Why does compactMap return a nil result?

up vote
9
down vote

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Consider this code snippet:

var a: String? = "abc"
var b: String?

let result = [a, b].compactMap $0

After executing it, result will be

["abc"]

which is the expected result. The element of result (ElementOfResult) here is 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

[Optional("abc"), nil]

The element of result (ElementOfResult) here is Any which makes sense, because Any is the common denominator of String and Int.

print(type(of: result))
Array<Any>

Why was a nil result returned by compactMap which contradicts its definition?

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.

Declaration

func compactMap(_ transform: (Self.Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

asked Aug 18 at 8:10

mrvincenzo

3,34932141

What is more puzzling is why the type of result is [Any]. I would have expected it to be [Any?]
â€“Â user1046037
Aug 18 at 8:50

1

@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?
â€“Â Purpose
Aug 18 at 9:02

add a commentÂ |Â

up vote
9
down vote

favorite

Consider this code snippet:

var a: String? = "abc"
var b: String?

let result = [a, b].compactMap $0

After executing it, result will be

["abc"]

which is the expected result. The element of result (ElementOfResult) here is 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

[Optional("abc"), nil]

The element of result (ElementOfResult) here is Any which makes sense, because Any is the common denominator of String and Int.

print(type(of: result))
Array<Any>

Why was a nil result returned by compactMap which contradicts its definition?

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.

Declaration

func compactMap(_ transform: (Self.Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

asked Aug 18 at 8:10

mrvincenzo

3,34932141

What is more puzzling is why the type of result is [Any]. I would have expected it to be [Any?]
â€“Â user1046037
Aug 18 at 8:50

1

@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?
â€“Â Purpose
Aug 18 at 9:02

add a commentÂ |Â

up vote
9
down vote

favorite

Consider this code snippet:

var a: String? = "abc"
var b: String?

let result = [a, b].compactMap $0

After executing it, result will be

["abc"]

which is the expected result. The element of result (ElementOfResult) here is 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

[Optional("abc"), nil]

The element of result (ElementOfResult) here is Any which makes sense, because Any is the common denominator of String and Int.

print(type(of: result))
Array<Any>

Why was a nil result returned by compactMap which contradicts its definition?

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.

Declaration

func compactMap(_ transform: (Self.Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

asked Aug 18 at 8:10

mrvincenzo

3,34932141

Consider this code snippet:

var a: String? = "abc"
var b: String?

let result = [a, b].compactMap $0

After executing it, result will be

["abc"]

which is the expected result. The element of result (ElementOfResult) here is 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

[Optional("abc"), nil]

The element of result (ElementOfResult) here is Any which makes sense, because Any is the common denominator of String and Int.

print(type(of: result))
Array<Any>

Why was a nil result returned by compactMap which contradicts its definition?

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.

Declaration

func compactMap(_ transform: (Self.Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

asked Aug 18 at 8:10

mrvincenzo

3,34932141

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

edited Aug 18 at 13:16

Peter Mortensen

12.9k1983111

asked Aug 18 at 8:10

mrvincenzo

3,34932141

asked Aug 18 at 8:10

mrvincenzo

3,34932141

asked Aug 18 at 8:10

mrvincenzo

3,34932141

What is more puzzling is why the type of result is [Any]. I would have expected it to be [Any?]
â€“Â user1046037
Aug 18 at 8:50

1

@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?
â€“Â Purpose
Aug 18 at 9:02

add a commentÂ |Â

What is more puzzling is why the type of result is [Any]. I would have expected it to be [Any?]
â€“Â user1046037
Aug 18 at 8:50

1

@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?
â€“Â Purpose
Aug 18 at 9:02

What is more puzzling is why the type of result is [Any]. I would have expected it to be [Any?]
â€“Â user1046037
Aug 18 at 8:50

@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?
â€“Â Purpose
Aug 18 at 9:02

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
5
down vote

accepted

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].

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.

The compiler can warn you about wrapping optionals in non-optional Anys:

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?]:

let result = ([a, b] as [Any?]).compactMap $0

So you kind of unwrap them from Any.

Or:

let result = [a as Any?, b as Any?].compactMap $0

Why can an optional type be wrapped inside an Any?

According to the docs (In the Type Casting for Any and AnyObject section):

Any can represent an instance of any type at all, including function types.

Thus, Optional<T> undoubtedly can be represented by Any.

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

2

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

add a commentÂ |Â

up vote
1
down vote

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.

answered Aug 18 at 8:18

Purpose

848414

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
5
down vote

accepted

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].

The compiler can warn you about wrapping optionals in non-optional Anys:

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?]:

let result = ([a, b] as [Any?]).compactMap $0

So you kind of unwrap them from Any.

Or:

let result = [a as Any?, b as Any?].compactMap $0

Why can an optional type be wrapped inside an Any?

According to the docs (In the Type Casting for Any and AnyObject section):

Any can represent an instance of any type at all, including function types.

Thus, Optional<T> undoubtedly can be represented by Any.

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

2

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

add a commentÂ |Â

up vote
5
down vote

accepted

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].

The compiler can warn you about wrapping optionals in non-optional Anys:

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?]:

let result = ([a, b] as [Any?]).compactMap $0

So you kind of unwrap them from Any.

Or:

let result = [a as Any?, b as Any?].compactMap $0

Why can an optional type be wrapped inside an Any?

According to the docs (In the Type Casting for Any and AnyObject section):

Any can represent an instance of any type at all, including function types.

Thus, Optional<T> undoubtedly can be represented by Any.

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

2

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

add a commentÂ |Â

up vote
5
down vote

accepted

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].

The compiler can warn you about wrapping optionals in non-optional Anys:

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?]:

let result = ([a, b] as [Any?]).compactMap $0

So you kind of unwrap them from Any.

Or:

let result = [a as Any?, b as Any?].compactMap $0

Why can an optional type be wrapped inside an Any?

According to the docs (In the Type Casting for Any and AnyObject section):

Any can represent an instance of any type at all, including function types.

Thus, Optional<T> undoubtedly can be represented by Any.

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

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].

The compiler can warn you about wrapping optionals in non-optional Anys:

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?]:

let result = ([a, b] as [Any?]).compactMap $0

So you kind of unwrap them from Any.

Or:

let result = [a as Any?, b as Any?].compactMap $0

Why can an optional type be wrapped inside an Any?

According to the docs (In the Type Casting for Any and AnyObject section):

Any can represent an instance of any type at all, including function types.

Thus, Optional<T> undoubtedly can be represented by Any.

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

edited Aug 18 at 8:59

answered Aug 18 at 8:51

Sweeper

54.5k960117

answered Aug 18 at 8:51

Sweeper

54.5k960117

answered Aug 18 at 8:51

Sweeper

54.5k960117

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

2

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

add a commentÂ |Â

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

2

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

Based on the original example, wondering how result can be of the type [Any] and still hold a nil
â€“Â user1046037
Aug 18 at 8:52

@user1046037 Because anything is Any. Any can hold every type.
â€“Â Sweeper
Aug 18 at 8:54

The following code throws a compilation error, let x : Any = nil
â€“Â user1046037
Aug 18 at 8:55

@user1046037 nil isnÃ¢Â€Â˜t inside the resulting array, but an optional which is nil which only prints nil.
â€“Â Purpose
Aug 18 at 8:56

My bad, Any can be nil, var y : Int? = nil; let x : Any = y as Any
â€“Â user1046037
Aug 18 at 8:58

add a commentÂ |Â

up vote
1
down vote

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.

answered Aug 18 at 8:18

Purpose

848414

add a commentÂ |Â

up vote
1
down vote

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.

answered Aug 18 at 8:18

Purpose

848414

add a commentÂ |Â

up vote
1
down vote

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.

answered Aug 18 at 8:18

Purpose

848414

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.

answered Aug 18 at 8:18

Purpose

848414

answered Aug 18 at 8:18

Purpose

848414

answered Aug 18 at 8:18

Purpose

848414

answered Aug 18 at 8:18

Purpose

848414

add a commentÂ |Â

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