辅助模式用法在不推断适当类型的情况下编译

问题描述

考虑以下涉及 Aux-pattern 的简单示例:

Consider the following simple example involving Aux-pattern:

sealed trait AdtBase

abstract case class Foo(){
  type T <: AdtBase
}

object Foo{
  type Aux[TT] = Foo { type T = TT }
}

abstract case class Bar(){
  type T <: AdtBase
  val foo: Foo.Aux[T]
}

object Bar {
  type Aux[TT] = Bar { type T = TT }

  def apply[TT <: AdtBase](f: Foo.Aux[TT]): Bar = new Bar() {
    override type T = TT
    override val foo: Foo.Aux[T] = f
  }
}

case class Baz(foo: Foo)

def testBaz(baz: Baz) = Bar(baz.foo) //Compiles fine
def testFoo(foo: Foo) = Bar(foo) //Error: Type mismatch

Scastie

我真的不明白为什么 testBaz 会编译.我也预计类型不匹配.

I don't really understand why testBaz compiles. I expected type mismatch as well.

推荐答案

似乎没有深层原因.

因为当您显式指定类型参数时，两种方法都会编译

Since when you specify type parameter explicitly both methods compile

def testBaz(baz: Baz) = Bar[baz.foo.T](baz.foo) //compiles
def testFoo(foo: Foo) = Bar[foo.T](foo)         //compiles

好像在

def testBaz(baz: Baz) = Bar(baz.foo) //compiles
//def testFoo(foo: Foo) = Bar(foo)   //doesn't compile

在第一种情况下，类型 baz.foo.T 是推断出来的，而在第二种情况下，类型 foo.T 只是没有推断出来

in the first case the type baz.foo.T is inferred while in the second case the type foo.T is just not inferred

// found   : Foo
// required: Foo.Aux[this.T]

在 Scala 中，总是有可能不会推断出某些类型参数，您必须明确指定它.

In Scala it's always possible that some type parameter will not be inferred and you'll have to specify it explicitly.

也许我找到了可能的原因.

Maybe I found a possible reason.

代码

class testFoo2(foo: Foo) {
  // Bar(foo) // doesn't compile
}

不编译，但如果你使 foo 成为 val

doesn't compile but if you make foo a val

class testFoo2(val foo: Foo) {
  Bar(foo) // compiles
}

然后就可以了.可能问题是当 foo 是 val 时，它更稳定".在这种情况下，它更容易"推断路径依赖类型 foo.T.

then it does. Probably the thing is that when foo is a val it's more "stable" and in such case it's "easier" to infer path-dependent type foo.T.

所以 testBaz 和 testFoo 的区别在于 Baz 是一个 case 类，所以 foo 是一个val 而在 testFoo 中 foo 只是一个方法参数，因此不太稳定".

So the difference between testBaz and testFoo can be that Baz is a case class so foo is a val while in testFoo foo is just a method parameter and therefore less "stable".

同理，相反

trait A[T]
def m[T](a: A[T]) = ???
m(??? : A[_]) // compiles

代码

trait A { type T }
def m[_T](a: A { type T = _T}) = ???
m(??? : A) // doesn't compile

不编译但如果我们提取一个变量

doesn't compile but if we extract a variable

val a: A = ???
m(a) // compiles

然后就可以了.问题是现在 a 是稳定的并且可以推断类型 a.T.

then it does. The thing is that now a is stable and type a.T can be inferred.

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