Swift instance variable with protocol

Question

I need to translate the following lines of Objective-c code into swift. This is a sample from the Objective-c JSONModel-Framework , where the Optional protocol provided by the Framework applies to an instance variable of type NSString . I found the link associated with the message , but I did not succeed. Using my MYModel.swift implementation of Xcode complains Cannot specialize non-generic type NSString

thanks for your help!

MYModel.swift

 @objc(MYModel) public class MYModel : JSONModel { ... public var name : NSString<Optional> ... } 

MYModel.h

 @interface MYModel : JSONModel ... @property (strong, nonatomic) NSString<Optional>* name; ... 

JSONModel.h

 ... /** * Protocol for defining optional properties in a JSON Model class. Use like below to define * model properties that are not required to have values in the JSON input: * * @property (strong, nonatomic) NSString<Optional>* propertyName; * */ @protocol Optional @end ... 

Answer 1

< and > do not match the protocol. This applies to types with types such as an array:

 Array<T> 

so you can write var a: Array<String> .

You want something else, the variable must be of type String and conform to the protocol

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You can extend String using the protocol and add the necessary functions yourself.

Since your Optional protocol is empty, just write:

 extension NSString: Optional {} // you can use String if you like 

To create a write protocol in Swift:

 protocol Optional {} 

You can also create an Objective-C protocol.

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You should not use the option because it is already gone, but since Swift has a namespace, it works. You could, of course, write something like this:

 extension NSString: JsonOptProtocol {} protocol JsonOptProtocol {} // or create that in Objective-C like you did 

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Documentation link .

Answer 2

The optional type declared in the Swift standard library is currently JSONModel incompatible with Swift because of this.