iphone - CF字符串!在 swift 1.2 中不能转换为 String

Question

Alamofire中有这个功能

func escape(string: String) -> String {
    let legalURLCharactersToBeEscaped: CFStringRef = ":/?&=;+!@#$()',*"

    return CFURLCreateStringByAddingPercentEscapes(nil, string, nil, legalURLCharactersToBeEscaped, CFStringBuiltInEncodings.UTF8.rawValue)

}

编译器在此函数中显示“CFString!不可转换为字符串”错误。我试过Convert CFString to NSString - Swift投这个，但没有运气。

Answer 1

来自Xcode 6.3 Release Notes :

The implicit conversions from bridged Objective-C classes (NSString/NSArray/NSDictionary) to their corresponding Swift value types (String/Array/Dictionary) have been removed, making the Swift type system simpler and more predictable.
...
In order to perform such a bridging conversion, make the conversion explicit with the as keyword.

您必须显式地将 CFString/NSString 转换为 Swift String:

func escape(string: String) -> String {
    let legalURLCharactersToBeEscaped: CFStringRef = ":/?&=;+!@#$()',*"

    return CFURLCreateStringByAddingPercentEscapes(nil, string,
        nil, legalURLCharactersToBeEscaped,
        CFStringBuiltInEncodings.UTF8.rawValue) as String
                                     // HERE ---^
}

另一个方向的转换(Swift String 到 NSString)仍然是自动完成的，这就是为什么你的 string 参数 函数可以直接传递给 CFURLCreateStringByAddingPercentEscapes() 函数 一个 CFString 参数。