Swift泛型类型的属性和方法 [英] Swift generic type property and method

问题描述

我如何在属性中存储泛型类型，然后使用该类型属性传递方法？

How do I store generic type in property and then use that type property to pass in method?

我有一个工厂，其方法接收视图控制器类型但返回的实例

I have factory whose method receives view controllers type but returns instance of that view controller (container takes care of that).

public protocol ViewControllerFactoryProtocol {
    func getViewController<T: UIViewController>(type: T.Type) -> UIViewController
}

public class ViewControllerFactory: ViewControllerFactoryProtocol {

private let container: Container

public init(container: Container) {
    self.container = container
}

public func getViewController<T: UIViewController>(type: T.Type) -> UIViewController {
    return self.container.resolve(type)!
}

}

我有这样的属性

var destinationViewController: UIViewController.Type { get }

现在，我想执行以下操作：

Now I would like to do something like:

factory.getViewController(self.destinationViewController)

我在其中声明 destinationViewController 作为 LoginViewController.self

但它不能那样工作。奇怪的是，如果我直接执行此操作，它将起作用：

But its not working like that. Weird thing is that it is working if I do it directly:

factory.getViewController(LoginViewController.self)

有帮助吗？？谢谢

推荐答案

没有看到解决的代码，就不可能说为什么会崩溃，但是我有个好主意。我怀疑您误解了通用类型参数和运行时类型参数之间的区别。请考虑以下简化代码。

Without seeing the code for resolve it's not possible to say why it's crashing, but I have a good idea. I suspect you're mistaking the difference between generic type parameters and runtime type parameters. Consider this simplified code.

func printType<T>(type: T.Type) {
    print("T is \(T.self)")
    print("type is \(type)")
}

class Super {}
class Sub: Super {}

printType(Super.self) // Super/Super. Good.
printType(Sub.self)   // Sub/Sub. Good.

let type: Super.Type = Sub.self
printType(type) // Super/Sub !!!!!!

为什么最后一种情况是Super / Sub？因为 printType< T> 在编译时已解决。它只看定义：

Why is the last case Super/Sub? Because printType<T> is resolved at compile time. It looks just at the definitions:

func printType<T>(type: T.Type)
let type: Super.Type

printType(type)

要完成这项工作，我需要 T ，这样 T.Type 与 Super.Type 。好吧，这就是超级。因此，将其编译为：

To make this work, I need a T such that T.Type is the same as Super.Type. Well, that's Super. So this gets compiled as:

printType<Super>(type)

现在在运行时，我们看到 type 等于 Sub.self ，它是 Super.type 的子类型，所以可以。我们将其传递给 printType< Super> 并获得您所看到的响应。

Now at runtime, we see that type is equal to Sub.self, which is a subtype of Super.type, so that's ok. We pass it along to printType<Super> and get the response you're seeing.

解决，您在要使用 type T $ c>，而您正在尝试解析 UIViewController ，它可能返回nil。

So probably internal to resolve, you're using T somewhere that you wanted to use type, and that you're trying to "resolve" UIViewController, which probably returns nil.

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