从另一个泛型类型继承的Swift泛型类型

Question

I am trying to get the code below working in a playground:

public protocol SomeTypeProtocol {
    associatedtype T
    func convertTo<TNew:SomeTypeProtocol>()-> TNew
}

public class SomeClass<T>: SomeTypeProtocol{

    var item:T
    public init(item:T)
    {
        self.item = item
    }
    public func convertTo<TNew:SomeTypeProtocol where TNew.T : T>()-> TNew
    {
        return SomeClass<TNew.T>(item: item  as! TNew.T)
    }
}

Basically, I have a protocol with an associated type T, and a class conforming to that protocol with generic type T, and I need to implement the convertTo function that simply converts the class's type T to another type TNew.T which should be a subclass from T. I have this implemented in other languages like C# and Java, so I am not sure why I can't get it working in Swift. I am getting the following errors:

1) error: type 'TNew.T' constrained to non-protocol type 'T'
    public func convertTo<TNew:SomeTypeProtocol where TNew.T : T>()-> TNew
2) error: cannot invoke initializer for type 'SomeClass<TNew.T>' with an argument list of type '(item: TNew.T)'
        return SomeClass<TNew.T>(item: item as! TNew.T)

3) note: expected an argument list of type '(item: T)'
        return SomeClass<TNew.T>(item: item as! TNew.T)

4)note: protocol requires nested type 'T'
    associatedtype T

Answer 1

TNew.T : T

This is where the problem lies.

The colon, : , is used to declare a conformance constraint, not equality. You need to rewrite this as TNew.T == T if you want to equate the two types TNew.T and T .

Answer 2

I believe the closest you can get to doing something like that is this:

public func convertTo<TNew where TNew : T>()-> SomeClass<TNew>
{
    return SomeClass<TNew>(item: item  as! TNew)
}

But I don't believe you can't enforce it being only a subclass and not the same class.

Be careful doing this type of thing with generics. If you're needing to constantly change the Type of your generic, you might want to consider using a different design.