编写惰性序列转换的 Swift 问题

Question

我将继续尝试 Functional Swift 并且非常享受挑战。我正在使用将元素转换为惰性序列的转换。

为了预先说明错误，我得到:无法转换类型的值“转换”(又名“(Int)-> LazySequence>”)到预期的参数类型'() -> LazySequence<[]>'

我的问题在于组合它们，但我需要提供一些上下文来说明问题。

这是转换:

typealias Transform<T, U> = (T) -> LazySequence<[U]>

我可以定义转发应用程序:

precedencegroup LazyForwardApplication {
  associativity: left
}

infix operator |~>: LazyForwardApplication

func |~> <T: LazySequenceProtocol, U>(
  input: T,
  transform: @escaping Transform<T.Elements.Element,U>
  ) -> LazySequence<FlattenSequence<LazyMapSequence<T.Elements, LazySequence<[U]>>>> {

  return input.flatMap(transform)
}

返回类型有点冗长，但它工作正常:

let start = [10,20,30].lazy

let add4_5_6: Transform<Int, Int> = {
  let result = [ $0 + 4, $0 + 5, $0 + 6]
  print("> add4_5_6(\($0)) -> \(result)")
  return result.lazy
}

//请注意，我将调试放在了部分中，这样我可以确定它是延迟发生的。

let result1 = start |~> add4_5_6
result1.forEach{ print($0) }
// 14, 15, 16, 24, 25, 26, 34, 35, 36

还有一个类似的例子:

let add7000_8000: Transform<Int, Int> = {
  let result = [ $0 + 7000, $0 + 8000]
  print("> add7000_8000(\($0)) -> \(result)")
  return result.lazy
}

let result2 = start |~> add7000_8000
result2.forEach{ print($0) }
// 7010, 8010, 7020, 8020, 7030, 8030

我可以将它们串联在一起:

// Double application
let result3 = start |~> add4_5_6 |~> add7000_8000
result3.forEach{ print($0) }
// 7014, 8014, 7015, 8015, 7016, 8016,
// 7024, 8024, 7025, 8025, 7026, 8026,
// 7034, 8034, 7035, 8035, 7036, 8036

但我也希望能够编写它们:

// Forward Composition
precedencegroup LazyForwardComposition {
  associativity: right
}
infix operator >~>: LazyForwardComposition

func >~> <T, U: Sequence, V: Sequence>(
  left:  @escaping Transform<T,U>,
  right: @escaping Transform<U,V>
  ) -> (T) -> LazySequence<FlattenSequence<LazyMapSequence<[U], LazySequence<[V]>>>> {

  return { input in
    let b: LazySequence<[U]> = left(input)
    let c = b.flatMap(right)
    return c
  }
}

这里是我遇到错误的地方:

let composed = add4_5_6 >~> add7000_8000
// ERROR IN ABOVE LINE: Cannot convert value of type
'Transform<Int, Int>' (aka '(Int) -> LazySequence<Array<Int>>')
to expected argument type
'(_) -> LazySequence<[_]>'

let result4 = start |~> composed
result4.forEach{ print($0) }

结果会和 result3 一样

我已经解决了这个问题几次，但总是卡住。任何关于如何解决的想法表示赞赏。

(我之前的问题是类似的领域但不同的问题: Swift: Lazily encapsulating chains of map, filter, flatMap)

对于 Playground :

typealias Transform<T, U> = (T) -> LazySequence<[U]>

// And I can define Forward Application:

precedencegroup LazyForwardApplication {
  associativity: left
}

infix operator |~>: LazyForwardApplication

func |~> <T: LazySequenceProtocol, U>(
  input: T,
  transform: @escaping Transform<T.Elements.Element,U>
  ) -> LazySequence<FlattenSequence<LazyMapSequence<T.Elements, LazySequence<[U]>>>> {

  return input.flatMap(transform)
}

// The return type is a bit of a mouthful but it works fine:

let start = [10,20,30].lazy

let add4_5_6: Transform<Int, Int> = {
  let result = [ $0 + 4, $0 + 5, $0 + 6]
  print("> add4_5_6(\($0)) -> \(result)")
  return result.lazy
}

// Note that I put the debug in partly so I can be sure that it's happening lazily.

let result1 = start |~> add4_5_6
result1.forEach{ print($0) }
// 14, 15, 16, 24, 25, 26, 34, 35, 36

// And another similar example:

let add7000_8000: Transform<Int, Int> = {
  let result = [ $0 + 7000, $0 + 8000]
  print("> add7000_8000(\($0)) -> \(result)")
  return result.lazy
}

let result2 = start |~> add7000_8000
result2.forEach{ print($0) }
// 7010, 8010, 7020, 8020, 7030, 8030

// And I can chain these together inline: 

// Double application
let result3 = start |~> add4_5_6 |~> add7000_8000
result3.forEach{ print($0) }
// 7014, 8014, 7015, 8015, 7016, 8016,
// 7024, 8024, 7025, 8025, 7026, 8026,
// 7034, 8034, 7035, 8035, 7036, 8036

// But I'd like to be able to compose them too:

// Forward Composition
precedencegroup LazyForwardComposition {
  associativity: right
}
infix operator >~>: LazyForwardComposition

func >~> <T, U: Sequence, V: Sequence>(
  left:  @escaping Transform<T,U>,
  right: @escaping Transform<U,V>
  ) -> (T) -> LazySequence<FlattenSequence<LazyMapSequence<[U], LazySequence<[V]>>>> {

  return { input in
    let b: LazySequence<[U]> = left(input)
    let c = b.flatMap(right)
    return c
  }
}

// And here's where I get an error:

let composed = add4_5_6 >~> add7000_8000
// ERROR IN ABOVE LINE: Cannot convert value of type 'Transform<Int, Int>' (aka '(Int) -> LazySequence<Array<Int>>') to expected argument type '(_) -> LazySequence<[_]>'

let result4 = start |~> composed
result4.forEach{ print($0) }

// The result would come out the same as result3

Answer 1

我有一个部分答案，但也许它可以帮助你到达某个地方。

首先，Transform<T, U>定义为 (T) -> LazySequence<[U]> , 所以 U和 V通用类型不能特化为 Sequence :

func >~> <T, U, V>(
    left:  @escaping Transform<T,U>,
    right: @escaping Transform<U,V>
    ) -> (T) -> LazySequence<FlattenSequence<LazyMapSequence<[U], LazySequence<[V]>>>> {

    return { input in
        let b = left(input)
        let c = b.flatMap(right)
        return c
    }
}

其次，你的|~>运算符(operator)接受 Transform作为右手参数，因此您不能将它与 >~> 的返回类型一起使用范围。我能够通过以下行获得结果:

let result4 = start.flatMap(composed)

您可能会使 |~> 过载运算符接受正确的类型，但它看起来不太好。或者也许它会，有足够的类型别名 :)