Swift Type Inference and Protocols with Associated Types

Question

EDIT: I can't understand why in the where clause below - where U.CacheType == T in AnyCacheable class Swift doesn't treat that statement as a constraint but simply sets T to U.CacheType. Type inference is the worst when things aren't readily apparent :-)

I am trying to follow Swift's Type erasure discussed here -

Specifically the code below:

protocol Cacheable {
    associatedtype CacheType
    func decode(_ data:Data) ->CacheType?
    func encode()->Data?

}

extension String:Cacheable {
    func decode(_ data:Data)->String? {
        let string = String(data: data, encoding: .utf8)
        return string
    }
    func encode()->Data? {
        return data(using: .utf8)
    }
}

class AnyCacheable<T>:Cacheable {
    private let _encode:()->Data?
    private let _decode:(_ data:Data)->T?

    init<U:Cacheable>(_ cacheable:U) where U.CacheType == T {
        self._encode = cacheable.encode
        self._decode = cacheable.decode
    }

    func decode(_ data:Data)->T? {
        return _decode(data)
    }

    func encode() -> Data? {
        return _encode()
    }
}

It works perfectly fine if I create a new instance of AnyCacheable as -

let cacheable:AnyCacheable = AnyCacheable("Swift")

I don't need to explicitly specify the concrete type of 'T' like let cacheable:AnyCacheable = AnyCacheable<String>("Swift")

How does Swift infer the concrete type for 'T'? From the initializer -

init<U:Cacheable>(_ cacheable:U) where U.CacheType == T {
    self._encode = cacheable.encode
    self._decode = cacheable.decode
}

I can see that Swift can infer the type for 'U' from the initializer argument (in this case a String type). In the where clause 'T' is on rhs. So how does that expression evaluate to true?

matt matt · Accepted Answer · 2018-02-06T19:01:40

String is a Cacheable and its decode returns a String, so its associated type CacheType must be String.

AnyCacheable is initialized with a String, which is a Cacheable as required; so its U is String. But U.CacheType is T. So T is String.

To see that this is so, change the definition of String's adoption of Cacheable to this:

extension String:Cacheable {
    func decode(_ data:Data)->Int? {
        return 42
    }
    func encode()->Data? {
        return data(using: .utf8)
    }
}

Now compile your code and look to see what type you get in the line

let cacheable:AnyCacheable = AnyCacheable("Swift")

It is AnyCacheable<Int>.

Swift Type Inference and Protocols with Associated Types

1 Answers