I am trying to implement a basic protocol extension like so:
protocol Value {
func get() -> Float
mutating func set(to:Float)
}
extension Value {
static func min(of a:Value, and b:Value) -> Float {
if a < b { //Expression type 'Bool' is ambiguous without more context
return a.get()
}else{
return b.get()
}
}
static func < (a:Value, b:Value) -> Bool {
return a.get() < b.get()
}
}
At the if clause the compiler says:Expression type 'Bool' is ambiguous without more context. Why doesn't this work?
As touched on in this Q&A, there's a difference between operator overloads implemented as static members and operator overloads implemented as top-level functions. static members take an additional (implicit) self parameter, which the compiler needs to be able to infer.
So how is the value of self inferred? Well, it has to be done from either the operands or return type of the overload. For a protocol extension, this means one of those types needs to be Self. Bear in mind that you can't directly call an operator on a type (i.e you can't say (Self.<)(a, b)).
Consider the following example:
protocol Value {
func get() -> Float
}
extension Value {
static func < (a: Value, b: Value) -> Bool {
print("Being called on conforming type: \(self)")
return a.get() < b.get()
}
}
struct S : Value {
func get() -> Float { return 0 }
}
let value: Value = S()
print(value < value) // Ambiguous reference to member '<'
What's the value of self in the call to <? The compiler can't infer it (really I think it should error directly on the overload as it's un-callable). Bear in mind that self at static scope in a protocol extension must be a concrete conforming type; it can't just be Value.self (as static methods in protocol extensions are only available to call on concrete conforming types, not on the protocol type itself).
We can fix both the above example, and your example by defining the overload as a top-level function instead:
protocol Value {
func get() -> Float
}
func < (a: Value, b: Value) -> Bool {
return a.get() < b.get()
}
struct S : Value {
func get() -> Float { return 0 }
}
let value: Value = S()
print(value < value) // false
This works because now we don't need to infer a value for self.
We could have also given the compiler a way to infer the value of self, by making one or both of the parameters take Self:
protocol Value {
func get() -> Float
}
extension Value {
static func < (a: Self, b: Self) -> Bool {
print("Being called on conforming type: \(self)")
return a.get() < b.get()
}
}
struct S : Value {
func get() -> Float { return 0 }
}
let s = S()
print(s < s)
// Being called on conforming type: S
// false
The compiler can now infer self from the static type of operands. However, as said above, this needs to be a concrete type, so you can't deal with heterogenous Value operands (you could work with one operand taking a Value; but not both as then there'd be no way to infer self).
Although note that if you're providing a default implementation of <, you should probably also provide a default implementation of ==. Unless you have a good reason not to, I would also advise you make these overloads take homogenous concrete operands (i.e parameters of type Self), such that they can provide a default implementation for Comparable.
Also rather than having get() and set(to:) requirements, I would advise a settable property requirement instead:
// Not deriving from Comparable could be useful if you need to use the protocol as
// an actual type; however note that you won't be able to access Comparable stuff,
// such as the auto >, <=, >= overloads from a protocol extension.
protocol Value {
var floatValue: Double { get set }
}
extension Value {
static func == (lhs: Self, rhs: Self) -> Bool {
return lhs.floatValue == rhs.floatValue
}
static func < (lhs: Self, rhs: Self) -> Bool {
return lhs.floatValue < rhs.floatValue
}
}
Finally, if Comparable conformance is essential for conformance to Value, you should make it derive from Comparable:
protocol Value : Comparable {
var floatValue: Double { get set }
}
You shouldn't need a min(of:and:) function in either case, as when the conforming type conforms to Comparable, it can use the top-level min(_:_:) function.
You can't write
if a < b {
because a and b have type Value which is NOT Comparable.
However you can compare the float value associated to a and b
if a.get() < b.get() {
If you want to be able to make types that can use operators such as >, <, ==, etc., they have to conform to the Comparable protocol:
protocol Value: Comparable {
func get() -> Float
mutating func set(to: Float)
}
This comes with more restrictions though. You will have to change all the Value types in the protocol extension to Self:
extension Value {
static func min(of a: Self, and b: Self) -> Float {
if a < b { //Expression type 'Bool' is ambiguous without more context
return a.get()
}else{
return b.get()
}
}
static func < (a: Self, b: Self) -> Bool {
return a.get() < b.get()
}
}
The Self types get replaced with the type that implements the protocol. So if I implemented Value on a type Container, the methods signatures would look like this:
class Container: Value {
static func min(of a: Container, and b: Container) -> Float
static func < (a: Container, b: Container) -> Bool
}
As a side note, if you want Value to conform to Comparable, you might want to also add the == operator to the Value extension:
static func <(lhs: Self, rhs: Self) -> Bool {
return lhs.get() < rhs.get()
}
How can I compactly write an extension of an Array in Swift 3.0 which works for both Float and Double Element types?
The following doesn't work:
extension Array where Element: FloatingPoint
{
public func multiply(mult: Double) -> [Double] {
return self.map{Double($0) * mult}
}
}
With the error being
Unable to infer closure type in the current context
Why can't the closure type be inferred? Is this a current limitation of the compiler, or is there a good reason why the closure type can't be inferred?
A more explicit version doesn't work either:
extension Array where Element: FloatingPoint
{
public func multiply(mult: Double) -> [Double] {
return self.map{(x: Element) -> Double in Double(v: x) * mult}
}
}
With the error this time being
Ambiguous reference to member '*'
Where again I'm not sure of the reason for this error.
Logically, your extension should work by multiplying an array of homogenous floating-point types with a value of the same type, and returning an array of that type. You can simply express this with an argument of type Element, and a return of [Element]:
// this could also just be an extension of Sequence
extension Array where Element : FloatingPoint {
public func multiply(by factor: Element) -> [Element] {
return self.map { $0 * factor }
}
}
Therefore for a [Float], this would accept an argument of type Float and return a [Float].
However, what if I indeed want to return an array of Double?
I do not believe it's possible to construct a Double from an arbitrary FloatingPoint (or even BinaryFloatingPoint) conforming instance, as neither protocol (although they do require implementation of various aspects of the IEEE 754 specification) actually defines the precise encoding of the conforming type.
However, if you really want this, you could just write two overloads – one for Float elements and one for Double elements:
extension Sequence where Iterator.Element == Float {
public func multiply(by factor: Double) -> [Double] {
return self.map { Double($0) * factor }
}
}
extension Sequence where Iterator.Element == Double {
public func multiply(by factor: Double) -> [Double] {
return self.map { $0 * factor }
}
}
Alternatively, if you plan on making this work with a broader range of types, you can use a protocol in order to define a requirement that allows conforming types to express their value as a Double:
protocol ConvertibleToDouble {
func _asDouble() -> Double
}
extension Float : ConvertibleToDouble {
func _asDouble() -> Double { return Double(self) }
}
extension Double : ConvertibleToDouble {
func _asDouble() -> Double { return self }
}
extension Sequence where Iterator.Element : ConvertibleToDouble {
func multiply(by factor: Double) -> [Double] {
return self.map { $0._asDouble() * factor }
}
}
I am attempting to create a function that can return any type. I do not want it to return an object of type Any, but of other types, i.e. String, Bool, Int, etc. You get the idea.
You can easily do this using generics in this fashion:
func example<T>(_ arg: T) -> T {
// Stuff here
}
But is it possible to do it without passing in any arguments of the same type? Here is what I am thinking of:
func example<T>() -> T {
// Stuff here
}
When I try to do this, everything works until I call the function, then I get this error:
generic parameter 'T' could not be inferred
is it possible to do it without passing in any arguments of the same type?
The answer is yes, but there needs to be a way for the compiler to infer the correct version of the generic function. If it knows what it is assigning the result to, it will work. So for instance, you could explicitly type a let or var declaration. The below works in a playground on Swift 3.
protocol Fooable
{
init()
}
extension Int: Fooable {}
extension String: Fooable {}
func foo<T: Fooable>() -> T
{
return T()
}
let x: String = foo() // x is assigned the empty string
let y: Int = foo() // y is assigned 0
I wrote an extension onto Int as below.
extension Int {
func squared () -> Int {
return self * self
}
}
print(10.squared()) // works
The above code works. Now I want to extend the IntegerType protocol so that Int, UInt, Int64, etc would all conform. My code is as below.
extension IntegerType {
func squared () -> IntegerType { // this line creates error
return self * self
}
}
I get error:
Protocol 'IntegerType' can only be used as a generic constraint because it has Self or associated type requirements
I already saw this question and its video & this question, still couldn't understand. I only understood that there is some associatedType which in this case is Self but couldn't connect the dots. I feel like also my lack of knowledge on the Generics subject is also a reason...
Can someone elaborate a bit on the subject and why does the extension create an error?
You just have to return Self
edit/update:
Note: You can extend all numeric types (Integer & FloatingPoint) in Swift 4 extending the Numeric Protocol
Swift 4
extension Numeric {
func squared() -> Self {
return self * self
}
}
Swift 3
extension Integer {
func squared() -> Self {
return self * self
}
}
A function return type can only be a concrete Type.
The point is Type. Anything struct, class or Protocols that are completely defined in themselves are pure Type. However when a protocol or struct depend on another Generic Type Placeholder such as T, then this is a partial type.
Type are a data construct that compiler has to allocate certain memory.
So something like this:
let a = Array<T>() or let b = T is not sufficient information for the compiler to deduce at compile time.
Hence, this wont work.
extension IntegerType {
func squared () -> IntegerType { // this line creates error
return self * self
}
}
Here, IntegerType is a partial type. It is a generic protocol that only when conformed can then we know the exact type. Similar to Array. Array itself is not a type. Its a generic container. Only when someone creates it with Array() or Array()... then it has a type.
The same happened with you.
public protocol IntegerType : _IntegerType, RandomAccessIndexType {
then again,
public protocol RandomAccessIndexType : BidirectionalIndexType, Strideable, _RandomAccessAmbiguity {
#warn_unused_result
public func advancedBy(n: Self.Distance) -> Self
then again,
public protocol _RandomAccessAmbiguity {
associatedtype Distance : _SignedIntegerType = Int
}
Hence, as RandomAccessIndexType has Self requirement meaning until and unless someone conforms to it, Self is unknown placeholder. It is partial Type.
Since IntegerType conforms to the RandomAccessIndexType and _RandomAccessAmbuiguity which requires Distance associated type too.
Hence you cant do this too
let a: IntegerType = 12
Again IntegerType needs to know Self and Distance (associatedType).
Int however provides the details like so
public struct Int : SignedIntegerType, Comparable, Equatable {
/// A type that can represent the number of steps between pairs of
/// values.
public typealias Distance = Int
Hence you can do such
let a:Int = 10
because it provides Self for SignedIntegerType and Distance for its other counterpart.
Simply put it:
A partial type cannot be used where a concrete type can be. A partial type are good for other generics and constraining them.
Swift lets you create an Array extension that sums Integer's with:
extension Array {
func sum() -> Int {
return self.map { $0 as Int }.reduce(0) { $0 + $1 }
}
}
Which can now be used to sum Int[] like:
[1,2,3].sum() //6
But how can we make a generic version that supports summing other Number types like Double[] as well?
[1.1,2.1,3.1].sum() //fails
This question is NOT how to sum numbers, but how to create a generic Array Extension to do it.
Getting Closer
This is the closest I've been able to get if it helps anyone get closer to the solution:
You can create a protocol that can fulfills what we need to do, i.e:
protocol Addable {
func +(lhs: Self, rhs: Self) -> Self
init()
}
Then extend each of the types we want to support that conforms to the above protocol:
extension Int : Addable {
}
extension Double : Addable {
}
And then add an extension with that constraint:
extension Array {
func sum<T : Addable>(min:T) -> T
{
return self.map { $0 as T }.reduce(min) { $0 + $1 }
}
}
Which can now be used against numbers that we've extended to support the protocol, i.e:
[1,2,3].sum(0) //6
[1.1,2.1,3.1].sum(0.0) //6.3
Unfortunately I haven't been able to get it working without having to supply an argument, i.e:
func sum<T : Addable>(x:T...) -> T?
{
return self.map { $0 as T }.reduce(T()) { $0 + $1 }
}
The modified method still works with 1 argument:
[1,2,3].sum(0) //6
But is unable to resolve the method when calling it with no arguments, i.e:
[1,2,3].sum() //Could not find member 'sum'
Adding Integer to the method signature also doesn't help method resolution:
func sum<T where T : Integer, T: Addable>() -> T?
{
return self.map { $0 as T }.reduce(T()) { $0 + $1 }
}
But hopefully this will help others come closer to the solution.
Some Progress
From #GabrielePetronella answer, it looks like we can call the above method if we explicitly specify the type on the call-site like:
let i:Int = [1,2,3].sum()
let d:Double = [1.1,2.2,3.3].sum()
As of Swift 2 it's possible to do this using protocol extensions. (See The Swift Programming Language: Protocols for more information).
First of all, the Addable protocol:
protocol Addable: IntegerLiteralConvertible {
func + (lhs: Self, rhs: Self) -> Self
}
extension Int : Addable {}
extension Double: Addable {}
// ...
Next, extend SequenceType to add sequences of Addable elements:
extension SequenceType where Generator.Element: Addable {
var sum: Generator.Element {
return reduce(0, combine: +)
}
}
Usage:
let ints = [0, 1, 2, 3]
print(ints.sum) // Prints: "6"
let doubles = [0.0, 1.0, 2.0, 3.0]
print(doubles.sum) // Prints: "6.0"
I think I found a reasonable way of doing it, borrowing some ideas from scalaz and starting from your proposed implementation.
Basically what we want is to have typeclasses that represents monoids.
In other words, we need:
an associative function
an identity value (i.e. a zero)
Here's a proposed solution, which works around the swift type system limitations
First of all, our friendly Addable typeclass
protocol Addable {
class func add(lhs: Self, _ rhs: Self) -> Self
class func zero() -> Self
}
Now let's make Int implement it.
extension Int: Addable {
static func add(lhs: Int, _ rhs: Int) -> Int {
return lhs + rhs
}
static func zero() -> Int {
return 0
}
}
So far so good. Now we have all the pieces we need to build a generic `sum function:
extension Array {
func sum<T : Addable>() -> T {
return self.map { $0 as T }.reduce(T.zero()) { T.add($0, $1) }
}
}
Let's test it
let result: Int = [1,2,3].sum() // 6, yay!
Due to limitations of the type system, you need to explicitly cast the result type, since the compiler is not able to figure by itself that Addable resolves to Int.
So you cannot just do:
let result = [1,2,3].sum()
I think it's a bearable drawback of this approach.
Of course, this is completely generic and it can be used on any class, for any kind of monoid.
The reason why I'm not using the default + operator, but I'm instead defining an add function, is that this allows any type to implement the Addable typeclass. If you use +, then a type which has no + operator defined, then you need to implement such operator in the global scope, which I kind of dislike.
Anyway, here's how it would work if you need for instance to make both Int and String 'multipliable', given that * is defined for Int but not for `String.
protocol Multipliable {
func *(lhs: Self, rhs: Self) -> Self
class func m_zero() -> Self
}
func *(lhs: String, rhs: String) -> String {
return rhs + lhs
}
extension String: Multipliable {
static func m_zero() -> String {
return ""
}
}
extension Int: Multipliable {
static func m_zero() -> Int {
return 1
}
}
extension Array {
func mult<T: Multipliable>() -> T {
return self.map { $0 as T }.reduce(T.m_zero()) { $0 * $1 }
}
}
let y: String = ["hello", " ", "world"].mult()
Now array of String can use the method mult to perform a reverse concatenation (just a silly example), and the implementation uses the * operator, newly defined for String, whereas Int keeps using its usual * operator and we only need to define a zero for the monoid.
For code cleanness, I much prefer having the whole typeclass implementation to live in the extension scope, but I guess it's a matter of taste.
In Swift 2, you can solve it like this:
Define the monoid for addition as protocol
protocol Addable {
init()
func +(lhs: Self, rhs: Self) -> Self
static var zero: Self { get }
}
extension Addable {
static var zero: Self { return Self() }
}
In addition to other solutions, this explicitly defines the zero element using the standard initializer.
Then declare Int and Double as Addable:
extension Int: Addable {}
extension Double: Addable {}
Now you can define a sum() method for all Arrays storing Addable elements:
extension Array where Element: Addable {
func sum() -> Element {
return self.reduce(Element.zero, combine: +)
}
}
Here's a silly implementation:
extension Array {
func sum(arr:Array<Int>) -> Int {
return arr.reduce(0, {(e1:Int, e2:Int) -> Int in return e1 + e2})
}
func sum(arr:Array<Double>) -> Double {
return arr.reduce(0, {(e1:Double, e2:Double) -> Double in return e1 + e2})
}
}
It's silly because you have to say arr.sum(arr). In other words, it isn't encapsulated; it's a "free" function sum that just happens to be hiding inside Array. Thus I failed to solve the problem you're really trying to solve.
3> [1,2,3].reduce(0, +)
$R2: Int = 6
4> [1.1,2.1,3.1].reduce(0, +)
$R3: Double = 6.3000000000000007
Map, Filter, Reduce and more
From my understanding of the swift grammar, a type identifier cannot be used with generic parameters, only a generic argument. Hence, the extension declaration can only be used with a concrete type.
It's doable based on prior answers in Swift 1.x with minimal effort:
import Foundation
protocol Addable {
func +(lhs: Self, rhs: Self) -> Self
init(_: Int)
init()
}
extension Int : Addable {}
extension Int8 : Addable {}
extension Int16 : Addable {}
extension Int32 : Addable {}
extension Int64 : Addable {}
extension UInt : Addable {}
extension UInt8 : Addable {}
extension UInt16 : Addable {}
extension UInt32 : Addable {}
extension UInt64 : Addable {}
extension Double : Addable {}
extension Float : Addable {}
extension Float80 : Addable {}
// NSNumber is a messy, fat class for ObjC to box non-NSObject values
// Bit is weird
extension Array {
func sum<T : Addable>(min: T = T(0)) -> T {
return map { $0 as! T }.reduce(min) { $0 + $1 }
}
}
And here: https://gist.github.com/46c1d4d1e9425f730b08
Swift 2, as used elsewhere, plans major improvements, including exception handling, promises and better generic metaprogramming.
Help for anyone else struggling to apply the extension to all Numeric values without it looking messy:
extension Numeric where Self: Comparable {
/// Limits a numerical value.
///
/// - Parameter range: The range the value is limited to be in.
/// - Returns: The numerical value clipped to the range.
func limit(to range: ClosedRange<Self>) -> Self {
if self < range.lowerBound {
return range.lowerBound
} else if self > range.upperBound {
return range.upperBound
} else {
return self
}
}
}