Which function does Swift call? Part 6: Tinkering with priorities

This is part of a series of posts on how Swift resolves ambiguity in overloaded functions. You can find Part 1 here. In the previous article, we looked at why you get a Range from the ... by default. The answer? Because Range has some extra validation that relies on the ... input being comparable, and this makes the Range version more specific.

Defaulting to Neither

So, suppose you didn’t want Range to be the default, but still wanted to benefit from this extra validation?

If your goal was to keep the ambiguity, and force the user to be explicit about whether they wanted a Range or a ClosedInterval, you could declare the following:

func ...
  <T: Comparable where T: ForwardIndexType>
  (start: T, end: T) -> ClosedInterval<T> {
    return ClosedInterval(start, end)

Now, unless you specify explicitly what type you want, you’ll get an ambiguous usage error when you use the ... operator.

(It might make you a bit nervous to see ForwardIndexType in the declaration of a function for creating intervals, given intervals have nothing to do wth indices. But don’t worry, it’s really only there to carve out an identical domain to the equivalent Range function to force the ambiguity for the same set of possible inputs.)

Defaulting to ClosedInterval

What if you actually wanted ClosedInterval to be the default? This is a little trickier.

If you just wanted to cover the most common case, integers, then you could do it like so:

func ...<T: IntegerType>(start: T, end: T) -> ClosedInterval<T> {
    return ClosedInterval(start, end)

// x will now be a ClosedInterval
let x = 1...5

This works because IntegerType conforms to RandomAccessIndex (which eventually comforms to ForwardIndexType).

Now, this would be a very specific carve-out for integers. You could leave it there, because integers are a special case, being as how they’re so fundamental and it’s a bit odd that they’re defined as an index type too.

But If you wanted an interval for all other type that are both comparable and an index, you’d need to handle each one on a case-by-case basis. For example, string indices are comparable:

// note, this doesn’t even have to be generic, since Strings aren’t generic
func ...(start: String.Index, end: String.Index) -> ClosedInterval<String.Index> {
    return ClosedInterval(start, end)

let s = "hello"

// y will now be a ClosedInterval
let y = s.startIndex...s.endIndex

Two problems become apparent with this.

First, this gets real old real quickly. Every time you create a new index type you have implement a whole new ... function for it.

Second, even then it’s still out of your control. You want other people to be able to create new index and comparable types and use them with ranges and intervals, and they aren’t necessarily going to do this. So instead of a nice predictable “you get a range by default”, we now have “you’ll probably get an interval, so long as the type has the appropriate ... overload”. That doesnt’t sound good at all.

Your best bet at this point would be to force anyone who wants to their type to be useable with an interval to tag it with a particular protocol. This is what stride does – types have to conform to the Strideable protocol. So let’s try defining an equivalent for ClosedInterval. We’ll call it, uhm, Intervalable.

protocol Intervalable: Equatable { }

extension Int: Intervalable { }

extension String.Index: Intervalable { }

// The following definintions of ... would need
// to REPLACE the current definitions.  So you
// can only do this if you're the author of
// ClosedInterval

func ...<T: Intervalable>
  (start: T, end: T) -> ClosedInterval<T> {
    return ClosedInterval(start, end)

func ...<T: Intervalable where T: ForwardIndexType>
  (start: T, end: T) -> ClosedInterval<T> {
    return ClosedInterval(start, end)

// x will be a ClosedInterval
let x = 1...5

let s = "hello"

// y will be a ClosedInterval
let y = s.startIndex...s.endIndex

Of course, this is only something the author of the original type can really implement, not something you can do yourself if you personally prefer the default to be the other way around.

And it’s a bit heavy-handed to force every type to conform to a protocol when really all it needs is to be comparable for the interval to work. But I can’t think of another option (even for the implementor) right now.1 If anyone else can think of a good solution, let me know.

  1. except possible tinkering with the Comparable hierarchy, which is even more restricting since only Apple could do that. 

Which function does Swift call? Part 5: Range vs Interval

This is part of a series of posts on how Swift resolves ambiguity in overloaded functions. You can find Part 1 here. In the previous article, we looked at how generics are handled.

So finally…

Now that we’ve covered how generics fit in, we can finally answer the original question – how come you get a Range back from 1...5 by default? To recap the sample code:

// 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

The infix ... function is defined three times in the standard library, with the following signatures:

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>

The first two are pretty similar. They are both generic, and have one generic placeholder that is constrained by a single protocol. There’s nothing favouring one of the two constraints – Comparable and ForwardIndexType both descend from Equatable, but neither is a descendent of the other. Left as just those two functions, you’d get an ambiguous call error.

It’s the third function above that is what makes Swift pick Range over ClosedInterval. In it, Pos is constrained not just by ForwardIndexType but also by Comparable. Constraining by two protocols wins over one, so this is the overload that’s picked.

Well, after 4 long lead-up posts, that was a bit of an anticlimax, eh?

So let’s ask another question – why is that last function there?

One possible explanation: the version with two protocol constrains was implemented purely to break the tie, since ambiguous call errors are annoying. Creating a Range doesn’t actually require comparable, but adding the comparable requirement has the effect of favouring ranges over intervals. If this was the desired outcome, that could be all the extra ... is there for.

But that’s probably not why. There’s another reason why there’s an overload that requires Comparable.

Factory Functions

A recurring pattern in the Swift standard library is to use free functions as factories to construct different types depending on the function name or arguments.

For example, the lazy function is defined 4 times for 4 different argument types: once each for random-access, bidirectional, and forward collections; and once for sequences.1 Each one returns a different matching lazy type (e.g. LazyRandomAccessCollection). When you call lazy, the compiler picks the most specific match (in the order just given). When you combine this with type inference, you get the most powerful type available declared for you, all determined at compile time.

// a will be a LazyRandomAccessCollection
// since arrays are random access
let a = lazy([1,2,3,4])

// s will be a LazyBidirectionalCollection,
// since strings can't be indexed randomly
let s = lazy("hello")

// r will be a LazySequence, since StrideTo
// isn't a collection, just a sequence
let r = lazy(stride(from: 1, to: 8, by: 2))

// there's nothing stopping you declaring these
// lazy objects manually directly:
let l = LazyRandomAccessCollection([1,2,3,4])

Each version of lazy might not even do anything other than initialize and return the relevant type – it just makes declaring different lazy types more convenient than giving the relevant class name in full. All this is made possible by the overloading priorities described in this series. Most everyday users of lazy don’t need to know any of this though – they just use the function and it does what you’d intuitively guess was the “right” thing.

In some cases, which factory function to call is controlled more directly by the caller. For example, the stride function creates either a StrideTo or a StrideThrough depending on whether the named middle argument is to: or through:.

// x will be a StrideThrough
let x = stride(from: 1, through: 10, by: 5)

// y will be a StrideTo
let y = stride(from: 1, to: 10, by: 5)

In the case of stride, this is the only way you can construct these objects, as their initializers are private (which means stride can call them since it’s declared inside Swift, but you can’t).

Using Factory Functions to Perform Validation

For Range, the ... operator performs two extra tasks on top of constructing the return value. First, if the input supports comparators, it can validate that the begin and end arguments aren’t inverted. If they are, you get a runtime assertion. You only get this validation when using the ... operator. If you try and declare a range like this: Range(start: 5, end: 1), it’ll work.2

By the way, you’ll even get this check with String ranges. Swift string indices aren’t random-access (because of variable-length characters ), but they are comparable. While you can’t know how many characters are between two indices, you can know that one index comes before the another.

The other purpose of the ... function is to normalize closed ranges into half-open ranges. Range is kind of the opposite way around to ClosedInterval and HalfOpenInterval. There is no closed version of Range, ranges are always half-open. When you call ..., it increments the second parameter to make it equivalent to a half-open range. If you type 1...5 into a playground, you’ll see what is actually created is displayed as 1..<6.

This is why you'll get a runtime assertion if you ever try and construct a closed range through the end index of a string, even though in theory it could be a valid thing to do:

let s = "hello"
let start = s.startIndex
let end = s.endIndex

// fatal error: can not increment endIndex
let r = start...end

// same as if you did this:

Could you put this logic into Range.init? In the case of the half-open conversion, maybe you could, by having two versions of init with to: and through: arguments. But it's cleaner to do it using operators.

(You could maybe say the same about stride, but then you'd need custom ternary operators.3)

But in the case of the inversion check that uses comparable, you'd have to make a generic version of init, maybe something like this:

extension Range {
    init<C: Comparable>(start: C, end: C) {
        assert(start <= end, "Can't form Range with end < start")
        self.init(start: start, end: end)

Well, this will compile but it won't do what you want. In fact your init will never be called, even if you pass in a comparable type. Why? Because it's generic, and the regular version of Range.init is not. As we've seen previously, generic functions never get called when there's a possible non-generic overload.

“Hey, no wait, the current Range.init is so generic!” you complain. “Look:”

// Excerpt from the Swift definition of Range
struct Range : Equatable ...etc {
    /// Construct a range with `startIndex == start` and `endIndex ==
    /// end`.
    init(start: T, end: T)

“See? T is a generic placeholder! And we’re puting more constraints on our function’s placeholder, so it should be the one that’s called.”

OK yes, T is a generic placeholder. But it’s not a placeholder for the function. It’s a placeholder for the struct. In the context of the function, T is already fixed in place as a specific type.4 To help understand this, try the following code:

struct S<T> {
    func f(i: Int) { print("f(Int)") }
    func f(t: T) { print("f(T)") }

    func g(i: Int) { print("g(Int)") }

    // note, here scoping rules mean the placeholder 
    // T is a _different_ T to the struct's T
    func g<T: IntegerType>(t: T) { print("g(T)") }

// fix T to be an Int
let s = S<Int>()

// error: Ambiguous use of f

// prints "g(Int)" not "g(T)"
// because generics lose...

Incidentally, it’s stuff like the potentially confusing scoping of T above (or that placeholders might get randomly renamed) that makes me nervous about re-using a struct’s placeholders when extending that struct. It doesn’t feel right. Usually, nicely-written generic classes typealias their placeholders in some fashion. For example, Range typealiases T as Index. Probably best to use that.

Anyway, for Range, declaring a generic function ..., that doesn’t have to compete with any non-generic alternatives, makes these problems go away. Plus operators look nice.

Next up: what if we didn’t want Range to be the default? What could we do about it?

  1. For more on Swift’s lazy types, see this article 
  2. well, depending on your definition of “work” I guess. 
  3. If you want to see how these are possible (if perhaps inadvisable) Nate Cook recently wrote a good article about them. 
  4. If you really want to tie your brain in knots, try to think about a secenario where the struct placeholder is fixed by which init is chosen, which relies on what T is… 

Which function does Swift call? Part 4: Generics

This is part 4 of a series on how overloaded functions in Swift are chosen. Part 1 covered overloading by return type, part 2 was about how different functions with simple single arguments were picked on a “best match” basis, and part 3 went into a little more detail about protocols.

We started the series with the question, why do you get a Range type back from 1...5 instead of a ClosedInterval? Today we cover how generics fit into the matching hierarchy, and we’ll finally have enough information to answer that question.

Generics are Lower Priority

Generics are lower down the pecking order. Remember, Swift likes to be as “specific” as possible, and generics are less specific. Functions with non-generic arguments (even ones that are protocols) are always preferred over generic ones:

// This f will match a single argument of any type,
// but is very low priority
func f<T>(t: T) { 

// This f takes a specific implemented type, so will be
// preferred over anything else when passed a String.
func f(s: String) {

// This f takes a specific protocol, so would be
// preferred over the generic version (though not
// over a version of f that took an actual Bool)
func f(b: BooleanType) {

// this prints "String"

// this prints "BooleanType"

// this prints "T"

OK that’s simple enough. But once we’re in generic territory, there’s a bunch of additional rules for picking one generic function over another.

Making Generic Placeholders More Specific with Constraints

Within a type parameter list, you can apply various different constraints to your generic parameters. You can require generic placeholders conform to protocols, and that their associated types be equal to a certain type, or conform to a protocol.

Just like before, the rule of thumb is: the more specific you make a function, the more likely it is to be picked in overload resolution.

This means a generic function with a placeholder that has a constraint (such as conforming to a protocol) will be picked over one that doesn’t:

func f<T>(t: T) {

func f<T: IntegerType>(t: T) {
    print("T: IntegerType")

// prints "T: IntegerType"

If the choice is between two functions that both have constraints on their placeholder, and one is more specific than the other, the more specific one is picked. These rules follow a familiar pattern – they mirror the rules for non-generic overloading.

For example, when one protocol inherits from another, the inheriting protocol constraint is preferred:

func g<T: IntegerType>(t: T) {

func g<T: SignedIntegerType>(t: T) {

// prints "SignedIntegerType"

Or if the choice is between adhering to one protocol or two, two is preferred:

func f<B: BooleanType>(b: B) {

func f<B: protocol<BooleanType,Printable>>(b: B) {
    print("BooleanType and Printable")

// prints "BooleanType and Printable"

The where clause introduces the ability to constrain placeholders by their associated types. These additional constraints make the match more specific, bumping it up the priority:

// argument must be an IntegerType
func f<T: IntegerType>(t: T) {

// argument must be an IntegerType 
// AND it's Distance must be an IntegerType
func f<T: IntegerType where T.Distance: IntegerType>(t: T) {
    print("T.Distance: IntegerType")

// prints "T.Distance: IntegerType"

Here, the extra where clause is analogous to a regular argument needing to conform to two protocols.

Swift will also distinguish between two where clauses, one of which is more specific than another. For example, requiring an associated type to conform to an inheriting protocol:

func f<T: IntegerType where T.Distance: IntegerType>(t: T) {
    print("T.Distance: IntegerType")

// where T.Distance: SignedIntegerType is more specific than
// where just T.Distance: IntegerType
func f<T: IntegerType where T.Distance: SignedIntegerType>(t: T) {
    print("T.Distance: SignedIntegerType")

// so this prints "T.Distance: SignedIntegerType"

Where clauses also allow you to check a value is equal to a specific type, not just that it conforms to a protocol:

func f<T: IntegerType where T.Distance == Int>(t: T) {
    print("Distance == Int")

func f<T: IntegerType where T.Distance: IntegerType>(t: T) {
    print("Distance: IntegerType")

// prints "Distance == Int"

It isn’t surprising that the version that specifies Distance is equal to an exact type is more specific than the version that just requires it conforms to a protocol. This is similar to the rule for non-generic functions that a specific type as an argument is preferred over a protocol.

One interesting point about checking equality in the where clause. You can check for equality to a protocol. But it probably isn’t what you want – equality to a protocol is not the same as conforming to a protocol:

// create a protocol that requires the
// implementor defines an associated type
protocol P {
    typealias L

// implement P and choose Ints as the
// associated type
struct S: P {
    typealias L = Int
let s = S()

// Ints are printable so this will match
func f<T: P where T.L: Printable>(t: T) {
    print("T.L: Printable")

// == is more specific than :, so this will
// be the one that's called, right?
func f<T: P where T.L == Printable>(t: T) {
    print("T.L == Printable")

// nope! prints "T.L: Printable"

// here is a struct where L really is == Printable
struct S2: P {
    typealias L = Printable
let s2 = S2()

// and this time, yes, it prints "T.L == Printable"

It’s actually pretty hard to do this by accident – you can only check for type equality if a protocol you’re equating to has no Self or associated type requirements (i.e. the protocol doesn’t require a typealias. or use Self as a function argument), same as you can’t use those protocols as non-generic arguments. This rules out most of the protocols in the standard library (Printable is one of only a handful that doesn’t).

Beware the Unexpected Overload

So, if given a choice between a generic overload and a non-generic one, Swift choses the non-generic one. If choosing between two generic functions, there’s a set of rules that are very similar to the ones for choosing between non-generic functions.

It’s not a perfect parallel, though – there are some differences. For example, if given a choice between an inheriting constraint or more constraints (“depth versus breadth”), unlike with non-generic protocols, Swift will actually chose breadth:

protocol P { }
protocol Q { }

protocol A: P { }

struct S: A, Q { }
let s = S()

// this f is for an inherited protocol
// (deeper than just P)
func f(p: A) {

// this f is for more protocols
// (broader than just P)
func f(p: protocol<P, Q>) {
    print("P and Q")

// error: Ambiguous use of 'f'

// however, if we define a similar situation
// with generics instead:

func g<T: A>(t: T) {
    print("T: A")

func g<T: protocol<P, Q>>(t: T) {
    print("T: P and Q")

// the second one will be picked
// prints "T: P and Q"

Let’s see that example again with some real-world types from the standard library:

func f<T: SignedIntegerType>(t: T) {

// IntegerType is less specific than SignedIntegerType,
// but depth beats breadth so this one should be called:
func f<T: protocol<IntegerType, Printable>>(t: T) {
    print("T: P and Q")

// prints "SignedIntegerType"
// wait, what?

If the example with P and Q worked one way, why the difference with SignedIntegerType and Printable?

Well, turns out IntegerType itself conforms to Printable (by way of _IntegerType). Since it already conforms to it, the Printable in protocol is redundant and can be ignored – it doesn’t make that version of f any more specific than T just conforming to IntegerType. Since SignedIntegerType inherits from IntegerType, it is more specific, so that’s the one that gets picked.

I point this out not to be fussy about the hierarchy of standard library protocols, but to point out how easy it is to get an unexpected version of an overloaded function. For this reason, it’s a good rule to never overload a function with different functionality. Overload for performance optimization (like a collection algorithm that extends), or to extend a function to cover your user-defined new type, or to provide the caller with a more powerful but basically equivalent type (like with Range and ClosedInterval). But overload to vary functionality based on the input and sooner or later you’ll be sorry.

Anyway, with that little mini-lecture out of the way, we finally have enough of the details of overloading to answer the original question – why does Range get picked? We’ll cover that in the next article.

Which function does Swift call? Part 3: Protocol Composition

Previously, we looked at how Swift functions can be overloaded just by return type, and how Swift picks between different possible overloads based on a best-match mechanism.

All this was working towards the reason why 1...5 returns a Range rather than a ClosedInterval. And to understand that, we’ll have to look at how generics fit into the matching criteria.

But before we do, a brief diversion into protocol composition.

Protocols can be composed by putting zero1 or more protocols between the angle brackets of a protocol<> statement. The Apple Swift book says:

Protocol compositions do not define a new, permanent protocol type. Rather, they define a temporary local protocol that has the combined requirements of all protocols in the composition.

Reading that, you might think that declaring protocol<P,Q> was essentially like declaring an anonymous protocol Tmp: P, Q { }, and then using Tmp, without ever giving it a name. But that’s not quite what this means, and a way to spot the difference is by declaring functions that take arguments declared with protocol.

First, a recap of some of the behaviours of regular protocols. Suppose we define 4 protocols: first P and Q, and then two more, A and B that both just conform to P and Q:

protocol P { }
protocol Q { }

protocol A: P, Q { }
protocol B: P, Q { }

Although A and B are functionally identical, they are two different protocols. This means you can write two functions, one that takes A and one that takes B, and they’ll be two different functions with equal priority, so if you try to call them and either one would work, you’ll get an ambiguous call error:

struct S: A, B { }
let s = S()

func f(a: A) {

func f(b: B) {

// error: Ambiguous use of 'f'

If, on the other hand, we use protocol<P,Q> instead of B, we get a different result:

func g(a: A) {

func g(p: protocol<P,Q>) {

// prints "A"

If protocol<P,Q> were really declaring a new anonymous protocol that conformed to P and Q, we’d get the same error as before. Instead, its like saying “an argument that conforms to both P and Q”.

As we saw in the last article, an inherited protocol (in this case A) always trumps its ancestors (in this case P and Q). So the A version of g is called. This is consistent with our rule of thumb, that Swift will always pick the function with more “specific” arguments – here, inheriting protocols are more specific than inherited ones.

Another way to be more specific is to require more protocols. Here’s an example of how three protocols is more specific than two:

protocol R { }
extension S: R { }

// you can assign an alias for any
// protocol<> definition if you prefer
typealias Two = protocol<P,Q>
typealias Three = protocol<P,Q,R>

func h(p: Two) {

func h(p: Three) {

// prints "Three"

// you can still force the other to be
// called if you want: this prints "Two"
h(s as Two)

Finally, if given a choice between an inheriting protocol, or more protocols (i.e. depth vs breadth), Swift won’t favour one over the other:

func o(p: protocol<A,B>) {
    print("A and B")

func o(p: protocol<P,Q,R>) {
    print("P, Q and R")

// error: Ambiguous use of 'o'

// prints "A and B"
o(s as protocol<A,B>)
// prints "P, Q and R"
o(s as protocol<P,Q,R>)

One of the conclusions here is that there are quite a few possible ways for you to declare functions that take the same argument type, and to tweak the declarations to favour one function being called over another. When you combine these with different return values, that gives you a tool to nudge Swift into inferring a specific type by default, but allowing the user to get a different one from the same function name if they prefer, while avoiding forcing the user to deal with ambiguity errors. This is what is happening with ... when it defaults to Range.

So now, on to generics, which will take this even further.

  1. zero being an interesting case we’ll look at some other time 

Which function does Swift call? Part 2: Single Arguments

In the previous article, we saw how Swift allows overloading by just the return type. With these kind of overloads, explicitly stating the type tells Swift which function you want to call:

// 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

But that still doesn’t explain why you get a Range back by default if you don’t specify what kind of return type you want. To answer that, we need to look at function arguments. We’ll start with just single-argument functions.

Unlike with return values, functions with the same name that take different types for their arguments don’t always need explicit disambiguation. Swift will try its best to pick one for you based on various best-match criteria.

As a rule of thumb, Swift likes to pick the most “specific” function possible based on the arguments passed. Swift’s fave thing, the thing that beats out all other single arguments, is a function that takes the exact type you’re passing.

If it can’t find a function that takes that exact type, the next preferred option is an inherited class or protocol. Child classes or protocols are preferred over parents.

To see this in action:

protocol P { }
protocol Q: P { }

struct S: Q { }

// this is the function that will be picked
// if called with type S
func f(s: S) { print("S") }

// if that weren't defined, this would be the next
// choice for a call passing in an S, since Q is
// more specialized than P:
func f(p: Q) { print("Q") }

// and then finally this
func f(p: P) { print("P") }

f(S())  // prints S

class C: P { }
class D: C { }
class E: D { }

// this is the function that will be picked
func f(d: D) { print("D") }

// if that weren't defined, this would be the next choice
func f(c: C) { print("C") }

// if neither were defined, the version above for the 
// protocol P would be called

f(E())  // prints D

This prioritized overloading allows you to write more specialized functions for specific types. For example, suppose you want an overloaded version of contains for ClosedInterval that used the fact that you can do the check in constant rather than linear time using comparators.

Well, that sounds like a familiar concept – it’s polymorphism. But don’t get too carried away. It’s important to understand that:

Overloaded functions are chosen at compile time

In all the cases above, which function is called is determined statically not dynamically, at compile-time not run-time. This means it’s determined by the type of the variable not what the variable points to:

class C { }
class D: C { }

// function that takes the base class
func f(c: C) { print("C") }
// function that takes the child class
func f(d: D) { print("D") }

// the object x references is of type D,
// but x is of type C
let x: C = D()

// which function to call is determined by the
// type of x, and not what it points to
f(x)    // Prints "C" _not_ "D"

// the same goes for protocols...
protocol P { }
struct S: P { }

func f(s: S) { print("S") }
func f(p: P) { print("P") }

let p: P = S()

// despite p pointing to an object of type S,
// the version that takes a P will be called:
f(p)    // Prints "P" _not_ "S"

If instead you need run-time polymorphism – that is, you want the function to be picked based on what a variable points to and not what the type of the variable is – you should be using methods not functions:

class C {
    func f() { print("C") }

class D: C {
    override func f() { print("D") }

// function that takes the child class

// the object x references is of type D,
// but x is of type C:
let x: C = D()

// which method to call is determined by 
// what x points to, and not what type it is
x.f()    // Prints "D" _not_ "C"

The two approaches aren’t mutually exclusive, mind you. You could for example overload a function to take a particular protocol, checking for protocol conformance statically, but then invoke a method on that protocol to get dynamic behaviour.

The downside to that approach is that:

Functions that take Protocols can be ambiguous

With protocols, it’s possible to use multiple inheritance to generate ambiguous situations.

protocol P { }
protocol Q { }

struct S: P, Q { }
let s = S()

func f(p: P) { print("P") }
func f(p: Q) { print("Q") }

// error: argument could be either a P or a Q

// so you need to disambiguate, either:
f(s as P)  // prints P

// or:
let q: Q = S()
f(q)       // prints Q

// the same happens with classes vs protocols:
class C  { }
class D: C, P { }

func f(c: C) { print("C") }

// error: argument could be either a C or a P
// you must specify, either:
f(D() as P)
// or:
f(D() as C)

Defaulting to ClosedInterval for Integers

So, knowing now that a function that takes a struct will be chosen ahead of a function that takes a protocol, we should be able to write a version of the ... operator that takes a struct (Int), that will beat the current versions that take protocols (Comparable to create a ClosedInterval, and ForwardIndex to create a Range):

func ...(start: Int, end: Int) -> ClosedInterval<Int> {
    return ClosedInterval(start, end)

// x is now of type ClosedInterval
let x = 1...5

This is not very appealing though. If you pass in other integer types, you still get a Range:

let i: Int32 = 1, j: Int32 = 5

// r will be of type Range still:
let r = i...j

It would be much better to write a generic version that took an IntegerType.

And it still doesn’t explain why ranges are preferred to closed intervals. Comparable and ForwardIndex don’t inherit from each other. Why isn’t it ambiguous, as in our protocol multiple inheritance example above?

Well, because I was lying when I said the current implementations of ... took a protocol. They actually take generic arguments constrained by protocols. The next article will look at how generic versions of functions fit into the matching criteria.

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

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.