swift - flatMap API 合约如何将可选输入转换为非可选结果？

Question

这是flatMap在Swift 3.0.2中的约定

public struct Array<Element> : RandomAccessCollection, MutableCollection {
    public func flatMap<ElementOfResult>(_ transform: (Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]
}

如果我采用 [String?] 数组，flatMap 返回 [String]

let albums = ["Fearless", nil, "Speak Now", nil, "Red"]
let result = albums.flatMap { $0 }
type(of: result) 
// Array<String>.Type

这里 ElementOfResult 变成了 String ，为什么不是 String? ？泛型类型系统如何能够从表达式中去除 Optional 部分？

Answer 1

当您使用身份转换 { $0 } 时，编译器将推断 ElementOfResult?(转换的结果)等同于 Element (转换的参数)。在这种情况下，Element 是 String?，因此 ElementOfResult? == String?。这里不需要可选的提升，所以ElementOfResult可以推断为String。

因此 flatMap(_:) 在这种情况下返回一个 [String]。

在内部，这种从闭包返回的 ElementOfResult? 到 ElementOfResult 的转换只是通过有条件地展开可选的来完成的，如果成功，展开的值将附加到结果。你可以看到 exact implementation here .

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作为附录，请注意 as Martin points out , 闭包主体只有在它们是单语句 闭包时才会参与类型推断(参见 this related bug report )。 Jordan Rose 给出了原因 in this mailing list discussion :

Swift's type inference is currently statement-oriented, so there's no easy way to do [multiple-statement closure] inference. This is at least partly a compilation-time concern: Swift's type system allows many more possible conversions than, say, Haskell or OCaml, so solving the types for an entire multi-statement function is not a trivial problem, possibly not a tractable problem.

这意味着对于传递给诸如 map(_:) 或 flatMap(_:) 之类的方法的具有多个语句的闭包(其中结果类型是通用占位符)，您必须显式注释闭包的返回类型或方法返回本身。

例如，这不会编译:

// error: Unable to infer complex closure return type; add explicit type to disambiguate.
let result = albums.flatMap {
    print($0 as Any)
    return $0
}

但是这些确实:

// explicitly annotate [ElementOfResult] to be [String] – thus ElementOfResult == String.
let result: [String] = albums.flatMap {
    print($0 as Any)
    return $0
}

// explicitly annotate ElementOfResult? to be String? – thus ElementOfResult == String.
let result = albums.flatMap { element -> String? in
    print(element as Any)
    return element
}