return protocol with associated type

Question

how to return protocol with associated type?

protocol AProtocol {

}

class A: AProtocol {

}

class Main {
    func sendA() -> AProtocol {
        return A()
    }
}

it works.

but

protocol BProtocol {
    associatedtype B
}

class B: BProtocol {
    typealias B = Int
}

class Main {
    func sendA() -> AProtocol {
        return A()
    }

    func sendB() -> BProtocol { // error
        return B()
    }

// function1
    func sendB_<T: BProtocol>() -> T{
        return B() as! T
    }
}

i want to return 'return B()' in function 1 is it possible?

Answer 1

In this function

func sendB_<T: BProtocol>() -> T{
    return B() as! T
}

you cannot return a B as a T because the person who uses the function defines what T is, not you and T can be any type that conforms to Protocol For example, I could do this:

class C: BProtocol 
{
    typealias B = Float
}

let c: C = Main().sendB_()

By doing that, I am setting T to be a C and the forced typecast in sendB_() will fail.

Unfortunately, protocols with associated types cannot, themselves, be treated like a concrete type, so the approach you took with AProtocol will not work.

As I see it, you have two options. Change your function return type to B. After all, you always do return a B

func sendB_() -> B {
    return B()
}

If you want to keep it generic, try

protocol BProtocol 
{
    associatedtype B

    init() // Needed to be able to do T() in generic function
}

func sendB_<T: BProtocol>() -> T{
    return T()
}

You need to add the initialiser to the protocol to make sure that an instance of type T always exists.