Filter function syntax? - swift

This works:
func removeObject<T : Equatable>(object: T, array: [T]) -> Array<T>
{
return array.filter() { $0 != object }
}
let threeThings = ["one", "two", "three"]
twoThings = removeObject("three", threeThings)
However, I'd like to check for inequality with this !==. Then I get error "Type 'T' does not conform to protocol 'AnyObject'"
How can this code be fixed? (I've see the code here but I'd like to learn how to use filter properly).

The identical operator === and its negation !== are only defined for
instances of classes, i.e. instances of AnyObject:
func removeObject<T : AnyObject>(object: T, array: [T]) -> Array<T>
{
return array.filter() { $0 !== object }
}
=== checks if two variables refer to the same single instance.
Note that your code
let threeThings = ["one", "two", "three"]
twoThings = removeObject("three", threeThings)
does still compile and run, but gives the (perhaps unexpected) result
[one, two, three]
The Swift strings (which are value types and not class types) are automatically
bridged to NSString, and the two instances of NSString representing "three"
need not be the same.

If you want to use !== instead of !=, then, instead of the type constraint <T : Equatable> say <T : AnyObject>. All you have to do is listen to what the error message is telling you!
Note that this has nothing to do with using the filter function. It is simply a matter of types. You cannot use a method with an object of a type for which that method is not implemented. !== is implemented for AnyObject so if you want to use it you must guarantee to the compiler that this type will be an AnyObject. That is what the type constraint does.

!== checks for "identity", not "equality". Identity is a property of reference types which all support the AnyObject protocol, and it means that the two variables that you are comparing point to the same actual object, and not just another object with the same value.
That means you can't use === or !== with normal value types, like strings, integers, arrays, etc.
Try typing this into a playground:
let a = "yes"
let b = a
println(a === b)
You should get a message saying that String doesn't conform to the AnyObject protocol, because String is a value type (a struct) and doesn't support AnyObject, so you can't use === with it.
You can use === with any instance of a class, because classes are reference types, not value types.
You could put a constraint on T requiring that it conform to AnyObject.
func removeObject<T : Equatable where T: AnyObject>...
It should work, then, for reference types (including all class instances). But then your function won't work for value types (which include all structs).
Why do you need to check for identity (===) instead of equality (==)?

Related

Swift: binary operator '<' cannot be applied to two 'Comparable' operands

I have a protocol defined that has within it a function that returns an associated type that the user can define in their protocol implementation. The only requirement is that the associated type conform to Comparable.
I have another class that uses elements that conform to this protocol. The issue I'm running into is that the compiler is complaining that I can't compare elements of the associated type, even though I've specifically marked them as Comparable in the protocol definition. Anyone see what I'm doing wrong here?
protocol MyElement {
associatedtype T : Comparable
func getValue() -> T
}
class MyNode {
init(elements:[any MyElement]) {
// Sort the elements
let sortedElements = elements.sorted(by: { ( a:any MyElement, b:any MyElement ) -> Bool in
let aT = a.getValue()
let bT = b.getValue()
return aT < bT
})
}
}
The compiler complains at the line aT < bT:
binary operator '<' cannot be applied to two 'Comparable' operands
The < operator applies to two values of a single type that conforms to Comparable — not to two things designated merely as Comparable. A protocol does not conform to itself.
The use of any with MyElement doesn't affect this basic fact about the Comparable-constrained generic placeholder T. Since MyElement is generic over T, we would have to know that the two different MyElement-conforming objects you propose to sort resolve their T to the same type as one another — and in the code you wrote, we don't know that, so the compiler balks.

In Swift 5, what is a way to compare pointers to two closures?

In Swift 5, what is a way to compare pointers to two closures?
I was told here
In Swift 3, what is a way to compare two closures?
that this is somehow a different question.
So now you have it.
public typealias Action = () -> Void
let action1: Action = {}
let action2: Action = {}
assert(action1 == action1)
assert(action1 != action2)
About pointers: notably, objects (classes) are reference types, so the equality of two objects is not the same as the identity of two objects. For equality (if defined within the class), you have the == operator, and for identity, the === operator:
class Foo:Equatable {
var i:Int = 0
static func == (lhs: Foo, rhs: Foo) -> Bool {
return lhs.i == rhs.i
}
}
let foo1 = Foo()
let foo2 = Foo()
let referenceToFoo1 = foo1
foo1 == foo2 // equality: true, since foo1.i == foo2.i, which is the criteria for equality
foo1 === foo2 // identity: false, since they are two different instances
foo1 === referenceToFoo1 // identity: true, both variables point to the same instance
foo1.i = 2
print(referenceToFoo1.i) // 2, since it refers to the same object
"Reference" can also be called a "pointer," although in Swift, unlike in C, we don't have to delve into pointer arithmetic, which is a pretty low-level way of dealing with memory addresses of pointers to objects and other data.
Just like classes, closures are also reference types in Swift, so in addition to their "equality" (do they contain the same code and captured information etc.) we can also look into their identity (i.e. do these two variables refer to the same closure or two different ones, even if they look the same etc.).
Problem is, Swift doesn't really seem to want to help us there.
Just like == doesn't work for closures, neither does ===. Neither do closure-types seem to conform to any protocol (and note that there's no single "closure type" but rather an infinite multitude of them depending on parameters and return types).
Furthermore, even casting a closure to AnyObject doesn't work:
foo1 as AnyObject === referenceToFoo1 as AnyObject // true, as expected
// with the `action1` closure from the question:
action1 as AnyObject === action1 as AnyObject // false; a bit of a surprise
Looks like every time Swift casts a closure to AnyObject, it creates a new AnyObject instance for some reason… So comparing these also reveals nothing.
So… I don't think we can reason about the identity or equality of closures in Swift. Maybe there is some other way, possibly via unsafe pointers… well, let me know if anyone had any luck going down that rabbit hole!

Compare two objects in Swift without casting to a specific type

I'm trying to compare two objects whose type is known at run time but not at compile time. So far I have the following function which seems to work as written:
/// Compares two objects of the given type.
/// Returns -1 if a is "less than" b, 1 if a is "greater than" b, 0 if they are equal, and nil if no comparison could be made.
func compare<T: Comparable>(type: T.Type, a: Any?, b: Any?) -> Int? {
guard let at = a as? T, let bt = b as? T else { return nil }
return at < bt ? -1 : at > bt ? 1 : 0
}
The problem is, the type is not necessarily known to comply with the Comparable protocol at compile time; I really need to be able to pass in any type (Any.Type), not just Comparable ones. I'd then like the function to return nil if the passed-in type does not conform to Comparable. How can I do this?
Edit (30/08/2018): Some more context. I'm using this function to sort various arrays of strings, integers, and other Comparable types. However, because these arrays are retrieved via reflection the element types are not known at compile time. I know that they will always be Comparable but the compiler doesn't. To resolve this, I'm passing in the type as a separate parameter as shown. However, because I'd like to keep the logic as general as possible I'm performing this sort function inside a conditional statement which chooses the necessary type from an array. The type of this array must be [Any.Type] to hold the required types, even though all its contents conform to Comparable (String.Type, Date.Type, etc.).

How flatMap API contract transforms Optional input to Non Optional result?

This is the contract of flatMap in Swift 3.0.2
public struct Array<Element> : RandomAccessCollection, MutableCollection {
public func flatMap<ElementOfResult>(_ transform: (Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]
}
If I take an Array of [String?] flatMap returns [String]
let albums = ["Fearless", nil, "Speak Now", nil, "Red"]
let result = albums.flatMap { $0 }
type(of: result)
// Array<String>.Type
Here ElementOfResult becomes String, why not String? ? How is the generic type system able to strip out the Optional part from the expression?
As you're using the identity transform { $0 }, the compiler will infer that ElementOfResult? (the result of the transform) is equivalent to Element (the argument of the transform). In this case, Element is String?, therefore ElementOfResult? == String?. There's no need for optional promotion here, so ElementOfResult can be inferred to be String.
Therefore flatMap(_:) in this case returns a [String].
Internally, this conversion from the closure's return of ElementOfResult? to ElementOfResult is simply done by conditionally unwrapping the optional, and if successful, the unwrapped value is appended to the result. You can see the exact implementation here.
As an addendum, note that as Martin points out, closure bodies only participate in type inference when they're single-statement closures (see this related bug report). The reasoning for this was given by Jordan Rose in this mailing list discussion:
Swift's type inference is currently statement-oriented, so there's no easy way to do [multiple-statement closure] inference. This is at least partly a compilation-time concern: Swift's type system allows many more possible conversions than, say, Haskell or OCaml, so solving the types for an entire multi-statement function is not a trivial problem, possibly not a tractable problem.
This means that for closures with multiple statements that are passed to methods such as map(_:) or flatMap(_:) (where the result type is a generic placeholder), you'll have to explicitly annotate the return type of the closure, or the method return itself.
For example, this doesn't compile:
// error: Unable to infer complex closure return type; add explicit type to disambiguate.
let result = albums.flatMap {
print($0 as Any)
return $0
}
But these do:
// explicitly annotate [ElementOfResult] to be [String] – thus ElementOfResult == String.
let result: [String] = albums.flatMap {
print($0 as Any)
return $0
}
// explicitly annotate ElementOfResult? to be String? – thus ElementOfResult == String.
let result = albums.flatMap { element -> String? in
print(element as Any)
return element
}

How to use Equatable protocol on two objects of an unknown type?

Say I have two variables, and I don't know what type they are (and it's not possible to know what type they are until runtime):
var a: Any
var b: Any
How can I test if they are equal, using the Equatable protocol? I can't just do a == b because that requires that both of the items are the same, Equatable type, and the compiler can't prove that because they could be different types (and one or both might not even be Equatable).
So, is it possible to tell the compiler to check if they both have the same type, and if that type conforms to Equatable, then to use the == operator on them and return the result, otherwise returning false?
If there is no way to do this, is there a good reason Swift prevents this, or is it a current limitation of Swift that could be fixed in the future?
Under the assumption that if a and b differ in type then they are never equal, you can use a generic function with constraint to achieve the goal.
func isEqual<T : Equatable>(a: T, b: T) -> Bool {
return a == b;
}
You cannot have a and b differ in type, as the Equatable protocol assumes that the LHS and RHS of the comparison are of the same type. This seems a reasonable constraint, but one can certainly write a notion of equality that doesn't require this. In these cases, you'll need your own equality protocol.
You use generics and function overload:
func isEqual<T: Equatable>(a: T, b: T) -> Bool {
return a == b
}
func isEqual<T, U>(a: T, b: U) -> Bool {
return false
}
If both variables have the same type, as that type conforms to Equatable, then the compiler will choose the first function, otherwise will go the second one.
This will work for Objective-C objects too, providing you cast to NSObject before calling the function: isEqual(var1 as? NSObject, var2 as? NSObject)