[FIXED] Why don't associated types for protocols use generic type syntax in Swift?

Issue

I'm confused about the difference between the syntax used for associated types for protocols, on the one hand, and generic types on the other.

In Swift, for example, one can define a generic type using something like

struct Stack<T> {
    var items = [T]()
    mutating func push(item: T) {
        items.append(item)
    }
    mutating func pop() -> T {
        return items.removeLast()
    }
}

while one defines a protocol with associated types using something like

protocol Container {
    associatedtype T
    mutating func append(item: T)
    var count: Int { get }
    subscript(i: Int) -> T { get }
}

Why isn't the latter just:

protocol Container<T> {
    mutating func append(item: T)
    var count: Int { get }
    subscript(i: Int) -> T { get }
}

Is there some deep (or perhaps just obvious and lost on me) reason that the language hasn't adopted the latter syntax?

Solution

This has been covered a few times on the devlist. The basic answer is that associated types are more flexible than type parameters. While you have a specific case here of one type parameter, it is quite possible to have several. For instance, Collections have an Element type, but also an Index type and a Generator type. If you specialized them entirely with type parameterization, you'd have to talk about things like Array<String, Int, Generator<String>> or the like. (This would allow me to create arrays that were subscripted by something other than Int, which could be considered a feature, but also adds a lot of complexity.)

It's possible to skip all that (Java does), but then you have fewer ways that you can constrain your types. Java in fact is pretty limited in how it can constrain types. You can't have an arbitrary indexing type on your collections in Java. Scala extends the Java type system with associated types just like Swift. Associated types have been incredibly powerful in Scala. They are also a regular source of confusion and hair-tearing.

Whether this extra power is worth it is a completely different question, and only time will tell. But associated types definitely are more powerful than simple type parameterization.

Answered By - Rob Napier

Answer Checked By - Senaida (PHPFixing Volunteer)