Swift generic type conforming to two protocols

2019-04-06 10:30发布

I have a generic method in one of my classes where I want to have a generic type conforming to UIViewController and UIPickerViewDelegate. How can I do that? I thought of doing this:

func foo<T: UIViewController, UIPickerViewDelegate> (#viewController: T) {}

But this code doesn't "recognize" the UIPickerViewDelegate. I also thought of using the pipe | instead of the comma but this is even worse, the compiler doesn't accept that. Is it possible to do this or do I have to do 2 parameters for the class and the protocol? Or is there a nicer workaround?

Thanks for your help and Merry Christmas :]

3条回答
欢心
2楼-- · 2019-04-06 10:46

Since Swift 4 you can use the power of Protocol Composition. Here you go:

func foo<T: UIViewController & UIPickerViewDelegate> (viewController: T) {}
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趁早两清
3楼-- · 2019-04-06 11:10

Your code:

func foo<T: UIViewController, UIPickerViewDelegate> (#viewController: T) {}

declares 2 generics parameters:

  • T which is UIViewController. And it's used as viewController parameter type.
  • UIPickerViewDelegate which is Any. And it's not used.

Instead, you should use "Where Clause", like:

func foo<T: UIViewController where T:UIPickerViewDelegate> (#viewController: T) {}
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爷的心禁止访问
4楼-- · 2019-04-06 11:12

Things have been changed in Swift 4: func foo<T: UIViewController> (viewController: T) where T:UIPickerViewDelegate {}

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