Scala 类型成员不编译

Question

我写了下面这个简单的例子，希望它能编译好：

abstract class TestObject extends App{
  type Type

  def method[F[_]](ft: F[Type], t: Test[F]{
    type Type = TestObject#Type
  }) = t.doSomeAction(ft) //ERROR
}

trait Test[F[_]]{
  type Type
  def doSomeAction(t: F[Type]) = println(t)
}

ideone demo

但是编译器打印出以下错误信息：

Error:(8, 23) type mismatch;
 found   : ft.type (with underlying type F[TestObject.this.Type])
 required: F[t.Type]
    (which expands to)  F[TestObject#Type]
Note: TestObject.this.Type <: t.Type, but type F is invariant in type _.
You may wish to define _ as +_ instead. (SLS 4.5)
  }) = t.doSomeAction(ft)

从Test#Type = TestObject#Type开始我就不太明白了。

Answer 1

所以问题是 TestObject#Type 捕获了一个不是你想要的存在类型。您要确保 Type of specific 实例对齐。你可以这样做：

https://scalafiddle.io/sf/wpI8iGg/0

或更常见的辅助模式

https://scalafiddle.io/sf/wpI8iGg/1

Scala 类型成员不编译

Scala type members do not compile

scala

type-members