swift – 如何在OS X目标中创建CFDictionary？

根据 documentation CFDictionaryCreate用于在swift中实例化CFDictionary.

func CFDictionaryCreate(_ allocator: CFAllocator!,_ keys: UnsafeMutablePointer<UnsafePointer<Void>>,_ values: UnsafeMutablePointer<UnsafePointer<Void>>,_ numValues: CFIndex,_ keyCallBacks: UnsafePointer<CFDictionaryKeyCallBacks>,_ valueCallBacks: UnsafePointer<CFDictionaryValueCallBacks>) -> CFDictionary!

如何创建键和值参数？
到目前为止,我已经尝试使用swift的String类型,希望它会自动转换为适当的类型：

import Foundation

var keys : [String] = ["key1","key2"]
var values : [String] = ["value1","value2"]
var keyCallbacks = kcfTypeDictionaryKeyCallBacks
var valueCallbacks = kcfTypeDictionaryValueCallBacks
var dict : CFDictionary = CFDictionaryCreate(kcfAllocatorDefault,&keys,&values,2,&keyCallbacks,&valueCallbacks)

不幸的是我收到一个错误,说String不是键和值数组元素的正确类型：

main.swift:41:2: error: 'String' is not identical to 'UnsafePointer<Void>'
    &configKeys,&configValues,3,&valueCallbacks)

如何制作UnsafePointer< Void>来自String？

解决方法

对于Swift 3,您需要额外投射到CFDictionary.
否则,上下文类型’CFDictionary’不能与字典文字一起使用.

let options: CFDictionary = [kSecImportExportPassphrase as String : "certificateKey"] as CFDictionary

有关更多信息,请参见https://bugs.swift.org/browse/SR-2388