SecAccessControlCreateWithFlags(
        allocator,
        protection,
        flags,
        accessControlError // <- Notice the lack of '&'
)

来源:this answer 中的示例代码.

Answer 1

(好的，所以一段时间内没有人添加任何新见解，所以我将用我上次编辑的内容回答我自己的问题:)

回答:忘记错误参数。我似乎两次使用地址(即指向指针的指针)。相反，我应该这样做:

let accessControlError:UnsafeMutablePointer<Unmanaged<CFError>?> = nil
// ^ Already a 'pointer'

let allocator:CFAllocator!         = kCFAllocatorDefault
let protection:AnyObject!             = kSecAttrAccessibleWhenUnlockedThisDeviceOnly
let flags:SecAccessControlCreateFlags = SecAccessControlCreateFlags.userPresence 

let accessControlRef = SecAccessControlCreateWithFlags(
        allocator,
        protection,
        flags,
        accessControlError // <- Notice the lack of '&'
)