Fulfilling an array of "existenstials" requirement with an array of "concrete types"

Using Swift
,
1

Hi,
Here I'm trying to get Bar to conform to both BarProtocol and Codable:

// ModuleA
public protocol FooProtocol {
    var fooValue: Int { get }
}
public protocol BarProtocol {
    var foos: [FooProtocol] { get }
}
public func load(bar: BarProtocol) {
    for foo in bar.foos { /* ... */ }
}

// ModuleB
import ModuleA
struct Foo: Codable, FooProtocol {
    var fooValue: Int
}
struct Bar: Codable, BarProtocol { // ❌ Type 'Bar' does not conform to protocol 'BarProtocol'
    var foos: [Foo]
}
func test() throws {
    let data = """
    {"foos" : [ {"fooValue" : 42} ] }
    """.data(using: .utf8)!
    let bar = try JSONDecoder().decode(Bar.self, from: data)
    ModuleA.load(bar: bar)
}

Unfortunately the var foos: [FooProtocol] requirement cannot be fulfilled with var foos: [Foo].
Also if I change Bar to this, then I lose conformance to Codable:

struct Bar: Codable, BarProtocol { // ❌ Type 'Bar' does not conform to protocol 'Decodable'
    var foos: [FooProtocol]
}

I found one way to get Bar to conform to both protocols like this:

struct Bar: Codable, BarProtocol {
    var _foos: [Foo]
    var foos: [FooProtocol] { return _foos }
}

Is there another way to achieve that without using that intermediary _foo property?
Maybe by changing something inside Bar implementation, or maybe inside the BarProtocol declaration?

Thank you.

2

One way to make it work is to not use existentials and give BarProtocol an associated type instead:

public protocol BarProtocol {
    associatedtype FooType: FooProtocol
    var foos: [FooType] { get }
}

This then requires your load function to become generic (at least until SE-0309: Unlock existentials for all protocols lands):

public func load<BarType: BarProtocol>(bar: BarType) {
    for foo in bar.foos { /* ... */ }
}

This makes your code compile. Whether it’s a workable solution for you is another question and depends on how you intend BarProtocol to be used.

2 Likes
3

I followed your advice and made it generic. Thank you for your help!

1 Like