Swift - Generic function parameter

Question

I am trying to do this:

protocol Fly {
}

class Bird: Fly {
}

func fetch<T: Fly>(model: T) {
    print("Done")
}

let bird: Fly = Bird()
fetch(model: bird)

However I get this error:

Cannot invoke 'fetch' with an argument list of type '(model: Fly)'

I set let bird: Fly = Bird() to be of type Fly, shouldn't it work since the function fetch takes any object that conforms to that protocol?

Any thoughts?

If you are using protocol the syntax to do this is `func fetch(model: T) where T: Fly`. What you have typed is for subclassing I believe. But still it shows same error. — adev, Aug 03 '17 at 05:22
@adev Thanks for pointing that out! Yeah still same error thought : ( — JEL, Aug 03 '17 at 05:24
Any reason why you are not doing this -> `func fetch(model: Fly)` That will work fine, — adev, Aug 03 '17 at 05:24
I will be using a library and might need to combine this function with it (unless not necessary but not sure at the moment). Wanted to see ideas as to what I'm doing wrong since I don't have experience with generics — JEL, Aug 03 '17 at 05:26

score 5 · Accepted Answer · answered Aug 03 '17 at 05:01

5

You are creating with Fly object with Bird instance

Replace code

protocol Fly {
}

class Bird: Fly {
}

func fetch<T: Fly>(model: T) {
    print("Done")
}

let bird: Bird = Bird() // Here is a problem 
fetch(model: bird)

answered Aug 03 '17 at 05:01

Prashant Tukadiya

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2

I know about that, but why can't I have this `let bird: Fly = Bird()` and pass that into the function if the function accepts any instance that conforms to `Fly`? – JEL Aug 03 '17 at 05:06
1

`Fly` Protocol doesn't confirm `Fly` it self . `Bird` does – Prashant Tukadiya Aug 03 '17 at 05:08
Oh that makes sense. Is there a way then to make the function `fetch` accept any type of type `Fly`? Or that doesn't make sense - not sure? – JEL Aug 03 '17 at 05:10
You can pass any class object that confirms protocol `Fly` as argument. – Prashant Tukadiya Aug 03 '17 at 05:11
2

`let bird: Fly = Bird()` is perfectly fine. I don't think that is the issue here. – adev Aug 03 '17 at 05:12
I just tried following and it works fine. `protocol Fly { func done() } class Bird: Fly { func done() { print("done") } } let bird: Fly = Bird() bird.done()` – adev Aug 03 '17 at 05:13
@adev I need to pass `Bird` into the function `func fetch(model: T)` which you are not doing – JEL Aug 03 '17 at 05:14
@adev In this case `func fetch(model: T)` is not part of protocol. – Prashant Tukadiya Aug 03 '17 at 05:15
@JEL, I was just showing `let bird: Fly = Bird()` is fine to do. I know this is not your case. So issue is not with that. My understanding is `func fetch(model: T)` accepts only if `Bird` is subclass of `Fly`. Not if it is implemented from protocol. – adev Aug 03 '17 at 05:16
@adev Oh ok, yeah I'm aware that is ok - thank you for mentioning too – JEL Aug 03 '17 at 05:17

score 2 · Answer 2 · answered Aug 03 '17 at 07:45

I set let bird: Fly = Bird() to be of type Fly, shouldn't it work since the function fetch takes any object that conforms to that protocol?

The overload resolution of the fetch(model: bird) call is done statically at compile time. The bird instance has been explicitly annotated to be of type Fly (which happens to be a protocol which can hold, dynamically, instances conforming to it). Since protocols doesn't conform to themselves, a call to fetch(model: bird) will not be eligable to use the constrained generic method func fetch<T: Fly>(model: T), since the type of bird, namely Fly, does not fulfill the type constraint T: Fly.

Swift - Generic function parameter

2 Answers2