Convert Abstract Type to Concrete Type in Swift

Question

I am trying to make Data Model for my app. here is the scenario:

my app has Customer Model which contains customer's info, and also contain his/her Payment Source. the API gives me two kind of payment sources: card and bank account which they have completely different fields.

So, here is my problem, I want to have abstract type which is PaymentSource then within each PaymentSource have a function to return object casted to it's type. some how I am type erasure.

I needed to put my abstract type in a box and use it as Concrete type (AnyPaymentSource).

So, I've done as following:

protocol PaymentSource {
    associatedtype Kind
    func cast() -> Kind
}

struct AnyPaymentSource<PS: PaymentSource> {
    private var paymentSource: PS
    init(paymentSource: PS) {
        self.paymentSource = paymentSource
    }
    func cast() -> PS.Kind {
        return paymentSource.cast()
    }
}

struct Card: PaymentSource {
    func cast() -> Card {
        return self
    }
}

struct BankAccount: PaymentSource {
    func cast() -> BankAccount {
        return self
    }
}

struct Customer { 
    var firstName: String
    var lastName: String
    var email: String
    var paymentSource : AnyPaymentSource<PaymentSource> 
}

but Customer gives me error with following description:

Using 'PaymentSource' as a concrete type conforming to protocol 'PaymentSource' is not supported

where am I doing wrong?

Answer 1

Swift is statically typed language . That means the type of a variable must be known at compile time.

When i was faced with this problem, i solved it something like this

protocol PaymentSource {
    associatedtype Kind
    func cast() -> Kind
}

struct AnyPaymentSource<PS: PaymentSource> {
    private var paymentSource: PS
    init(paymentSource: PS) {
        self.paymentSource = paymentSource
    }
    func cast() -> PS.Kind {
        return paymentSource.cast()
    }
}

struct Card: PaymentSource {
    func cast() -> Card {
        return self
    }
}

struct BankAccount: PaymentSource {
    func cast() -> BankAccount {
        return self
    }
}

struct Customer<T:PaymentSource> {
    var firstName: String
    var lastName: String
    var email: String
    var paymentSource : AnyPaymentSource<T>
}
func test(){
    let customerWithCard = Customer<Card>(
        firstName: "",
        lastName: "",
        email: "",
        paymentSource: AnyPaymentSource(paymentSource: Card())
    )
    let customerWithBankAccount = Customer<BankAccount>(
        firstName: "",
        lastName: "",
        email: "",
        paymentSource: AnyPaymentSource(paymentSource: BankAccount())
    )
    print(customerWithCard.paymentSource.cast())
    print(customerWithBankAccount.paymentSource.cast())
    return
}

Answer 2

If what are you trying to achieve is what @Andrew Ashurov mentioned in his answer, there is no need to implement AnyPaymentSource . As mentioned in Swift Protocols Documentation :

Protocols do not actually implement any functionality themselves. Nonetheless, any protocol you create will become a fully-fledged type for use in your code .

Meaning that are already able to treat a protocol as a type .

It might be:

protocol PaymentSource {
    func cast() -> Self
}

struct Card: PaymentSource {
    func cast() -> Card {
        return self
    }
}

struct BankAccount: PaymentSource {
    func cast() -> BankAccount {
        return self
    }
}

struct Customer {
    var firstName: String
    var lastName: String
    var email: String
    var paymentSource : PaymentSource?
}

Creating Customers:

let cardCustomer = Customer(firstName: "Card Fname", lastName: "Card Lname", email: "cardemail@example.com", paymentSource: Card())

let bankAccountCustomer = Customer(firstName: "Bank Account Fname", lastName: "Bank Account Lname", email: "bankaccountemail@example.com", paymentSource: BankAccount())

Note that in Customer struct, paymentSource property of type PaymentSource which means it can assigned as any type that conforms to PaymentSource protocol ( Card and BankAccount in your case).