Consider this code...
考虑这段代码…
import Foundation
let source = ["A", "B", nil, "D"]
print(type(of:source))
let result1 = source.flatMap{
Consider this code...
考虑这段代码…
import Foundation
let source = ["A", "B", nil, "D"]
print(type(of:source))
let result1 = source.flatMap{ $0 }
print(type(of:result1))
print(result1)
extension Array
{
func sameThing() -> Array
{
return self.flatMap{ $0 }
}
}
let result2 = source.sameThing()
print(type(of:result2))
print(result2)
result1 is an Array<String> while result2 is an Array<Optional<String>>. But why?
resulis数组 <可选的
>。但是为什么呢?
I've tried using a Sequence instead of an array, but no luck there either.
我尝试过使用序列而不是数组，但也没有运气。
1 个解决方案
#1
1
Your return type preserves generic type
返回类型保留泛型类型
func sameThing(separator:String = " ") -> Array
Therefore on optional elements signature of this operation is equivalent to [T?] -> [T?]
因此，该操作的可选元素签名与[T?]- >[T ?]
Sequence.flatMap has two overloads with specifically crafted signatures
序列。flatMap有两个特别精心制作的签名重载
func flatMap<SegmentOfResult : Sequence>(
_ transform: (${GElement}) throws -> SegmentOfResult
) rethrows -> [SegmentOfResult.${GElement}]
func flatMap<ElementOfResult>(
_ transform: (${GElement}) throws -> ElementOfResult?
) rethrows -> [ElementOfResult] {
The second one wins. But since you asked for array of optionals, ElementOfResult becomes Optional<String>, and transform becomes Optional<String> -> Optional<Optional<String>>.
第二个赢。但是由于您要求选择的数组，ElementOfResult变成可选的
，转换变成可选的
->可选的
>。
Thus when closure { $0 } returns nil, it gets lifted to Optional(nil), i.e. having .some(nil) value, not .none. Then it will pass the nil guard and appear in result.
因此，当闭包{$0}返回nil时，它将被提升到可选的(nil)，即具有.some(nil)值，而不是.none。然后它将通过nil保护并显示结果。
}
print(type(of:result1))
print(result1)
extension Array
{
func sameThing() -> Array
{
return self.flatMap{
Consider this code...
考虑这段代码…
import Foundation
let source = ["A", "B", nil, "D"]
print(type(of:source))
let result1 = source.flatMap{ $0 }
print(type(of:result1))
print(result1)
extension Array
{
func sameThing() -> Array
{
return self.flatMap{ $0 }
}
}
let result2 = source.sameThing()
print(type(of:result2))
print(result2)
result1 is an Array<String> while result2 is an Array<Optional<String>>. But why?
resulis数组 <可选的
>。但是为什么呢?
I've tried using a Sequence instead of an array, but no luck there either.
我尝试过使用序列而不是数组，但也没有运气。
1 个解决方案
#1
1
Your return type preserves generic type
返回类型保留泛型类型
func sameThing(separator:String = " ") -> Array
Therefore on optional elements signature of this operation is equivalent to [T?] -> [T?]
因此，该操作的可选元素签名与[T?]- >[T ?]
Sequence.flatMap has two overloads with specifically crafted signatures
序列。flatMap有两个特别精心制作的签名重载
func flatMap<SegmentOfResult : Sequence>(
_ transform: (${GElement}) throws -> SegmentOfResult
) rethrows -> [SegmentOfResult.${GElement}]
func flatMap<ElementOfResult>(
_ transform: (${GElement}) throws -> ElementOfResult?
) rethrows -> [ElementOfResult] {
The second one wins. But since you asked for array of optionals, ElementOfResult becomes Optional<String>, and transform becomes Optional<String> -> Optional<Optional<String>>.
第二个赢。但是由于您要求选择的数组，ElementOfResult变成可选的
，转换变成可选的
->可选的
>。
Thus when closure { $0 } returns nil, it gets lifted to Optional(nil), i.e. having .some(nil) value, not .none. Then it will pass the nil guard and appear in result.
因此，当闭包{$0}返回nil时，它将被提升到可选的(nil)，即具有.some(nil)值，而不是.none。然后它将通过nil保护并显示结果。
}
}
}
let result2 = source.sameThing()
print(type(of:result2))
print(result2)
import Founda
