如何访问孩子的makeIterator功能？

Question

我有一个实现协议BaseModels的父类Sequence：

class BaseModel { 
    var dict : Dictionary
    init (json: AnyObject?) { dict = json as? Dictionary }
}
class BaseModels : Sequence {
    var arr : Array<Dictionary>

    init (json: AnyObject?) { arr = json as? Array<Dictionary> }
    func makeIterator() -> AnyIterator<BaseModel> {
        var nextIndex = 0;

        return AnyIterator {
            guard let arr = self.arr, nextIndex < arr.count else { return nil }
            let curIndex = nextIndex; nextIndex += 1;
            return BaseModel(json: arr[curIndex] as AnyObject?)
        }
    }
}

我有从这个父母继承的孩子班：

class Child : BaseModel { ... }
class Childs : BaseModels {
    func makeIterator() -> AnyIterator<Child> {
        var nextIndex = 0;

        return AnyIterator {
            guard let arr = self.arr, nextIndex < arr.count else { return nil }
            let curIndex = nextIndex; nextIndex += 1;
            return Child(json: arr[curIndex] as AnyObject?)
        }
    }
}

但是当我这样做时：

let childs = Childs(json: someJsonArray);
for (idx, el) in childs.enumerate() {
    // when I check the el's data type in here, its type is actually BaseModel, not Child. 
}

为什么el类型是BaseModel，而不是Child？如何正确实现这一点，以便在调用enumerate或for ... in ...时，它返回Child类型而不是BaseModel类？

PS：当我在override前面添加func makeIterator() -> AnyIterator<Child>时，Xcode说＆＃34;方法不会覆盖其超类中的任何方法＆＃34;。