swift - 当 ElementOfResult 被推断为 Optional 时，flatMap 不会过滤掉 nil

Question

flatMap 的 Swift 文档如下:

Returns an array containing the non-nil results of calling the given transformation with each element of this sequence.

在以下示例中，当返回类型为 ElementOfResult 时留给编译器推断 flatMap按记录工作，但在第 5 行时 ElementOfResult被指定，因此推断为 Optional<String>输入似乎是 flatMap停止过滤掉 nil的。

为什么要这样做？

~ swift
Welcome to Apple Swift version 3.0.2 (swiftlang-800.0.63 clang-800.0.42.1). Type :help for assistance.
  1> let words = ["1989", nil, "Fearless", nil, "Red"]
words: [String?] = 5 values {
  [0] = "1989"
  [1] = nil
  [2] = "Fearless"
  [3] = nil
  [4] = "Red"
}
  2> words.flatMap { $0 }
$R0: [String] = 3 values {
  [0] = "1989"
  [1] = "Fearless"
  [2] = "Red"
}
  3> let resultTypeInferred = words.flatMap { $0 }
resultTypeInferred: [String] = 3 values {
  [0] = "1989"
  [1] = "Fearless"
  [2] = "Red"
}
  4> let resultTypeSpecified: [String?] = words.flatMap { $0 }
resultTypeSpecified: [String?] = 5 values {
  [0] = "1989"
  [1] = nil
  [2] = "Fearless"
  [3] = nil
  [4] = "Red"
}

Answer 1

这是 flatMap() 的定义

public func flatMap<ElementOfResult>(_ transform: (Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]

当您设置 resultTypeSpecified 的类型时至 [String?] ，你告诉编译器 ElementOfResult是Optional<String> .

你的转换闭包的类型是 (String?) -> Optional<Optional<String>> .

flatMap将带走 1“层”的 optional ，但不会带走 2 层。

希望这个例子能让事情变得更清楚:

let input: [String??] = [
    Optional.some(Optional.some("1989")),
    Optional.some(Optional.none),
    Optional.some(Optional.some("Fearless")),
    Optional.some(Optional.none),
    Optional.some(Optional.some("Red"))
]

let output = input.flatMap({ $0 })