函数数组<Optional<T>> -> 可选<Array<T>>

Question

这是我正在尝试做的事情：

extension Array<Optional<T>> {   
  func unwrap() -> Optional<Array<T>> {
    let a = self.flatMap() { a in
      switch a {
      case Optional.Some(let x): return [x]
      case Optional.None: return []
      }
    }
    if a.count == self.count {
      return Optional.Some(a)
    } else {
      return Optional.None
    }   
  } 
}

但是，它没有编译错误 Use of undeclared type T。以下是我想如何使用它：

let a = [Optional.Some(2), Optional.Some(3), Optional.Some(4)]
let b = [Optional.Some(1), Optional.None]
a.unwrap() // Optional[2, 3, 4]
b.unwrap() // nil

我该如何解决这个问题？或者swift没有这种可能？

Answer 1

试试这个：

protocol OptionalType {
    typealias W
    var optional: W? { get }
}

extension Optional: OptionalType {
    typealias W = Wrapped
    var optional: W? { return self }
}

extension Array where Element: OptionalType {
    func unwrap() -> [Element.W]? {
        return reduce(Optional<[Element.W]>([])) { acc, e in
            acc.flatMap { a in e.optional.map { a + [[=10=]] } }
        }
    } 
}

然后，

let a: [Int?] = [1,   2, 3]
let b: [Int?] = [1, nil, 3]

a.unwrap() // ==> [1, 2, 3]
b.unwrap() // ==> nil

Answer 2

findall 的解决方案有效，尽管我认为在这种情况下避免泛型更具可读性：

func unwrap<Element>(optionalArray : [Element?]) -> [Element]? {
    let unwrappedArray = optionalArray.flatMap { (a) -> [Element] in
        switch a {
        case Optional.Some(let x): return [x]
        case Optional.None: return []
        }
    }

    return unwrappedArray.count == optionalArray.count ? Optional.Some(unwrappedArray) : Optional.None
}

用法：

let a = [Optional.Some(2), Optional.Some(3), Optional.Some(4)]
let b = [Optional.Some(1), Optional.None]

// Both are [Int]?
let unwrappedA = unwrap(a) // [2, 3, 4]
let unwrappedB = unwrap(b) // nil

另请参阅：How to determine if a generic is an optional in Swift?

Answer 3

Swift 4

受@findall 解决方案的启发，这适用于 Swift 4:

protocol OptionalType {
    associatedtype Wrapped
    var optional: Wrapped? { get }
}

extension Optional: OptionalType {
    var optional: Wrapped? { return self }
}

extension Sequence where Iterator.Element: OptionalType {
    func removeNils() -> [Iterator.Element.Wrapped] {
        return self.flatMap { [=10=].optional }
    }
}

测试：

class UtilitiesTests: XCTestCase {
    
    func testRemoveNils() {
        let optionalString: String? = nil
        let strings: [String?] = ["Foo", optionalString, "Bar", optionalString, "Baz"]
        XCTAssert(strings.count == 5)
        XCTAssert(strings.removeNils().count == 3)
        let integers: [Int?] = [2, nil, 4, nil, nil, 5]
        XCTAssert(integers.count == 6)
        XCTAssert(integers.removeNils().count == 3)
    }
}

函数数组<Optional<T>> -> 可选<Array<T>>

Function Array<Optional<T>> -> Optional<Array<T>>

arrays

generics

optional

swift

swift2

Swift 4

测试：