防止scala函数组合推断？类型

Question

我正在尝试在其类型参数中组合一些具有复合类型的函数：

trait Contains[T]
trait Sentence
trait Token
def sentenceSegmenter[T] = (c: Contains[T]) => null: Contains[T with Sentence]
def tokenizer[T <: Sentence] = (c: Contains[T]) => null: Contains[T with Token]

我的主要目标是能够用以下简单的东西来组合它们：

val pipeline = sentenceSegmenter andThen tokenizer

但是，这会产生编译错误，因为 Scala 推断 tokenizer 的类型需要是 Contains[? with Sentence] => ?：

scala> val pipeline = sentenceSegmenter andThen tokenizer
<console>:12: error: polymorphic expression cannot be instantiated to expected type;
 found   : [T <: Sentence]Contains[T] => Contains[T with Token]
 required: Contains[? with Sentence] => ?

       val pipeline = sentenceSegmenter andThen tokenizer
                                                ^

我尝试了一个稍微不同的 tokenizer 定义，它更接近于 Scala 推断的类型，但我得到了类似的错误：

scala> def tokenizer[T] = (c: Contains[T with Sentence]) => null: Contains[T with Sentence with Token]
tokenizer: [T]=> Contains[T with Sentence] => Contains[T with Sentence with Token]

scala> val pipeline = sentenceSegmenter andThen tokenizer
<console>:12: error: polymorphic expression cannot be instantiated to expected type;
 found   : [T]Contains[T with Sentence] => Contains[T with Sentence with Token]
 required: Contains[? with Sentence] => ?

       val pipeline = sentenceSegmenter andThen tokenizer
                                                ^

如果我指定几乎任何类型以及sentenceSegmenter，或者如果我创建一个没有类型参数的虚假初始函数，我就可以编译：

scala> val pipeline = sentenceSegmenter[Nothing] andThen tokenizer
pipeline: Contains[Nothing] => Contains[Nothing with Sentence with Sentence with Token] = <function1>

scala> val pipeline = sentenceSegmenter[Any] andThen tokenizer
pipeline: Contains[Any] => Contains[Any with Sentence with Sentence with Token] = <function1>

scala> val begin = identity[Contains[Any]] _
begin: Contains[Any] => Contains[Any] = <function1>

scala> val pipeline = begin andThen sentenceSegmenter andThen tokenizer
pipeline: Contains[Any] => Contains[Any with Sentence with Sentence with Token] = <function1>

我不介意Any 或Nothing 类型被推断出来，因为我并不真正关心T 是什么。 （我主要关心with XXX 部分。）但我希望它被推断出来，而不是必须明确指定它，或者通过虚假的初始函数提供它。

Answer 1

您不能绑定 (val) 类型参数。您必须改用def，这样它才能在使用时绑定类型：

  def pipeline[T] = sentenceSegmenter[T] andThen tokenizer

请注意，您可以使用推断类型调用管道：

scala> new Contains[Sentence] {}
res1: Contains[Sentence] = $anon$1@5aea1d29

scala> pipeline(res1)
res2: Contains[Sentence with Sentence with Token] = null