Why does compactMap return a nil result?
Why does compactMap return a nil result?
Consider this code snippet:
var a: String? = "abc"
var b: String?
let result = [a, b].compactMap $0
After executing it, result will be
result
["abc"]
which is the expected result. The element of result (ElementOfResult) here is String.
ElementOfResult
String
print(type(of: result))
Array<String>
Now to the interesting part. After changing the snippet to
var a: String? = "abc"
var b: Int?
let result = [a, b].compactMap $0
and executing it, result will be
result
[Optional("abc"), nil]
The element of result (ElementOfResult) here is Any which makes sense, because Any is the common denominator of String and Int.
ElementOfResult
Any
Any
String
Int
print(type(of: result))
Array<Any>
Why was a nil result returned by compactMap which contradicts its definition?
nil
compactMap
From Apple's compactMap documentation
compactMap
Returns an array containing the non-nil results of calling the given transformation with each element of this sequence.
func compactMap(_ transform: (Self.Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]
result
[Any]
[Any?]
@user1046037
compactMap is there to remove nil. What sense does it make to return [Any?] if compactMap already made sure none of them is nil?– Purpose
18 hours ago
compactMap
compactMap
2 Answers
2
This is because [a, b] is considered a [Any]. When the element types in an array literal are entirely unrelated (Int? and String?), the array type is inferred to be [Any].
[a, b]
[Any]
Int?
String?
[Any]
In the closure passed to compactMap, you returned $0, which is of type Any. This means that $0 can never be nil. All the optionals inside the array are all wrapped in an Any the moment you put them in the array. Because you never return a nil in the closure, all the elements stay in the result array.
compactMap
$0
Any
$0
nil
Any
nil
The compiler can warn you about wrapping optionals in non-optional Anys:
Any
var a: String? = "abc"
let any: Any = a // warning!
But unfortunately it doesn't warn you when you create arrays.
Anyway, you can get the expected behaviour by specifying that you want a [Any?]:
[Any?]
let result = ([a, b] as [Any?]).compactMap $0
So you kind of unwrap them from Any.
Any
Or:
let result = [a as Any?, b as Any?].compactMap $0
Why can an optional type be wrapped inside an Any?
Any
According to the docs (In the Type Casting for Any and AnyObject section):
Any
AnyObject
Any can represent an instance of any type at all, including function types.
Any
Thus, Optional<T> undoubtedly can be represented by Any.
Optional<T>
Any
Based on the original example, wondering how result can be of the type
[Any] and still hold a nil– user1046037
18 hours ago
[Any]
nil
@user1046037 Because anything is
Any. Any can hold every type.– Sweeper
18 hours ago
Any
Any
The following code throws a compilation error,
let x : Any = nil– user1046037
18 hours ago
let x : Any = nil
@user1046037
nil isn‘t inside the resulting array, but an optional which is nil which only prints nil.– Purpose
18 hours ago
nil
nil
My bad, Any can be nil,
var y : Int? = nil; let x : Any = y as Any– user1046037
18 hours ago
var y : Int? = nil; let x : Any = y as Any
You create an Any-array and compactMap over its elements gives compactMap only Any-elements, no Optional<Any> which it could think about being nil or not, so all emenents stay.
compactMap
Optional<Any>
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What is more puzzling is why the type of
resultis[Any]. I would have expected it to be[Any?]– user1046037
18 hours ago