T. *_*sen 5 nsdictionary toll-free-bridging swift cfdictionary swift2
好吧,这是我在使用CGImageSource时遇到的情况,并注意到CFDictionary和NSDictionary之间的免费桥接在某些情况下似乎遇到了问题.我已设法构建以下示例以显示我的意思:
func optionalProblemDictionary() -> CFDictionary? {
let key = "key"
let value = "value"
var keyCallBacks = CFDictionaryKeyCallBacks()
var valueCallBacks = CFDictionaryValueCallBacks()
let cfDictionary = CFDictionaryCreate(kCFAllocatorDefault, UnsafeMutablePointer(unsafeAddressOf(key)), UnsafeMutablePointer(unsafeAddressOf(value)), 1, &keyCallBacks, &valueCallBacks)
return cfDictionary
}
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相当简单(有点傻)但它的功能返回和可选的CFDictionary.尝试从此函数创建NSDictionary时,"fun"开始:
以下为什么不工作?
if let problemDictionary = optionalProblemDictionary() as? NSDictionary {
print(problemDictionary) // never enters, no warnings, compiles just fine
}
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虽然这很好吗?
if let cfDictionary = optionalProblemDictionary() {
let problemDictionary = cfDictionary as NSDictionary
print(problemDictionary)
}
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XCode 7.0(7A220)
原因似乎是函数返回一个可选的
CFDictionary?,并且不能转换为(非可选)
NSDictionary.
这是一个更简单的例子,展示了与CFStringvs 相同的问题NSString:
let cfString = "foobar" as CFString?
if let s1 = cfString as? NSString {
print("s1 = \(s1)") // not executed
}
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(问题仍然是为什么这不会给编译器错误或至少编译器警告,因为这个可选的强制转换 永远不会成功.)
但是对可选NSString?作品的投射:
if let s2 = cfString as NSString? {
print("s2 = \(s2)") // prints "s2 = foobar"
}
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在您的情况下,如果您将"有问题的案例"更改为
if let problemDictionary = cfDict as NSDictionary? {
print(problemDictionary)
}
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然后执行if-block.
请注意,CFDictionary在Swift中构建a的方法不正确,实际上导致程序在我的测试中崩溃.一个原因是字典回调设置为空结构.另一个问题是unsafeAddressOf(key)将Swift字符串桥接到NSString可以立即解除分配的字符串.
我不知道CFDictionary在Swift中构建一个最好的方法是什么,但这在我的测试中有效:
func optionalProblemDictionary() -> CFDictionary? {
let key = "key" as NSString
let value = "value" as NSString
var keys = [ unsafeAddressOf(key) ]
var values = [ unsafeAddressOf(value) ]
var keyCallBacks = kCFTypeDictionaryKeyCallBacks
var valueCallBacks = kCFTypeDictionaryValueCallBacks
let cfDictionary = CFDictionaryCreate(kCFAllocatorDefault, &keys, &values, 1, &keyCallBacks, &valueCallBacks)
return cfDictionary
}
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