> On May 2, 2017, at 16:53, David Sweeris via swift-users
> <swift-users@swift.org> wrote:
>
>>
>> On May 2, 2017, at 4:39 PM, Nevin Brackett-Rozinsky via swift-users
>> <swift-users@swift.org <mailto: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.
You can always throw away infromation by making a new generic context:
func callF<T>(_ x: T) -> Void {
f(x)
}
g(callF)
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