How to pass a class type as a function parameter

I have a similar use case in swift5:

class PlistUtils {

    static let shared = PlistUtils()

    // write data
    func saveItem<T: Encodable>(url: URL, value: T) -> Bool{
        let encoder = PropertyListEncoder()
        do {
            let data = try encoder.encode(value)
            try data.write(to: url)
            return true
        }catch {
            print("encode error: \(error)")
            return false
        }
    }

    // read data

    func loadItem<T: Decodable>(url: URL, type: T.Type) -> Any?{
        if let data = try? Data(contentsOf: url) {
            let decoder = PropertyListDecoder()
            do {
                let result = try decoder.decode(type, from: data)
                return result
            }catch{
                print("items decode failed ")
                return nil
            }
        }
        return nil
    }

}

You are approaching it in the wrong way: in Swift, unlike Objective-C, classes have specific types and even have an inheritance hierarchy (that is, if class B inherits from A, then B.Type also inherits from A.Type):

class A {}
class B: A {}
class C {}

// B inherits from A
let object: A = B()

// B.Type also inherits from A.Type
let type: A.Type = B.self

// Error: 'C' is not a subtype of 'A'
let type2: A.Type = C.self

That's why you shouldn't use AnyClass, unless you really want to allow any class. In this case the right type would be T.Type, because it expresses the link between the returningClass parameter and the parameter of the closure.

In fact, using it instead of AnyClass allows the compiler to correctly infer the types in the method call:

class func invokeService<T>(service: String, withParams params: Dictionary<String, String>, returningClass: T.Type, completionHandler handler: ((T) -> ())) {
    // The compiler correctly infers that T is the class of the instances of returningClass
    handler(returningClass())
}

Now there's the problem of constructing an instance of T to pass to handler: if you try and run the code right now the compiler will complain that T is not constructible with (). And rightfully so: T has to be explicitly constrained to require that it implements a specific initializer.

This can be done with a protocol like the following one:

protocol Initable {
    init()
}

class CityInfo : NSObject, Initable {
    var cityName: String?
    var regionCode: String?
    var regionName: String?

    // Nothing to change here, CityInfo already implements init()
}

Then you only have to change the generic constraints of invokeService from <T> to <T: Initable>.

Tip

If you get strange errors like "Cannot convert the expression's type '()' to type 'String'", it is often useful to move every argument of the method call to its own variable. It helps narrowing down the code that is causing the error and uncovering type inference issues:

let service = "test"
let params = ["test" : "test"]
let returningClass = CityInfo.self

CastDAO.invokeService(service, withParams: params, returningClass: returningClass) { cityInfo in /*...*/

}

Now there are two possibilities: the error moves to one of the variables (which means that the wrong part is there) or you get a cryptic message like "Cannot convert the expression's type () to type ($T6) -> ($T6) -> $T5".

The cause of the latter error is that the compiler is not able to infer the types of what you wrote. In this case the problem is that T is only used in the parameter of the closure and the closure you passed doesn't indicate any particular type so the compiler doesn't know what type to infer. By changing the type of returningClass to include T you give the compiler a way to determine the generic parameter.

you can get the class of AnyObject via this way:

Swift 3.x

let myClass: AnyClass = type(of: self)

Swift 2.x

let myClass: AnyClass = object_getClass(self)

and you can pass it as paramater later, if you'd like.