[swift-users] How to reference a generic function when a concrete version also exists?

[Jordan Rose]

> On May 2, 2017, at 16:53, David Sweeris via swift-users <swift-users at swift.org> wrote:
> 
>> 
>> On May 2, 2017, at 4:39 PM, Nevin Brackett-Rozinsky via swift-users <swift-users at swift.org <mailto:swift-users at 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)

Jordan
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