Swift协变泛型

问题描述：

以下是我想达到的一个例子：

Here is an example of what I'd like to achieve:

protocol SomeType {}

class SomeClass: SomeType {}

struct SomeGenericStruct<A> {
    typealias E = A
}

func take(someType: SomeGenericStruct<SomeType>) {}

let concreteGenericStruct1: SomeGenericStruct<SomeType> = SomeGenericStruct<SomeType>()
let concreteGenericStruct2: SomeGenericStruct<SomeClass> = SomeGenericStruct<SomeClass>()

take(concreteGenericStruct1)
take(concreteGenericStruct2) // much no work, very repair. wow.

甚至更简单：

Or even simpler:

let concreteGenericStruct3: SomeGenericStruct<SomeType> = SomeGenericStruct<SomeClass>() as SomeGenericStruct<SomeType> // still no work

我如何设法提供 take with concreteGenericStruct2 ？

How can I manage to provide take with concreteGenericStruct2?

答

为此：

You can use generic method for this:

func take<T where T: SomeType>(someType: SomeGenericStruct<T>) { }

唯一的问题是您无法传递 SomeGenericStruct< SomeType> 。它必须是一个具体类型的泛型。如果完全有必要的话，你可以只有两个函数做同样的事情：

The only problem with this is that you can not pass SomeGenericStruct<SomeType> to it. It must be a generic of some concrete type instead. If totally necessary, you can just have two functions doing the same thing essentially:

func take(someInput: SomeGenericStruct<SomeType>) { /* do stuff */ }
func take<T where T: SomeType>(someType: SomeGenericStruct<T>) { /* do same stuff */ }

Swift协变泛型

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