swift – flatMap API契约如何转换可选输入到非可选结果？

这是Swift 3.0.2中flatMap的契约

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

如果我取一个[String？]数组,则flatMap返回[String]

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

这里ElementOfResult变成String,为什么不是String？？泛型类型系统如何从表达式中删除Optional部分？

当您使用身份转换{$0}时,编译器会推断出ElementOfResult？ (转换的结果)等效于Element(转换的参数).在这种情况下,Element是String？,因此ElementOfResult？ ==字符串？这里不需要可选的促销,因此可以推断ElementOfResult是String.

因此,在这种情况下flatMap(_ :)返回[String].

在内部,这个转换是从闭包返回的ElementOfResult？通过有条件地展开可选项来完成ElementOfResult,如果成功,则将未展开的值附加到结果中.你可以看到exact implementation here.

作为附录,请注意as Martin points out,闭包只有在单语句闭包时才会参与类型推断(见this related bug report).其原因是Jordan Rose 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.

这意味着对于具有多个语句的闭包,这些语句传递给map(_ :)或flatMap(_ :)(其中结果类型是通用占位符)等方法,您必须显式地注释闭包的返回类型,或方法返回自己.

例如,这不编译：

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

但这些做到：

// 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
}

swift – flatMap API契约如何转换可选输入到非可选结果？

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