Which function does Swift call? Part 1: Return Values

ClosedInterval is shy. You have to coax it out from behind its friend, Range.

// r will be of type Range<Int>
let r = 1...5

// if you want a ClosedInterval<Int>, you
// have to be explicit:
let integer_interval: ClosedInterval = 1...5

// if you use doubles though, you don’t need
// to declare the type explicitly.
// floatingpoint_interval is a ClosedInterval<Double>
let floatingpoint_interval = 1.0...5.0

Ever since Swift 1.0 beta 5, Range has supposed to be only for representing collection index ranges. If you’re not operating on indices, ClosedInterval is probably what you want. It has methods like contains, which determines in constant time if an interval contains a value. Range can only use the non-member contains algorithm, which will turn your range into a sequence and iterate over it – don’t accidentally put a loop in your loop!

But Range is not going quietly. If you use the ... operator with an integer, it elbows ClosedRange out the way and dashes onto the stage. Why? Because integers are forward indexes (as used by Array), and the Swift ... operator has 3 declarations:

func ...<T : Comparable>(start: T, end: T) -> ClosedInterval<T>

func ...<Pos : ForwardIndexType>(minimum: Pos, maximum: Pos) -> Range<Pos>

func ...<Pos : ForwardIndexType where Pos : Comparable>(start: Pos, end: Pos) -> Range<Pos>

When you write let r = 1...5, Swift calls the last one of these, and you get back a Range. To understand why, we need to run through the different ways Swift decides which overloaded function to call. There are a lot of them.

Let’s start with:

Differing Return Values

In Swift, you can’t declare the exact same function twice:

func f() {  }

// error: Invalid redeclaration of f()
func f() {  }

What you can do in Swift, unlike in some other languages, is define two versions of a function that differ only by their return type:

struct A { }
struct B { }

func f() -> A { return A() }

// this second f is fine, even though it only differs
// by what it returns:
func f() -> B { return B() } 

// note, typealiases are just aliases not different types,
// so you can't do this:
typealias L = A
// error: Invalid redeclaration of 'f()'
func f() -> L { return A() }

When calling these only-differing-by-return-value functions, you need to give Swift enough information at the call site for it to unambiguously determine which one to call:

// error: Ambiguous use of 'f'
let x = f()

// instead you must specify which return value you want:
let x: A = f()    // calls the A-returning version
                  // x is of type A

// or if you prefer this syntax:
let y = f() as B  // calls the B-returning version
                  // y is of type B

// or, maybe you're passing it in as the argument to a function:
func takesA(a: A) { }

// the A-returning version of f will be called
takesA(f())

Finally, if you want declare a reference to a function f, you need to specify the full type of the function. Note, once you have assigned it, the reference is to a specific function. It doesn’t need further disambiguation:

// g will be a reference to the A-returning version
let g: ()->A = f

// h will be a reference to the B-returning version
let h = f as ()->B

// calling g doesn't need further info, it references 
// a specific f, and z will be of type A:
let z = g()

OK, so ... is a function that differs by return type – it returns either a ClosedInterval or a Range. By explicitly specifying the result needs to be a ClosedInterval, we can force Swift to call the function that returns one.

But if we don’t specify which ... explicitly, we don’t get an error about ambiguity as seen above. Instead, Swift picks the Range version by default. How come?

Because the different versions of ... also differ by the arguments they can take. In the next article, we’ll take a look at how that works.

8 thoughts on “Which function does Swift call? Part 1: Return Values

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