[Bug?] Generic overload resolution with arguments of function type

[hooman (Hooman Mehr)]

It seems that more specific function signatures are not selected when overloading on function arguments. For example:

let f = Double.advanced(by:)  // Signature: (Double) -> (Double) -> Double
let g = Double.squareRoot     // Signature: (Double) -> () -> Double

// Overloaded generic functions:
func test<T,U>  (_ f: (T)->U )      { print("Chose (T)->U") }
func test<T,U>  (_ f: (T)->()->U )  { print("Chose (T)->()->U") }
func test<T,U,V>(_ f: (T)->(U)->V ) { print("Chose (T)->(U)->V") }

test(f) // Prints: "Chose (T)->U"
test(g) // Prints: "Chose (T)->U"

Update: It turns out to be a bug in Xcode 9.2/Swift 4.0. It is fixed in Xcode 9.3/Swift 4.1.

[Jens]

I've run your code in Xcode 9.3 beta (in a Command Line App prj) and with both the default toolchain and the most recent dev snapshot, it prints:

Chose (T)->(U)->V
Chose (T)->()->U

What version of Swift are you using?
/Jens

@hooman

let f = Double.advanced(by:) // Signature: (Double) -> (Double) -> Double
let g = Double.squareRoot // Signature: (Double) -> () -> Double

// Overloaded generic functions:
func test<T,U> (_ f: (T)->U ) { print("Chose (T)->U") }
func test<T,U> (_ f: (T)->()->U ) { print("Chose (T)->()->U") }
func test<T,U,V>(_ f: (T)->(U)->V ) { print("Chose (T)->(U)->V") }

test(f) // Prints: "Chose (T)->U"
test(g) // Prints: "Chose (T)->U"

[hooman (Hooman Mehr)]

Interesting. I am getting this result on the current release version 9.2 (9C40b). I will try the beta to see what happens.

[hooman (Hooman Mehr)]

Result of further testing:

Xcode 9.2 / Swift 4.0 exhibits this bug, but it seems to be fixed in Xcode 9.3 / Swift 4.1.