> On May 2, 2017, at 4:39 PM, Nevin Brackett-Rozinsky via swift-users
> <swift-users@swift.org> wrote:
>
> If I write a generic function like this:
>
> func f<T>(_ x: T) { print("Generic: \(x)") }
>
> I can pass it to another function like this:
>
> func g(_ fn: (Int) -> Void) { fn(0) }
> g(f) // Prints “Generic: 0”
>
> However if I *also* write a non-generic function like this:
>
> func f(_ x: Int) { print("Int: \(x)") }
>
> Then when I make the same call as before:
>
> g(f) // Prints “Int: 0”
>
> It passes in the new, non-generic version.
>
> Is there something I can do, with both versions of f defined, to pass the
> generic f into g?
Not that I know of. Once the code path gets to g(), the compiler knows that the
function is specifically an (Int)->Void, as opposed to the generic (T)->Void,
and it’ll always go for the most specific overload. AFAIK, anyway.
- Dave Sweeris
_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users
