通用类型，这样一个方法的结果可以再次用作另一个方法的参数

Question

我的代码归结为工厂初始化对象，然后再次使用该对象执行其他操作：

trait Factory[T] {
  def initialize(): T;

  def finish(t: T): Unit;
}

据我了解，对于任何一个 Factory 实例，initialize 的结果应该始终适合传递给 finish，而不考虑 T。

工厂本身在不知道 T 是什么的地方被调用：

object Minimal {
  object StringFactory extends Factory[String] {}
  val factories = Map[Int, Factory[_]](0 -> StringFactory)

  val factory = factories(0)

  // (1)
  val obj = factory.initialize()
  factory.finish(obj)

  // (2)
  def wrapper[T](factory: Factory[T]): Unit = {
    val obj = factory.initialize()
    factory.finish(obj)
  }
  wrapper(factory)
}

变体 (2) 有效，变体 (1) 无效：

type mismatch; found : Minimal.obj.type (with underlying type Any) required: _

但我不知道如何解决这个问题。有可能吗？

编译器调用自己搞不清楚的wrapper方法得到了什么？在我看来，obj 的类型应该是 _，因为编译器似乎命名了 _ 的捕获。我怎样才能让编译器意识到这一点而不必为其引入一个全新的方法？

Answer 1

存在类型在将其实例分配给 val 本身后失去其存在性并成为上限，因此任何不进行此类分配的方式都可以工作，包括：

 scala> trait Factory[T] { type TT = T; def initialize(): TT; def finish(t: TT): Unit;}
 defined trait Factory

 scala> val factory: Factory[_] = new Factory[Int] {def initialize = 5; def finish(t: Int) {}}
 factory: Factory[_] = $anon@31d0ca61

 scala> factory.finish(factory.initialize())

这行不通：

scala> val obj = factory.initialize()
obj: Any = 5

scala> factory.finish(obj)
<console>:11: error: type mismatch;
 found   : Any
 required: factory.TT
    (which expands to)  _
              factory.finish(obj)
                             ^

这是因为 scala 不会将它们的类型视为相等（除非两者是同一类型的成员），因为存在性意味着 intialize() 可以 return Any 的任何子类，当 finish() 可以接受 Any:

的任何（但不总是相同的）子类

scala> trait Factory[T] { def initialize(): T; def finish(t: T): Unit;}
defined trait Factory

scala> val factory: Factory[_] = new Factory[Int] {def initialize = 5; def finish(t: Int) {}}
factory: Factory[_] = $anon@6e5da49

scala> factory.finish(factory.initialize())
<console>:10: error: type mismatch;
 found   : (some other)_(in value factory)
 required: _(in value factory)
              factory.finish(factory.initialize())
                                               ^

所以在这里绑定输入和输出的唯一方法是在它们之间共享类型成员。

Answer 2

一个解决方案是用抽象类型完全替换类型参数：

trait Factory {
  type T
  def initialize(): T;

  def finish(t: T): Unit;
}

object Minimal {
  object StringFactory extends Factory { 
    type T = String
    def initialize(): T = ???
    def finish(t: T): Unit = ??? 
  }
  val factories = Map[Int, Factory](0 -> StringFactory)

  val factory: Factory = factories(0)

  // (1)
  val obj: factory.T = factory.initialize()
  // Or simply (relying on inference): val obj = factory.initialize()      
  factory.finish(obj)

  // (2)
  def wrapper(factory: Factory): Unit = {
    val obj = factory.initialize()
    factory.finish(obj)
  }
  wrapper(factory)
}

Answer 3

基于 Régis 的 , I found out that the compiler infers obj: Factory.T. From there, it was a small step to combine this with dk14's 使用 type TT = T。结果是这样的，但是是通用的和静态类型检查的，没有引入包装器方法。向两位致敬！

从字面上回答原问题

From my point of view, obj's type should be _, as the compiler seems to name that capture of _. How can I make the compiler realize that without having to introduce a whole new method for it?

通过给 _ 显式名称 TT。当然，这些方法也需要使用该名称。

trait Factory[T] {
  type TT = T
  def initialize(): TT;

  def finish(t: TT): Unit;
}

object Minimal {
  object StringFactory extends Factory[String] {
    def initialize(): TT = ""
    def finish(t: TT): Unit = {}
  }
  val factories = Map[Int, Factory[_]](0 -> StringFactory)

  val factory = factories(0)

  // (1)
  val obj: factory.TT = factory.initialize()
  factory.finish(obj)

  // (2)
  def wrapper[T](factory: Factory[T]): Unit = {
    val obj = factory.initialize()
    factory.finish(obj)
  }
  wrapper(factory)
}

通用类型，这样一个方法的结果可以再次用作另一个方法的参数

Generic typing so that one method's result can be used as other method's parameter again

scala

generics

generic-variance