如何在OS X目标中创建CFDictionary？

Question

根据documentation的说法，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keys和values参数？到目前为止，我已经尝试使用swift的String类型，希望它能自动转换为合适的类型：

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keys和values数组元素的正确类型：

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

如何从String生成UnsafePointer<Void>

Answer 1

对于Swift 3，你需要额外的CFDictionary类型转换。否则上下文类型“”CFDictionary“”不能与字典文字一起使用，将引发。“”

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

有关详细信息，请参阅https://bugs.swift.org/browse/SR-2388。

Answer 2

rintaro的注释适用于字符串文字，但当我使用CFStringRef和Swift String时就不行了

// Error: Contextual type 'CFDictionary' cannot be used with dictionary literal
let options: CFDictionary = [kSecImportExportPassphrase : "certificateKey"]

我不得不将CFStringRef转换为Swift字符串：

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